calculus from rice university

RMT 2014 Calculus Test Solutions February 15, 2014

1. Let f(x) = x4 and let g(x) = x4. Compute f (2)g(2).

Answer: 15

Solution: We note that f (x) = 12x2 and g(x) = 20x6. Then f (x)g(x) = 20 12 x4.Plugging in x = 2 we get f (2)g(2) = 122016 = 3 5 = 15 .

2. There is a unique positive real number a such that the tangent line to y = x2 + 1 at x = a goesthrough the origin. Compute a.

Answer: 1

Solution: The slope of the tangent line is 2a. The equation for the tangent line is (y(a2+1)) =2a(x a). Setting x = y = 0 gives us a2 1 = 2a2, which has solution a = 1 .

3. Moor has $1000, and he is playing a gambling game. He gets to pick a number k between 0and 1 (inclusive). A fair coin is then flipped. If the coin comes up heads, Moor is given 5000kadditional dollars. Otherwise, Moor loses 1000k dollars. Moors happiness is equal to the logof the amount of money that he has after this gambling game. Find the value of k that Moorshould select to maximize his expected happiness.

Answer: 25

Solution: Suppose that Moor chooses a value of k. We write down the expected value of Moorshappiness.

If the coin comes up heads, Moor now has 1000 + 1000(5k) = 1000(5k + 1) dollars. If the coincomes up tails, Moor now has 10001000k = 1000(1k) dollars. Therefore, the expected valueof Moors happiness is

H(k) =1

2log(1000(5k + 1)) +

1

2log(1000(1 k)).

We want to maximize this. To do this, we differentiate, set the derivative equal to zero, andlook for critical values. Here,

H (k) =1

2

(5000

1000(5k + 1) 1000

1000(1 k))

=1

2

(5

5k + 1 1

1 k)

= 0

when 5k + 1 = 5(1 k), so 10k = 4, and hence k = 25 is the only critical value.The maximal value of H(k) for k [0, 1] must occur either at a critical value or an endpoint.Observe that among the three values H(0), H(1), and H(25), the largest is H(

25). Therefore,

Moor maximizes his happiness by selecting k =2

5.

4. The set of points (x, y) in the plane satisfying x2/5 + |y| = 1 form a curve enclosing a region.Compute the area of this region.

Answer: 87

Solution: The set of points satisfying the equation form a closed curve that encloses a region.Observe that this curve is preserved if we transform x 7 x or y 7 y, so it is symmetric inall 4 quadrants. In particular, we can find the area in the first quadrant, where x, y > 0. In thequadrant, we can rewrite our equation as y = 1x2/5. This curve intersects the coordinate axesat (0, 1) and (1, 0), and it is continuous, so the area is

A =

10

1 x2/5 dx = 27.


The total area is therefore 4A = 8/7 .

5. Compute the improper integral 20

(4 xx

x

4 x

)dx.

Answer: 4

Solution 1: First of all, we note the many symmetries of the given expression. Specifically, we

have

4xx and we subtract its reciprocal. We also recall that square roots, when we take their

derivative, give us their reciprocal. This inspires the guess that the function f(x) =x

4 xis somehow important to our integral. Indeed, we find that f (x) = 12

(4xx

x

4x)

so that 20

4xx

x

4x dx = 2x

4 x20

= 2(2 0) = 4 .Solution 2: Although solution 1 is, perhaps, the prettiest way of solving this problem, it isnot necessarily easy to notice. A more direct approach uses a trig substitution. Specifically,noting the importance of 4xx and remembering the Pythagorean identity sin

2 x + cos2 x = 1,

it makes sense to try the substitution x = 4 sin2 . Then

4xx =

cos2 xsin2 x

= cotx. Also,

dx = 8 sin cos d, 4 sin2 = 0 when = 0 and 4 sin2 = 2 when = pi4 . The integral becomes pi4

0(cot tan ) 8 sin cos d = 8

pi4

0cos2 sin2 d = 8

pi4

0cos 2 d.

This last integral may be easily computed by the substitution 2 7 :

8

pi4

0cos 2 d = 4

pi2

0cos d = 4(sin )

pi20

= 4(1 0) = 4 .

Solution 3: The simplest way to solve this problem is perhaps to write the integrand with acommon denominator. This gives 2

0

4 xxx

4 x dx = 20

(4 x) xx

4 x dx = 20

4 2x4x x2 dx.

Substitute u = 4x x2, du = (4 2x) dx. Then our integral becomes 20

4 2x4x x2 dx =

40

1udu = 2

u40

= 4 .

6. Compute

limx

[x x2 ln

(1 + x

x

)].

Answer: 12


Solution: We rewrite this limit in a form that allows us to apply LHopitals Rule. That is,

limx

[x x2 ln

(1 + x

x

)]= lim

x

1x ln

(1+xx

)1x2

= limx

1x2 x1+x( 1x2 )2x3

by LHopitals Rule

= limx

1

2

(x x

2

1 + x

)= lim

x1

2

(x

1 + x

)= lim

x1

2

(1 1

1 + x

)=

1

2(1 0) = 1

2.

7. For a given x > 0, let an be the sequence defined by a1 = x for n = 1 and an = xan1 for n 2.

Find the largest x for which the limit limn an converges.

Answer: e1/e

Solution: In order for limn an to have a limit L, it must be that x

L = L, so that x = L1/L.

Otherwise, we would be able to extend the recurrence and converge to a different limiting value.Thus, we seek the maximum of the function f(L) = L1/L. To do this, we solve dfdL = 0. Since

df

dL=

d

dLelnLL = L1/L

(1

L2 lnL

L2

),

we see that L = e. Thus, the maximum value for x is f(e) = e1/e . To be sure that this is amaximum, we check as follows:

d2f

dL2

e

= L1L4 (3L+ ln2(L) + 2(L 1) ln(L) + 1)

e= e 1e3 < 0.

8. Evaluate 22

1 + x2

1 + 2xdx.

Answer: 143

Solution: We substitute the variable x by x and add the resulting integral to the originalintegral to get

2I =

22

1 + x2

1 + 2xdx+

22 1 + x

2

1 + 2xdx =

22

1 + x2

1 + 2x+

1 + x2

1 + 2xdx

=

22

1 + x2

1 + 2x+

(1 + x2)2x

1 + 2xdx =

22

(1 + x2) (1 + 2x)1 + 2x

dx =

22

1 + x2 dx = 4 +16

3=

28

3.

So the given integral is I =14

3.

Note that more generally, for even functions f , aa

f(x)1+bx dx =

12

aa f(x) dx.


9. Let f satisfy x = f(x)ef(x). Calculate e0 f(x) dx.

Answer: e 1Solution 1: First, we compute the antiderivative. Make the substitution u = f(x), so hencedu = f (x) dx. Note that f (x) = ddxxe

f = ef f (x)xef , so f (x) = 1ef+x

= fx(1+f) . Thus,

f(x) dx = u

(duu

x(1+u)

)= x(1 + u) du = ueu(1 + u) du

f(x)dx =

ueu(1 + u)du = eu

(u2 u+ 1) = ef(x) (f(x)2 f(x) + 1)

To conclude, when x = 0, f(x) = 0 and when x = e, f(x) = 1. Thus, e0 f(x) dx = e

1(1 1 +1) e0(0 0 + 1) = e 1 .Solution 2: Note that f is monotonically increasing and is the inverse of the function g(y) = yey.Since f(e) = 1, the area under f(x) from 0 to e is the area of the rectangle with vertices(0, 0), (e, 0), (0, 1), (e, 1) minus the area to the left of f(x) from 0 to 1, and the latter is just theintegral of g(y) from 0 to 1. So we have e

0f(x)dx = e

10g(y)dy = e

10yeydy

= e [yey]10 + 10eydy =

10eydy = [ey]10 = e 1.

10. Given that

n=11n2

= pi2

6 , compute the sum

n=1

1

2nn2.

Answer: pi2

12 ln2 2

2

Solution: First of all, for the sake of clarity, I omit details about certain calculations whichare justifiable so there is a more clear focus on the actual computation. Specifically, I takederivatives and integrals of series without explaining, and I integrate functions with removablesingularities, but the ordinary student would not pay attention to these technical issues anyway.I proceed now:

The first step to obtaining any insight on this problem is to replace 12n with xn. This allows us to

take derivatives, getting rid of powers of n in the denominator. Thus, we write f(x) =

n=1xn

n2

and what we want to find is f(12) given that f(0) = 0 and f(1) =pi2

6 . As mentioned before, we

first take f (x) =

n=1xn1n and then we take (xf

(x)) =

n=1 xn1 = 11x . By reintegrating,

xf (x) = ln (1 x) + C but by plugging in x = 0 it is easy to check that C = 0. Thenf (x) = ln (1x)x . Because f(0) = 0, f(x) =

x0 ln (1t)t dt. Thus, the answer we are looking for

is equal to 1

20 ln (1t)t dt. This completes the first part of the solution. The second part consists

of computing this integral.

We denote I = 1

20 ln 1tt dt. There are two things we know about this integral: that finding

the antiderivative, if it even exists, would be extremely challenging, and also a related formula


10 ln (1t)t dt = pi

2

6 which is given. Noting that12 is the midpoint of the interval [0, 1] in which the

integral formula is relevant, we note that there are several transformations which give integralson the interval [12 , 1]. Specifically, the substitution of x 7 1 x yields I =

112 ln t1tdt. In

addition, I = 10 ln (1t)t dt

112 ln (1t)t dt = pi

2

6 112 ln (1t)t dt. Thus, we may write 2I =

pi2

6 + 112

ln 1tt ln t1tdt. The apparent symmetry of the integrand immediately brings to mind the

function g(x) = lnx ln (1 x) as a potential antiderivative: indeed, when we apply the productrule, we easily get g(x) = ln 1xx lnx1x . Thus, 2I = pi

2

6 + ln t ln (1 t)112. Because plugging in

t = 1 is undefined, we resort to using limits and easily obtain 0. Thus, 2I = pi2

6 ln2 12 = pi2

6 ln2 2

and I =pi2

12 ln

2 12

2=

pi2

12 ln

2 2

2are both correct and equally valid answers.

Note, with only a little more work (and some formalizing), we can obtain the more general result

that

n=1xn

n2+

n=1(1x)nn2

= pi2

6 lnx ln (1 x) when x (0, 1).

RMT 2013 Calculus Test February 2, 2013

1. Compute limx3

x2 + 2x 15x2 4x + 3 .

2. Compute all real values of b such that, for f(x) = x2 + bx 17, f(4) = f (4).3. Suppose a and b are real numbers such that

limx0

sin2 x

eax bx 1 =1

2.

Determine all possible ordered pairs (a, b).

4. Evaluate

40ex dx.

5. Evaluate limx0

sin2(5x) tan3(4x)

(log(2x + 1))5.

6. Computek=0

pi3

0sin2k x dx.

7. The function f(x) has the property that, for some real positive constant C, the expression

f (n)(x)

n + x + C

is independent of n for all nonnegative integers n, provided that n + x + C 6= 0. Given thatf (0) = 1 and

10f(x) dx = C + (e 2), determine the value of C.

Note: f (n)(x) is the n-th derivative of f(x), and f (0)(x) is defined to be f(x).

8. The function f(x) is defined for all x 0 and is always nonnegative. It has the additionalproperty that if any line is drawn from the origin with any positive slope m, it intersects thegraph y = f(x) at precisely one point, which is 1

munits from the origin. Suppose further that

f has a unique maximum value at some real number x = a. Find

a0f(x) dx.

9. Evaluate

pi/20

dx(sinx +

cosx

)4 .10. Evaluate lim

n

[(nk=1

2k

2k 1

) 1

(cosx)2n

2xdx

].


1. Answer: 4

Solution: Note thatx2 + 2x 15x2 4x+ 3 =

(x 3)(x+ 5)(x 3)(x 1) =

x+ 5

x 1. limx3x+ 5

x 1 =3 + 5

3 1 = 4 .

2. Answer: 3

Solution: We have that f(4) = 4b 1 and f (4) = 2(4) + b = b+ 8. Setting these equal to eachother, we see that b = 3 .

3. Answer: (2, 2) and (2,2)Solution: Since this is in an indeterminate form, we can use LHospitals Rule to obtain

limx0

sin (2x)

aeax b =1

2.

However, the numerator goes to zero, so the denominator must also go to zero to give us anotherindeterminate form. This implies that a = b. Using LHospitals Rule again, we have that

limx0

2 cos (2x)

a2eax=

1

2.

The numerator goes to 2, so the denominator must go to 4. Therefore, a = b = 2, giving us(a, b) = (2, 2) and (2,2) .

4. Answer: 2e2 + 2

Let w =x so that w2 = x and dx = 2w dw. Then the integral becomes 2

20wew dw.

To find this integral, use integration by parts:

u = w du = dw; dv = ew dw v = ewwew dw = uv

v du

= wew ewdw

= (w 1)ew.

Evaluating 2(w 1)ew at our limits of integration yields 2e2 + 2 .5. Answer: 50

Solution 1: For any function f with f(0) = 0, we know that

limx0

f(x)

x= lim

x0f(x) f(0)

x= f (0).

sin(5x), tan(4x), and log(2x + 1) are all 0 at x = 0, and their derivatives at 0 are 5, 4, and 2,respectively. So, divide numerator and denominator by x5 and re-arrange to get

limx0

sin2(5x) tan3(4x)

(log(2x+ 1))5= lim

x0

(sin(5x)x

)2 ( tan(4x)x )3(log(2x+1)

x

)5 = 52 4325 = 50 .


Solution 2: Recall from Taylor series that if f(0) = 0, then f(x) f (0)x when x is small.This allows us to write

limx0

sin2(5x) tan3(4x)

(log(2x+ 1))5= lim

x0(5x)2(4x)3

(2x)5= 50 .

6. Answer:3

Bring the sum into the integral, so we have pi3

0

k=0

sin2k x dx.

The integrand is a geometric series, so the answer is pi3

0

1

1 sin2 x dx = pi

3

0sec2 x dx = tan

(pi3

) tan(0) =

3 .

7. Answer:3 e

Solution: Since f (n)(x)/(n + x + C) is independent of n, we can say that it is equal to g(x).Multiplying by (n+ x+ C), we have that

f (n)(x) = (n+ x+ C)g(x).

Taking a derivative with respect to x, we obtain

f (n+1)(x) = (n+ x+ C)g(x) + g(x).

However, this is equal to (n + 1 + x + C)g(x) by the problem statement. Canceling terms, weobtain that g(x) = g(x). The only class of functions that is its own derivative is aex, so we havethat g(x) = aex (for some constant a). Now, f (x) = (x+ C + 1)aex, so f (0) = 1 gives us thata = 1/(C + 1). We also have that 1

0f(x) dx =

10

x+ C

C + 1 ex dx = C + (e 2).

Integration by parts gives us(e 1)C + 1

C + 1= C + (e 2),

which simplifies toC2 = 3 e,

from which it follows that the answer is

3 e .

8. Answer: 1+log(2)4

Solution 1: First, express x and y as functions parametrized by m. We have the system

y = mx

x2 + y2 =1

m.


Solving for y, we get y =

m1+m2

. Hence, maximizing y is equivalent to maximizing m1+m2

. By

differentiating with respect to m, we see that the maximum occurs when m = 1, at the point( 1

2, 1

2).

Now we just need to compute the integral. However, this parametric form is not convenient.Instead, by drawing the line y = x, we notice that the integral splits up into a right isoscelestriangle and a region between the line y = x and the y-axis. This suggests that we shouldconvert to polar coordinates. In fact, f(x) is equivalent to the graph r() = 1

tan(), since a line

at angle to the x-axis has slope tan(). The area we wish to compute is pi/2pi/4

1

2r()2 d =

1

2

pi/2pi/4

cot() d

=1

2[log(sin())]

pi/2pi/4

=1

2(0 log(1/

2))

=1

4log(2).

We add this area to the area of the triangle, which is 12

(12

)2= 14 , so our final answer is

1 + log(2)

4.

Solution 2: We begin as before to find a, but present a different method of computing theintegral.

Solving for x in terms of y, we get that

x2 + y2 = x/y = yx2 x+ y3 = 0 = x = 1

1 4y42y

.

We only care about the region where 1

1 4y4 = 2xy 1, since x, y 12. Hence, we take

x =114y42y .

Notice that we can compute the desired quantity as(12

)2 1

2

0

1

1 4y42y

dy,

since within the square bounded by the coordinate axes and x, y 12, the area between the

curve and the x-axis plus the area between the curve and the y-axis sum to the area of the wholesquare.


Now, using the substitution u =

1 4y4, we get 12

0

1

1 4y42y

dy =

01

(1 u)u4u2 4 du

=1

4

10

(1 u)u(1 u)(1 + u) du

=1

4

10

u

1 + udu

=1

4

10

1 11 + u

du

=1

4[u log(1 + u)]10

=1 log(2)

4.

The answer is 12 minus this quantity, so report1 + log(2)

4.

9. Answer: 1/3

Solution 1: Observe that by pulling a factor of cos2 x out of the denominator, we can write thegiven integral as pi/2

0

dx

(1 +

tanx)4 cos2 x=

pi/20

sec2 x dx

(1 +

tanx)4.

We now substitute u =

tanx+ 1:

du =sec2 x

2

tanxdx =

sec2 x

2(u 1) dx.

Thus, our integral is equal to 1

2u 2u4

du =

1

2u3 2u4 du,

which simplifies to [u2 + 2

3u3

]1

=1

3.

Solution 2: Let I be the value of the given integral. Note that

1

2I =

1

2

pi/20

((sin(x) +

cos(x)

)2)2dx,

which is the polar area bounded by the curve r() =(

sin() +

cos())2

and the x and y

axes for [0, pi/2]. Converting to Cartesian coordinates, we get

1 = r(

sin() +

cos())2

=(

r sin() +r cos()

)2= x+y = 1= y = (1x)2 = 1 + x 2x.


Therefore,

1

2I =

10

1 + x 2x dx

=

[x+

x2

2 4

3x3/2

]10

= 1 +1

2 4

3=

1

6

= I = 13

10. Answer: pi 2pi

2pi1Solution 1: Observe that (cosx)2n looks like a bunch of spikes, centered at 0, pi, 2pi, . . . , each

with mass In = pi/2pi/2(cosx)

2n dx.

We can integrate by parts to see that

Ik =

pi/2pi/2

(cosx)2k dx =[(cosx)2k1 sinx

]pi/2pi/2

+ (2k 1) pi/2pi/2

(cosx)2k2 sin2 x dx

= (2k 1) pi/2pi/2

(cosx)2k2(1 cos2 x) dx = (2k 1)(Ik1 Ik).

Therefore,

Ik =2k 1

2kIk1 = In =

(nk=1

2k 12k

)I0 = pi

nk=1

2k 12k

.

As n , the spikes get sharper and sharper; this means that the denominator 2x of theintegrand gets concentrated at x = 0, pi, 2pi, . . . . Therefore, we expect that as n,(

nk=1

2k

2k 1

) 1

(cosx)2n

2xdx

(nk=1

2k

2k 1

) k=0

In2kpi

= pi1

1 2pi = pi2pi

2pi 1 .

Solution 2: We present a more rigorous approach here. First, rewrite the problem into thefollowing form:

For a positive integer n, let an =1n(cosx)2n

2x dx. Also let c = limnnnk=1(1

12k ), which is a positive finite constant. Evaluate

1c limn an.

Let B = {0, pi, 2pi, . . . }. The idea is that the numerator of the integrand approaches a functionwith cpi area concentrated infinitely closely to each point in B. Therefore the limit should be

limn

1

n(cosx)2n

2xdx =

xB

cpi

2x= cpi

2pi

2pi 1 ,

where the last equality follows by the formula for summing geometric series.

We will soon get to a more precise way of thinking about the area being concentrated infinitelyclosely to points in B, but first lets see why the numerator should have cpi area around each


point in B. Since the area around each point in B is the same (cos2n is periodic), we need onlyconsider the area around 0. We can apply integration by parts to find a formula for the areaaround 0 in each term of the sequence. The recurrence is pi/2

pi/2

n(cosx)2ndx = (1 1

2n)

pi/2pi/2

n(cosx)2(n1)dx.

Repeatedly applying this formula, we get pi/2pi/2

n(cosx)2ndx = pi

n

nk=1

(1 12k

).

Taking the limit as n, the area around 0 goes to cpi. So the answer makes sense. Now wewill prove it more rigorously.

For x / B, the integrandn(cosx)2n

2x goes to 0 as n because cosx < 1. So for any open setS sufficiently disjoint from B, we might guess that

limn

S

n(cosx)2n

2xdx = 0.

If we require that every point in S is at least > 0 away from any point in B, then this is indeedtrue. There are two ways to see this.

The fanciest way to see it is to use the dominated convergence theorem which says that if asequence of functions fn converges pointwise to a funciton f and if there is some function with|fn(x)| < (x) for all x S and

S < , then limn

S fn =

S f . To apply this theorem,

we let fn be the integrand of the n-th term of the sequence. To construct , notice that sinceevery point in S is at least away from every point in B, there is some < 1 so that | cosx| < for all x S. So n(cosx)2n is bounded by n2n for all x S. Since n2n has a finitelimit as n , n(cosx)2n is bounded by some finite number B for all x S. So we can let(x) = B/2x. Then |fn(x)| < (x) for all x S and

S < , just as we need to apply the

theorem. So we apply the theorem to get

limn

S

n(cosx)2n

2xdx =

S

limn

n(cosx)2n

2xdx =

S

0dx = 0.

But we of course dont expect you to know the dominated convergence theorem, so we can alsoprove this using a bare hands method that is actually easier. (Bare hands is usually muchharder than the dominated convergence theorem proof. That is why people use the dominatedconvergence theorem. But we have arranged for this problem to work with bare hands.) As weargued above, there is some < 1 so that cosx < for all x S. Then n(cosx)2n < n2nfor all x S. So we have a bound

S

n(cosx)2n

2xdx n2n

S

1

2xdx.

Since the integral

12x converges, this goes to 0 as n so we again have the desired result.

The upshot of all this is that we can now define B to be the points in [1,) that are within of B and have

limn

1

n(cosx)2n

2xdx = lim

n

B

n(cosx)2n

2xdx.


To calculate the integral on the right, notice that it is just the sum over all integers k 0 of kpi+kpi

n(cosx)2n

2xdx =

1

2kpi

n(cosx)2n

2xdx.

By the formula for summing geometric series, the sum of this over all integers k 0 is

limn

1

n(cosx)2n

2xdx =

2pi

2pi 1

n(cosx)2n

2xdx. (1)

So we have reduced the problem to calculating the following limit:

limn

n(cosx)2n

2xdx.

To do this, bound the limit above and below by taking the highest and lowest possible values of2x out of the integral:

2 limn

n(cosx)2ndx lim

n

n(cosx)2n

2xdx 2 lim

n

n(cosx)2ndx.

By a very similar argument as above, nothing outside (, ) contributes to the integrals in ourbounds and therefore

limn

n(cosx)2ndx = lim

n

pi/2pi/2

n(cosx)2ndx.

We have already calculated the right hand side: it is cpi. So we can plug this back into ourbounds to get

2cpi limn

n(cosx)2n

2xdx 2cpi

Plugging this bound into (1) gives

2cpi2pi

2pi 1 limn 1

n(cosx)2n

2xdx 2cpi 2

pi

2pi 1 .

Since was arbitrary, taking 0 forces

limn

1

n(cosx)2n

2xdx = cpi

2pi

2pi 1 ,

as desired.

Finally, you might be interested in knowing why c is a positive finite constant. (This is notnecessary to solve the problem, but it is necessary to be sure that the problem makes sense.And it is interesting.) To see this, let bn =

nnk=1(1 12k ). Then

log bn =1

2log n+

nk=1

log(1 12k

) =1

2log n+

nk=1

( 12k

+O(1

k2)).

Here O( 1k2

) denotes some function of k whose absolute value is always less than C 1k2

for someC big enough. The fact that log(1 12k ) = 12k +O( 1k2 ) follows from the taylor expansion of log


around 1. It is well known thatn

k=11k = log n+ +O(

1n) where is some constant. Plugging

this in gives

log bn =1

2log n 1

2log n 1

2 +O(

1

n) +

nk=1

O(1

k2) = 1

2 +O(

1

n) +

nk=1

O(1

k2).

Since

k=11k2

converges to some constant,n

k=1O(1k2

) converges to some finite constant asn. Therefore log bn 12+ as n. This is some finite number, so c = exp(12+)is a positive finite number as desired.


1. What is 100

(x 5) + (x 5)2 + (x 5)3 dx?2. Find the maximum value of 3pi/2

pi/2sin(x)f(x) dx

subject to the constraint |f(x)| 5.3. Calculate 35

25

1

x x3/5 dx.

4. Compute the x-coordinate of the point on the curve y =x that is closest to the point (2, 1).

5. Let

f(x) = x+x2

2+x3

3+x4

4+x5

5,

and set g(x) = f1(x). Compute g(3)(0).

6. Compute

limx0

(sinx

x

) 11cos x

.

7. A differentiable function g satisfies x0

(x t+ 1)g(t) dt = x4 + x2

for all x 0. Find g(t).8. Compute

0

lnx

x2 + 4dx.

9. Find the ordered pair (, ) with non-infinite 6= 0 such that limn

n2

1!2! n!n

= holds.

10. Find the maximum of 10

f(x)3 dx

given the constraints

1 f(x) 1, 10

f(x) dx = 0.


1. Answer: 2503

Solution: This integral is equal to 55x+ x2 + x3 dx =

55x2 dx =

(53

3 (5)

3

3

)=

250

3.

2. Answer: 20

Solution: Clearly we want to maximize f(x) when sin (x) 0 and minimize f(x) when sin (x) < 0. Wedo this by setting f(x) = 5 in the first case and f(x) = 5 in the second case. Noting that the boundsof integration cover precisely one full period of sin, we see that the integral becomes equivalent to twicethe integral of 5 sin (x) over the half period where sin (x) 0. This results in 20 .

3. Answer: 52ln 8

3

Solution: Note that we can write the integral as 3525

1

x3/5(x2/5 1) dx.

We solve via u-substitution. Let u = x2/5 1:

du =2

5x3/5 dx = dx = 5

2x3/5 du.

The integral becomes

5

2

321221

x3/5

x3/5 u du =5

2

83

1

udu,

which evaluates to5

2(ln 8 ln 3) = 5

2ln

8

3.

4. Answer: 2+3

2

Solution: We want to minimize the distance between the points (a2, a) and (2, 1). We can equivalentlyminimize the square of the distance between those two points, which is

(2 a2)2 + (1 a)2 = a4 3a2 2a+ 5.The derivative of this function is 4a3 6a 2, which can be factored as 2(a+ 1)(2a2 2a 1). The rootsof this cubic are therefore a = 1, 1

3

2 . Two of the roots are negative and therefore invalid, so therefore

a = 1+3

2 and

a2 = x =2 +

3

2

5. Answer: 1

Solution: We begin by observing that due to the inverse function rule, the first three derivatives of g aredetermined by the first three derivatives of f . Additionally, f(0) = g(0) = 0. Let f(x) and g(x) = f1(x)be new functions whose first three derivatives at zero equal those of f and g respectively.

By Taylor Series expansion, we see that f(x) = ln(1 x) is a suitable choice. Then g(x) = 1 ex andg(3)(0) = g(3)(0) = e0 = 1 .

6. Answer: e1/3

Solution: We can approximate sinx and cosx by their Taylor series.

sinx x x3

6, cosx 1 x

2

2


Substituting the limit becomes

limx0

(1 x

2

6

)2/x2= limx

(1 +

1

6x2

)2x2= limx

(1 +

1

x2

)x2/3= limx

(1 +

1

x

)x/3= e1/3

because e = limx(1 + 1x

)x.

7. Answer: 12x2 24x+ 26 26exSolution: First differentiate the equation with respect to x:

g(x) +

x0

g(t) dt = 4x3 + 2x.

Differentiate again to obtaing(x) + g(x) = 12x2 + 2.

A particular solution 12x2 24x+ 26 can be found using the method of undetermined coefficient, so thegeneral solution will be

g(x) = 12x2 24x+ 26 + Cex

for some constant C. By substituting x = 0 into the first equation, we see that g(0) = 0. We therefore

find that C = 26, making the answer 12x2 24x+ 26 26ex .

8. Answer: pi ln 24

Solution: Substitute x = 2 tan to get 0

lnx

x2 + 4dx =

1

2

pi/20

ln(2 tan ) d =1

2 pi

2ln 2 +

1

2

pi/20

ln(tan ) d.

We will now show that this final integral is zero by substituting = pi/2 to yield pi/20

ln(tan ) d = 0pi/2

ln(

tan(pi

2

))d

=

pi/20

ln

(1

tan

)d =

pi/20

ln(tan) d,

which gives us what we wanted, so the answer ispi ln 2

4.

9. Answer: (1/2, e3/4)

Solution: Take the logarithm and approximate using Stirlings approximation1.

Stirlings approximation says that ln(n!) n lnn n in the limit of large n. Using this, we have

ln

(n2

1!2! n!n

)=

1

n2ln(1!2! n!) lnn = 1

n2

nk=1

ln(k!) lnn

1n2

nk=1

(k ln k k) lnn = 1n2

nk=1

(k ln k) 1n2n(n+ 1)

2 lnn

Approximate n(n+ 1) with n2 and approximate the infinite sum by an integral

1n2

n1

x lnx dx 12 lnn

1http://en.wikipedia.org/wiki/Stirlings_approximation


Integrating by parts

1n2

(n2

2lnn n

2

4+

1

4

) 1

2 lnn 1

2lnn 3

4 lnn.

Therefore,

limn ln

(n2

1!2! n!n

)= limn

[(1

2

)lnn 3

4

],

which is finite only when 12 = 0, in which case = 12 and the limit evaluates to 34 . Therefore, we wishto compute

limn

n2

1!2! n!n1/2

= exp

[limn ln

(n2

1!2! n!n

)]= e3/4 .

10. Answer: 1/4

Solution: Consider the following expression 10

(f(x) 1)(f(x) +

1

2

)2dx.

Since f(x) 1 this expression is less than or equal to 0. Meanwhile expanding the integrand gives

(f(x) 1)(f(x) +

1

2

)2= f(x)3 3

4f(x) 1

4,

so its integral is 10

(f(x) 1)(f(x) 1

2

)2dx =

10

f(x)3 dx 34

10

f(x) dx 14

10

dx

=

10

f(x)3 dx 14,

proving that the answer is at most 1/4. Equality occurs when

f(x) =

{1/2 if 0 x 2/31 if 2/3 < x 1 ,

so 1/4 is indeed the maximum.


1. Find

x+2(x1)2(x2)dx.

2. Tangent lines are drawn at the points of inflection for the function f(x) = cosx on [0, 2pi]. The linesintersect with the x-axis so as to form a triangle. What is the area of this triangle?

3. Let f be one of the solutions to the differential equation

f (x) 2xf (x) 2f(x) = 0.

Supposing that f has Taylor expansion

f(x) = 1 + x+ ax2 + bx3 + cx4 + dx5 +

near the origin, find (a, b, c, d).

4. What is the value of the alternating harmonic series 1 12 + 13 14 + . . .?5. Solve the integral equation

f(x) =

x0

exyf (y)dy (x2 x+ 1)ex

for f(x).

6. Evaluate the integral pi0

|sin(2x) sin(3x)| dx

and express your answer in the a+bc

d , where a, b, c, and d are integers.

7. Let f(x) = x3e(x

2)

1x2 . Find f(7)(0), the 7th derivative of f evaluated at 0.

8. For the curve sin(x) + sin(y) = 1 lying on the first quadrant, find the constant such that

limx0

xd2y

dx2

exists and is nonzero.

9. Evaluate the integral

pi2

0

dx

1 + (tanx)pi.

10. Evaluate the following integral: 10

dx

x(x+ 1)(ln(1 + 1x

))2011

RMT 2011 Calculus Solutions February 19, 2011

1. Answer: 4 ln(x 2) 4 ln(x 1) + 3/(x 1) or equivalent formsExpand by partial fractions:

x+ 2

(x 1)2(x 2) =A

x 2 +B

x 1 +C

(x 1)2 .

Then x+ 2 = A(x 1)2 + B(x 2)(x 1) + C(x 2). If x = 2, then 4 = A. If x = 1, then C = 3.If x = 0, then 2 = 4 + 2B + 6 B = 4. So

x+ 2

(x 1)2(x 2)dx = [

A

x 2 +B

x 1 +C

(x 1)2]dx

= 4 ln(x 2) 4 ln(x 1) + 3/(x 1).

2. Answer: pi2

4

By taking the second derivative, we see that the points of inflection occur when cosx = 0, sox = pi2 ,

3pi2 . At

pi2 , the slope of the tangent line is the derivative evaluated at this point, which is sin (pi2 ) = 1. So the tangent line has equation y = x + b, and since cos (pi2 ) = 0, 0 = pi2 + b

and so the tangent line is given by y = x + pi2 . Similarly, we can find that the tangent line at 3pi2 isy = x 3pi2 . The intersection of the two tangent lines forms the vertex of the triangle, which occurswhen x + pi2 = x 3pi2 , or at x = pi. At this point, both tangent lines equal pi2 . Hence the triangleformed has height pi2 . The tangent lines cross the x-axis when y = 0, giving

pi2 and

3pi2 , which are

separated by pi. Hence the area of the triangle is 12 pi2 pi = pi2

4 .

3. Answer: (1, 2/3, 1/2, 4/15)

Just compute the Taylor expansion of f (x) 2xf (x) 2f(x) to the third term, which is

(2a+ 6bx+ 12cx2 + 20dx3 + ) (2x+ 4ax2 + 6bx3 + ) (2 + 2x+ 2ax2 + 2bx3 + ).

All coefficients should be zero, so 2a 2 = 0, 6b 4 = 0, 12c 6a = 0 and 20d 8b = 0. Solving theseequations gives the answer.

4. Answer: ln(2)

Note that 11+x = 1 x + x2 x3 + . . . and that ln(x + 1) =

11+xdx = C + x x

2

2 +x3

3 x3

4 + . . ..

Since ln(1) = 0 = C, we set C = 0. Evaluating at x = 1, we get that ln(2) = 1 12 + 13 14 + . . ..5. Answer: f(x) = (2x 1)ex

Differentiate both sides to get

f (x) = f (x) + x0

exyf (y)dy (x2 + x)ex.

But x0

exyf (y)dy = f(x) + (x2 x+ 1)ex

so by substituting it we get

f(x) + (x2 x+ 1)ex (x2 + x)ex = 0,

and f(x) = (2x 1)ex.

6. Answer: 55+46

First, we eliminate the absolute value by finding the intervals on which the expression inside is positiveand negative. Sketching sin(2x) and sin(3x) suggests that there are two points 0 < x < pi such thatsin(2x) = sin(3x). The identity sin(x) = sin(pix) gives us the solution x = pi5 , and sin(x) = sin(3pix)


gives us x = 3pi5 . Then we have sin(3x) > sin(2x) on the intervals (0,pi5 ) and (

3pi5 , pi) but sin(2x) >

sin(3x) on (pi5 ,3pi5 ). So the integral becomes: pi

0

|sin(2x) sin(3x)|dx = pi

5

0

[sin(3x) sin(2x)]dx

+

3pi5

pi5

[sin(2x) sin(3x)]dx+ pi3pi5

[sin(3x) sin(2x)]dx

Next, we let F (x) equal cos(2x)2 cos(3x)3 , so that F (x) = sin(3x) sin(2x). Then the integral simplifiesto: [

F(pi

5

) F (0)

][F

(3pi

5

) F

(pi5

)]+

[F (pi) F

(3pi

5

)]= F (pi) + 2F

(pi5

) 2F

(3pi

5

) F (0)

=1

2+

1

3+ cos

2pi

5 2

3cos

3pi

5 cos 6pi

5+

2

3cos

9pi

5 1

2+

1

3

=2

3+

5

3cos

2pi

5+

5

3cos

pi

5.

Now we use the identities cos pi5 =5+14 and cos

2pi5 =

514 to simplify this further:

2

3+

5

3cos

2pi

5+

5

3cos

pi

5=

2

3+

5

3

(5 14

+

5 + 1

4

)

=2

3+

5

5

6

=5

5 + 4

6.

7. Answer: 12600

Since f (n)(0) = ann!, where an is the nth Taylor series coefficient, we just need to find the Taylor seriesof f and read off the appropriate coefficient. The Taylor series is given by

f(x) = x3(

1 +x2

1!+x4

2!+

)(1 + x2 + x4 + ) .

The coefficient of x7 is 12! +11! + 1 =

52 , so f

(7)(0) = 7! 52 = 12600.8. Answer: 3

2

Differentiate the equation to get

cos(x) +dy

dxcos(y) = 0

and again

sin(x) + d2y

dx2cos(y)

(dy

dx

)2sin(y) = 0.

By solving these we havedy

dx= cos(x)

cos(y)

andd2y

dx2=

sin(x) cos2(y) + sin(y) cos2(x)

cos3(y).


Let sin(x) = t, then sin(y) = 1 t. Also cos(x) = 1 t2 and cos(y) = 1 (1 t)2 = t(2 t).Substituting it gives

d2y

dx2=t2(2 t) + (1 t)(1 t2)

t3/2(1 2t)3/2 = t3/2 t

2 t+ 1(1 2t)3/2 .

Since limx0 xt = 1, =32 should give the limit limx0 x

d2ydx2 =

24 . All other values of will make

this limit undefined or zero.

9. Answer: pi4

We make the substitution, x = pi2 y (note that the actual variable of integration is irrelevant so weleave it as x). Then we have: pi

2

0

dx

1 + (tanx)pi=

0pi2

dx1 + tan (pi2 x)pi

=

pi2

0

dx

1 + (cotx)pi=

pi2

0

(tanx)pi dx

(tanx)pi + 1.

Then we add the original integral to both sides:

2

pi2

0

dx

1 + (tanx)pi=

pi2

0

1

1 + (tanx)pi+

(tanx)pi

1 + (tanx)pidx =

pi2

0

1 + (tanx)pi

1 + (tanx)pidx =

pi2

0

dx =pi

2.

So the integral we want is pi4 .

10. Answer: 12010(ln 2)2010

Make the substitution 1x = et 1, so that et dt(et1)2 = dx. This transforms the original integral into 1

0

dx

x(x+ 1)(ln(1 + 1x

))2011 = ln 2

et(et 1)2 dtet(et 1)2t2011 dt =

ln 2

dt

t2011dt

= 12010t2010

ln 2

=1

2010 (ln 2)2010 .


1. Find the exact value of 1 13!

+15! . . ..

2. At RMT 2009 is a man named Bill who has an infinite amount of time. This year, he is walkingcontinuously at a speed of 11+t2 , starting at time t = 0. If he continues to walk for an infinite amountof time, how far will he walk?

3. Evaluate limx0

10x2

sin2(3x).

4. Compute 10

tan1(x)dx

5. Let a(t) = cos2(2t) be the acceleration at time t of a point particle traveling on a straight line. Supposeat time t = 0, the particle is at position x = 1 with velocity v = 2. Find its position at time t = 2.

6. Find n=2

dn

dxn(eax)

for |a| < 1.7. Compute

limn

nk=1

n kn2

cos(

4kn

).

8. Evaluate0

4bx + 7ce2xdx. Remember to express your answer as a single fraction.

9. Computen=0

n

(15

)n.

10. Evaluaten=1

150 + n2/80000

, as a decimal to the nearest tenth.


1. Answer: sin 1

By Taylor Expansion, sinx = x x33! + x5

5! x7

7! + . . . . Let x=1, and the desired value equals sin1.

2. Answer: pi2

d =0

11+t2 dt = tan

1(t)|0 = limt tan1(t) tan1(0) = pi2 .

3. Answer: 109

By lHopitals rule,

limx0

10x2

sin2(3x)= lim

x020x

6 sin(3x) cos(3x)

= limx0

20x3 sin(6x)

= limx0

2018 cos(6x)

=109.

4. Answer: pi2 ln(2)4

or equivalent expression

We integrate by parts: 10

1 tan1(x)dx = [x tan1(x)]10 10

x

x2 + 1dx

=pi

4 0

[12

ln(x2 + 1)]10

=pi

4 ln(2)

2.

5. Answer: cos(8)32

+ 3332

v(t) =a(t)dt =

cos2(2t)dt =

1 + cos(4t)

2dt =

sin(4t)8

+t

2+ c1,

where c1 is a constant. Plug in t = 0 to find v(0) = c1 = 2. So v(t) = sin(4t)8 + t2 2.

x(t) =v(t)dt =

sin(4t)

8+t

2 2dt = cos(4t)

32+t2

4 2t+ c2.

Plug in t = 0 to get x(0) = 116 + c2 = 1, so c2 = 3332 . Thus,

x(2) = cos(8)16

+3332.

6. Answer: a2

1+aeax

Since dn

dxn (eax) = (a)neax,

n=2

dn

dxn(eax) = eax

n=2

(a)n.

This forms a geometric series with common ratio a and first element a2, which converges since |a| < 1.Thus the answer is a

2

1+aeax.


7. Answer: 1cos(4)16

Define a partition on [0, 1] with n elements by setting xi = in for 0 i n. Then xi xi1 = 1n forall i. If we let f(y) = (1 y) cos(4y) and put yk = kn for 1 k n, then we have

nk=1

n kn2

cos(

4kn

)=

nk=1

f(yk)(xi xi1).

Thus, we may conclude that

limn

nk=1

f(yk)(xi xi1) = 10

f(y)dy

= 10

(1 y) cos(4y)

=[(

1 y4

)sin(4y) cos(4y)

16

]10

= cos(4)

16+

116.

8. Answer: 14e212

e21 , or1412e21e2

To evaluate the floor function, split the integral into unit intervals: 0

4bx+ 7ce2xdx =k=0

k+1k

4(k + 7)e2xdx

= (14e0 14e2) + (16e2 16e4) + (18e4 18e6) + . . .= 12 + 2(e0 + e2 + e4 + . . .)

= 12 +2

1 e2 =14 12e2

1 e2 =14e2 12e2 1 .

9. Answer: 516

Let S =n=0

n5n . Then

15S =

n=0

n

5n+1=n=0

n+ 1 15n+1

=n=0

n+ 15n+1

n=0

15n+1

=n=1

n

5n

15

1 15= S 1

4

S5

= S 14 4S

5=

14 S = 5

16.

10. Answer: 62.8

This sum is difficult to evaluate exactly. However, it can be closely approximated by the improperintegral of the same function, which is easily evaluated using u-substitution.


0

dx

50 + x2/80000=

150

0

dx

1 + (x/20000)2

=150

0

2000du1 + u2

=200050

[tan1 u

]0

= 40 limb

(tan1(b) tan1(0))= 40 lim

btan1(b)

= 40pi

2= 20pi 62.83.

To see that this integral is correct to the nearest tenth, we observe that since the integrand is a mono-tonic function, we can bound it above and below by Riemann sums. More precisely:

n=1

150 + n2/80, 000

20pi n=0

150 + n2/80, 000

.

By rearranging terms, this implies that:

20pi 150n=1

150 + n2/80, 000

20pi.

From this it follows that 62.8 is indeed correct to the nearest tenth.


1. Compute pi/20

sinx cosx dx.

2. Evaluate:

limx0

10x2 12x3e13x

2 13. Find the area enclosed by the graph given by the parametric equations

y = sin(2t)x = sin(t)

4. Find the value of the nth derivative of f(x) = sinn(x) at x = 0.

5. Water flows into a tank at 3 gallons per minute. The tank initially contains 100 gallons of water, with50 pounds of salt. The tank is well-mixed, and drains at a rate of 2 gallons per minute. How manypounds of salt are left after one hour?

6. Evaluatee3x sin(x)dx.

7. Computen=0

2n1

n!.

8. Find f(x) such that limh0

h2

f(x+2x)2f(x+h)+f(x) = x3

2 x 12x .

9. Suppose x(t) + x(t) = t5x(t). Let the power series representation of x be x(t) =ant

n. Find an interms of an1 and an7, where n > 7.

10. Evaluate: x

t2tetdt


1. Answer: 12

Using the product to sum trigonometric identity, we get: pi/20

sinx cosx dx = pi/20

12

(sin 2x sin 0) dx

= 12

pi/20

sin 2x dx

= 12( 12 cos 2x)pi/20

= 12 (1/2 (1/2))= 12 (1)= 1/2

2. Answer: 30

Two applications of LHopitals rule are necessary to obtain a fraction which does not divide by zero;the result is

20 3x23ex2 + 49x

2ex2=

20 023 + 0

= 30

3. Answer: 83

The graph is a figure-8. By symmetry, the area it encloses is four times the area under the curve that weget from just considering 0 t pi2 . For these values of t, we can write y = 2 sin(t) cos(t) = 2x

1 x2

for 0 x 1. But this is integrable easily enough by letting u = 1 x2. The antiderivative that weget is 23u

32 , which gives an area of 23 . Thus the total area enclosed by the parametric graph is

83 .

4. Answer: n!

You can think of taking the derivatives this way, in an extended version of the product rule: For eachj n, let fj(x) = sin(x). Then f(x) = f1(x)f2(x)...fn(x). Each time you take the derivative, youpick one of the fj s and replace it with its derivative. Some fj s might be differentiated more thanonce in this process. To find the derivative of f , take the results from each way of choosing the fj s todifferentiate and add them all up.

That said, since sin(0) = 0, if any fj doesnt get differentiated, itll make the whole product zerowhen x = 0, so these terms dont matter. But were only taking the nth derivative, so if every fjgets differentiated at least once, then they all have to be differentiated exactly once. There are n!remaining terms (the number of ways to choose what order we differentiate the fj s in), each of whichis cosn(0) = 1, so the value we want is just n!.

5. Answer: 62532

Let w(t) and s(t) denote the amounts of water and salt, respectively, in the tank at time t. We canimmediately see that w(t) = t + 100. Since the tank is constantly mixed, we know that

ds

dt= 2 s(t)

w(t)ds

s= 2 dt

t + 100ln(s) = 2 ln(C(t + 100))

s =C

(t + 100)2

Since s(0) = 50, C = 500000, so s(60) = 625/32.

6. Answer: 110e3x (3 sin(x) cos(x))


Let I be the answer. Integrating by parts twice:

I =13e3x sin(x)

13e3x cos(x)dx

=13e3x sin(x) 1

9e3x cos(x) 1

9I

I =910

(13e3x sin(x) 1

9e3x cos(x)

)

7. Answer: 12e2

Multiply the sum by 2 and you getn=0

2n

n!.

Now notice that the nth derivative of e2x is 2ne2x, which gives 2n evaluated at x = 0. The Taylor

series for e2x around x = 0 is thusn=0

2n

n!xn. Evaluated at x = 1, this gives the sum above, so the sum

we want is equal to 12e2.

8. Answer: tan1 x or tan1(x) + ax+ b

The limit is, by definition, 1f (x) . Therefore:

1f (x)

= x4 + 2x2 + 1

2x= (1 + x

2)2

2x

f (x) = 2x(1 + x2)2

f (x) =1

1 + x2

f(x) = tan1 x

9. Answer: an7(n1)an1n(n1)

Plugging the power series into the differential equation gives:n(n 1)antn2 +

nant

n1 =

antn+5

n(n 1)antn2 +

(n 1)an1tn2 =

an7tn2

n(n 1)antn2 + (n 1)an1tn2 = an7tn2n(n 1)an + (n 1)an1 = an7

10. Answer: x(2e)x

1+ln 2 (2e)x)

(1+ln 2)2

x

t2tetdt =[t

1ln(2e)

(2e)t]x 0

1ln 2e

(2e)tdt

=x(2e)x

1 + ln 2[

(2e)t

(1 + ln 2)

]x

=x(2e)x

1 + ln 2 (2e)

x

(1 + ln 2)2

Calculus Test2007 Rice Math Tournament

February 24, 2007

1. Find limx0

1+cos x3x2+4x3 .

2. A line through the origin is tangent to y = x3 + 3x+ 1 at the point (a, b). What is a?

3. A boat springs a leak at time t = 0, with water coming in at constant rate. At a time t = > 0 hours,someone notices that there is a leak and starts to record distance the boat travels. The boats speed isinversely related to the amount of water in the boat. If the boat travels twice as far in the first houras in the second hour, what is ?

4. Let I(n) = pi0sin(nx)dx. Find

n=0

I(5n).

5. Let k(x) be 0 for x < k and 1 for x k. The Dirac delta function is defined to be k(x) = ddxk(x).(Its really called a distribution, and we promise it makes sense.) Suppose d

2

dx2 f(x) = 1(x)+ 2(x) andf(0) = f (0) = 0. What is f(5)?

6. Point A is chosen randomly from the circumference of the unit circle, while point B is chosen randomlyin the interior. A rectangle is then constructed using A,B as opposite vertices, with sides parallel orperpendicular to the coordinate axes. What is the probability that the rectangle lies entirely insidethe circle?

7. A balloon in cross-section has the equation y = 2x x2ex/2, with the x-axis beginning at the topof the balloon pointing toward the knot at the bottom. What is its volume?

8. Silas does nothing but sleep, drink coffee, and prove theorems, and he never more than one at a time.It takes 5 minutes to drink a cup of coffee. When doing math, Silas proves s+ ln c theorems per hour,where c is the number of cups of coffee he drinks per day, and s is the number of hours he sleeps perday. How much coffee should Silas get in a day to prove the most theorems?

9. Evaluate limn

2nk=n+1

1k .

10. Find the 10th nonzero term of the power series for f(x) = x(x21)2 (expanding about x = 0).

Calculus Solutions2007 Rice Math Tournament

February 24, 2007

1. Answer: 16

Use lHopitals rule:

limx0

1 + cosx3x2 + 4x3

= limx0

sinx6x+ 12x2

= limx0

cosx6 + 12x

2. Answer:342

y = 3a2 + 3 =a3 + 3a+ 1

a=

y

x2a3 1

a= 0

a3 =12

3. Answer:512

The speed will cancel out so assume it is 1. We then have: t+1

1tdt = 2

+2+1

1tdt

ln + 1

= 2 ln + 2 + 1

+ 1

=( + 2 + 1

)2 =

152

4. Answer: 52

For odd n, I(n) = cos(nx)npi0= 2/n, so

n=0

I(5n) =n=0

2/5n = 5/2

5. Answer: 7

We have f (x) =(1(x) + 2(x)) dx = 1(x) +2(x) +C, and f (0) = 0 so C = 0. Integrating up to

f is most easily accomplished graphically; the region under the curve from 0 to 5 is a 1 4 rectanglefrom x = 1 to x = 5 with a 1 3 rectangle from x = 2 to x = 5 on top.

6. Answer: 4pi2

Suppose A lies at polar coordinate 0 < < pi/2. For the rectangle to lie within the circle, B must lie inthe rectangle with vertices at A, A reflected over the x-axis, A reflected over the y-axis, and A reflectedover both axes. Thus for this fixed A, the probability is (2 sin )(2 cos )/pi = 2 sin(2)/pi. The totalprobability is then 2pi

pi/20

2pi sin(2)d. (Integrating over the circle requires taking the absolute value

of the expression for area, which then splits up into four sections identical to the one considered here.)

1

7. Answer: 4pie2

V = pi 20

(2x x2ex/2

)2dx = pi

20

(2x x2)exdx = pix2ex20

8. Answer: 12 cups of coffee

The number of theorems proven is (s + ln c)(24 s c/12). Differentiating with respect to s gives24 c12 2s ln c = 0, so s = 12 c24 12 ln c. This is a maximum in s since the second derivativeis 2. Plugging this back in and simplifying gives (12 c24 + ln c2 )2 = f(c)2 theorems proven. Thisdifferentiates to 2f (c)f(c), so the derivative will be zero when either f(c) or f (c) is zero. f(c) = 0is difficult to solve, involving both a logarithm and a binomial, but f (c) = 12c 124 , so c = 12 is asolution. It is a maximum in c since the second derivative is 2f (c)2 + 2f(c)f (c), with f (12) < 0,f(12) > 0, and f (12) = 0.

9. Answer: ln 2

2nk=n+1

1k=

n

n

nk=1

1k + n

=n

k=1

1n

11 + kn

This is a Riemann sum: 21

1xdx = ln 2.

10. Answer: 10x19

Note thatf(x)dx = 12(1x2) =

14

(1

1+x +1

1x). These are geometric sums, so we have

f(x)dx =

14

( k=0

xk +k=0

(x)k)

=12

k=0

x2k

f(x) =k=0

kx2k1

2


February 25, 2006

1. Evaluate:

limx0

ddx

sin xx

x

2. Given the equation 4y + 3y y = 0 and its solution y = et, what are the values of ?3. Find the volume of an hourglass constructed by revolving the graph of y = sin2(x) + 110 from pi2 to pi2

about the x-axis.

4. Evaluate

limx0

ln(x+ 1)x ((1 + x) 1x e)

5. Evaluate:(x tan1 x)dx

6. Evaluate pi/20

sin3 xsin3 x+ cos3 x

dx

.

7. Find Hn+1(x) in terms of Hn(x),H n(x),Hn(x), . . . for

Hn(x) = (1)nex2 dn

dxn

(ex

2)

8. A unicorn is tied to a cylindrical wizards magic tower with an elven rope stretching from the unicornto the top of the tower. The tower has radius 2 and height 8; the rope is of length 10. The unicornbegins as far away from the center of the tower as possible. The unicorn is startled and begins to runas close to counterclockwise as possible; as it does so the rope winds around the tower. Find the areaswept out by the shadow of the rope, assuming the sun is directly overhead. Also, you may assumethat the unicorn is a point on the ground, and that the elven rope is so light it makes a straight linefrom the unicorn to the tower.

9. Define the function tanhx = exexex+ex Let tanh

1 denote the inverse function of tanh. Evaluate andsimplify:

d

dxtanh1 tanx

10. Four ants Alan, Bill, Carl, and Diane begin at the points (0, 0), (1, 0), (1, 1), and (0, 1), respectively.Beginning at the same time they begin to walk at constant speed so that Alan is always movingdirectly toward Bill, Bill toward Carl, Carl toward Diane, and Diane toward Alan. An approximatesolution finds that after some time, Alan is at the point (0.6, 0.4). Assuming for the moment thatthis approximation is correct (it is, to better than 1%) and so the pont lies on Alans path, what isthe radius of curvature at that point. In standard Cartesian coordinates, the radius of curvature of afunction y(x) is given by:

R =

(1 +

(dydx

)2)3/2 d2ydx2

1


February 25, 2006

1. Answer: 13

limx0

ddx

sin xx

x= lim

x0x cosx sinx

x3

= limx0

cosx x sinx cosx3x2

= limx0

sinx3x

= limx0

cosx3

= 13

2. Answer: 14,1

You substitute the solution of y = et into the differential equation. You then get 42et+3etet =0 where you can divide through by et and end up with 42+3 = 0. You get (4 1)(+1) = 0where must be equal to 14 , 1.

3. Answer: 97pi2

200

The volume is pi2

pi2pi

(sin2 x+

110

)dx = pi

pi2

pi2

(sin4 x+

15sin2 x+

1100

)dx

=pi2

100+ pi

pi2

pi2

(sin2 x(1 cos2 x) + 1

5sin2 x

)dx

=pi2

100+ pi

pi2

pi2

(65sin2 x 1

4sin2 2x

)dx

=pi2

100+ pi

(65pi

2 14pi

2

)=

97pi2

200

4. Answer: 1e

Rearrange to get limx0ln(x+1)

1x

(1+x)12e

. Let y = ln(x + 1)1x . Using LHopitals rule, limx0

ln(y)ye =

limx0yy

y = limx01y . To evaluate the limit in the denominator,

limx0

(1 + x)1x = lim

x0eln

(1+x)1x

= elimx0ln(1+x)

x = elimx01

1+x1 = e

so the answer is 1e

5. Answer: x2+ 1+x

2

2tan1 x

(x tan1 x)dx = x

2+12

(1 + 2x tan1 x

)dx

= x2+12

(1 + x2

1 + x2+ 2x tan1 x

)dx

= x2+12

((1 + x2)

d

dx(tan1 x) +

d

dx(1 + x2) tan1 x

)dx

= x2+1 + x2

2tan1 x

1

6. Answer: pi4

Let u = pi2 x. Substituting then changing u to x gives pi/20


dx = pi/20

cos3 xcos3 x+ sin3 x

dx

Adding the two integrals

2 pi/20


dx = pi/20

dx

so pi/20


dx =pi

2 12=

pi

4

7. Answer: 2xHn(x)H n(x)

Hn+1(x) = ex2 ddx

(ex

2Hn(x)

)= ex2

(2xex2Hn(x) + ex2H n(x)

)= 2xHn(x)H n(x)

8. Answer: 18 + 9pi

The shadow of the rope has length 6 by the Pythagorean theorem; we can work the rest of the problempretending we have a horizontal rope of length 6 from the unicorn to the tower. The unicorn travels aquarter-circle before the rope begins to wrap around the tower; from then on, if the rope has wrappedaround an angle of the tower, the rope remaining is 6 2 long. Using the Pythagorean theorem,we find that if r is the unicorns distance from the center of the tower, r2 = 22 + (6 2)2. The areaswept out is the initial quarter-circle travelled by the unicorn plus the area swept out by the line fromthe unicorn to the center of the tower minus the area of the tower covered in the angle traversed:

pi62

4+12

6/20

(22 + (6 2)2)d pi22 6/22pi

6-2

2

6

2

9. Answer: sec 2x

First we solve for the inverse function; let x = tanh1 y.

y = tanhx =ex exex + ex

y =e2x 1e2x + 1(

e2x + 1)y = e2x 1

e2x(y 1) = y + 1x =

12lny + 1y 1

Now we return to the given expression:

d

dxtanh1 tanx =

d

dx

12lntanx+ 1tanx 1

=12d

dx(ln (tanx+ 1) ln (tanx 1))

=12

(sec2 x

tanx+ 1 sec

2 x

tanx 1)

=12sec2 x

(tanx 1) (tanx+ 1)tan2 x 1

=sec2 x

1 tan2 x =1

cos2 x

1 sin2 xcos2 x=

1cos2 x sin2 x = sec 2x

10. Answer: 15

By symmetry, Bills path must be the same as Alans, rotated 90 counter-clockwise and shifted tostart at (1, 0). If Alans position is given by (x, y), Bills is then given by (1y, x). The slope of Alanspath at a point (x, y) is then y = xy1yx . This is undefined at the given point, so noting that thesign of the derivatives in the radius of curvature does not matter, we switch to the function x(y), withx = 1yxxy =

10.60.40.60.4 = 0 at the given point. Differentiating, x

= (1x)(xy)(x1)(1yx)

(xy)2 =(1)(.6.4)(1)(1.6.4)

(.6.4)2 = 5. The radius of curvature is then

R =

(1 +

(dxdy

)2)3/2d2xdy2 =

(1 + 0)3/2

|5| =15

The answer can also found using the parametric form of the radius curvature with (x(t), y(t))

3


February 26, 2005

1. Mathisgreatco, Inc. can produce at most 24 spherical cow statues each week. Experience has shownthat the demand for spherical cows sets the price at D = 110 2n where n is the number of statuesproduced that week. Producing n statues costs 600 + 10n + n2 dollars. How many statues should bemade each week to maximize profit?

2. f(x) = 12f(2x+ 1) and f(1) = 2005. What is limx1+ f(x) assuming it exists?

3. Sammy the Owl wants to design a window that is a rectangle with a semicircle on top. If the totalperimeter is constrained to be 24 feet, what dimensions should Sammy pick so that the window admitsthe greatest amount of light? Give the radius of the semicircular region and the height of the rectangularportion.

4. Given f(x) = x5 + 2x3 + 2x, find (f1)(5).5. Find the average value of f(x) = |x3| on the interval [1, 4].6. Sammy the Owl wants to watch his home movie of a trip to Rice University on his movie screen. The

lower edge of the screen, which is 30 feet high, is 6 feet above eye level. How far from the screen shouldthe observer sit to obtain the most favorable view, i.e. to maximize the visual angle (the observedangle between the top and bottom of the screen)?

7. Find the value of x for which the function f(x) = xx achieves a minimum on the positive real line (i.e.for all real numbers x such that x > 0)?

8. Evaluate

limn

20

(1 +

t

n+ 1

)ndt

9. Define a new function (x) with the properties (x) = 0 for all x 6= 0 and

(x)f(x)dx = f(0)

for every function f(x). Compute

(x) sin(x)dx.

For the purposes of this problem, you may assume the existence of

(x) d (x)dx

.

10. Computei=0

i(i+ 1)pi2

for 0 < p < 1.

1


February 26, 2005

1. Answer: 17

The profit is P is revenue minus cost. Thus P = n D (600 + 10n + n2) = 3n2 + 100n 600.Consider the smoothed version of P , and we can find the critical point of 16 23 . Thus either n = 16 orn = 17 is the largest profit. Quick computation yields n = 17.

2. Answer: 0

f(0) = 12f(1), f( 12 ) = 12f(0), f( 34 ) = 12f( 12 ), and so on. So the limit is 0.3. Answer: ( 24

4+pi, 244+pi

)

We want to maximize the area of the window. Let x be the radius of the semicircular region and ythe height of the rectangular portion of the window. The area of the window is A = 12pix

2 + 2xy. Theperimeter is P = pix + 2x + 2y. Eliminating y in A yields A = 24x (2 + pi2 )x2. We can take thederivative and find the critical point of x = 244+pi . Solving for y yields the same answer.

4. Answer: 113

Applying the chain rule to f(f1(x)) = x yields f (f1(x)) (f1)(x) = 1.Rearranging yields (f1)(x) = 1f (f1(x)) .f1(5) is the solution to 5 = x5 + 2x3 + 2x which has the unique solution of x = 1.f (1) = 5 + 6 + 2 = 13.Thus 113 is the answer.

5. Answer: 25720

The average value isR 41 f(x)dx4(1) =

15

41 |x3|dx = 15 (

01x3dx+

40x3dx) = 25720 .

6. Answer: 66 feet

From the diagram below, we can see we want to maximize . Note that tan = tan (AB) =tanAtanB1+tanA tanB =

36x 6x

1+ 36x6x

= 42xx2+216 . Using implicit differentiation, we can find that = cos2 42x2+9072(x2+216)2 .

The maximum occurs at x = 66.

7. Answer: 1e

f(x) = xx

= ln f(x) = x lnx

= f(x)

f(x)= 1 + lnx

= f (x) = f(x)[1 + lnx] = xx[1 + lnx]Critical points at f

(x) = 0

= 0 = xx[1 + lnx]= 0 = 1 + lnx (xx always positive for x > 0)

= lnx = 1= x = e1 = 1

e(critical point)

1

Furthermore,

f (x) = f (x)[1 + lnx] +f(x)x

f = xx(1 + lnx)2 + xx1 > 0 (for x > 0)

So, f (x) > 0 for all x > 0.Hence, the function is concave up everywhere, and so x = 1e must be a minimum.

8. Answer: e2 1limn

20

(1 + tn+1

)ndt = limn

(1 + tn+1

)n+1|20

= limn(1 + 2n+1

)n+1 limn

(1 + 0n+1

)n+1= e2 1 (since limn

(1 + tn

)n = et)9. Answer: 1

Use integration by parts:

(x) sin(x)dx

= (x) sin(x)dx|

(x) cos(x)dx

= 0 cos(0) ((x) = 0at )= 1

10. Answer: 2p(1p)3

i=0

i(i+ 1)pi2 =i=0

i(i 1)pi2 + 2i=0

ipi2

i=0

i(i 1)pi2 = d2

dp2

i=0

pi =d2

dp21

1 p =d

dp

1(1 p)2 =

2(1 p)3

i=0

ipi2 =1p

i=0

ipi1 =1p

d

dp

11 p =

1p(1 p)2

Then 2(1p)3 +2

p(1p)2 =2

p(1p)3

2


February 28, 2004

1. Evaluate limx(4x2 + 7x 2x).

2. Suppose the function f(x) f(2x) has derivative 5 at x = 1 and derivative 7 at x = 2. Find thederivative of f(x) f(4x) at x = 1.

3. An object moves along the x-axis with its position at any given time t 0 given by x(t) = 5t4 t5.During what time interval is the object slowing down?

4. For x > 0, let f(x) = xx. Find all values of x for which f(x) = f (x).

5. The highway department of North Eulerina plans to construct a new road between towns Alpha andBeta. Town Alpha lies on a long abandoned road running east west. Town Beta lies 3 miles north and5 miles east of Alpha. Instead of building a road directly between Alpha and Beta, the departmentproposes renovating part of the abandoned road (from Alpha to some point P ) and then bulding a newroad from P to Beta. If the cost of restoring each mile of old road is $200,000 and the cost per mile ofa new road is $400,000, how much of the old road should be restored in order to minimize costs?

6. Consider the two graphs y = 2x and x2 xy + 2y2 = 28. What is the absolute value of the tangent ofthe angle between the two curves at the points where they meet?

7. A mouse is sitting in a toy car hooked to a spring launching device on a negligibly small turntable.The car has no way to turn, but the mouse can control when the car is launched and when the carstops (the car has brakes). When the mouse chooses to launch, the car will immediately leave theturntable on a straight trajectory at 1 m/s. Suddenly someone turns on the turntable; it spins at 30rpm. Consider the set of points the mouse can reach in his car within 1 second after the turntable isset in motion. What is the area of this set?

8. A spherical cow is being pulled out of a deep well. The bottom of the well is 100 feet down and thecow and all the water on him weighs 200 pounds. He is being hauled up at a constant rate with a chainwhich weighs 2 pounds per foot. The water on the cow drips off the cow at the rate of 12 pounds perfoot as he is being hauled up. How much work is required to rescue the cow in foot-pounds? Rememberwork is force times distance.

9. The base of a solid is the region between the parabolas x = y2 and 2y2 = 3 x. Find the volume ofthe solid if the cross-sections perpendicular to the x-axis are equilateral triangles.

10. Find the positive constant c0 such that the series

n=0

n!(cn)n

converges for c > c0 and diverges for 0 < c < c0.

1


February 28, 2004

1. Answer: 74

limx(4x2 + 7x 2x) = limx(

4x2 + 7x 2x) (

4x2+7x+2x)

(4x2+7x+2x)

= limx 7x(4x2+7x+2x) 1x1x

=

limx 7(

4+ 7x+2)= 74 .

2. Answer: 19

The derivative of f(x) f(2x) is f (x) 2f (2x). So f (1) 2f (2) = 5, f (2) 2f (4) = 7. Thus

f (1) 4f (4) = (f (1) 2f (2)) + 2(f (2) 2f (4)) = 5 + 2 7 = 19,

the answer.

3. Answer: [3,4] or (3,4) or from t=3 to t=4

The velocity of the object is given by v(t) = x(t) = 20t3 5t4, and the acceleration function isa(t) = v(t) = 60t220t3. The object is slowing down when the velocity is positive and the accelerationis negative, or vice versa. v(t) is positive from t = 0 to t = 4 and is negative after that. a(t) is positivefrom t = 0 to t = 3 and negative afterward. These only differ in sign from t = 3 to t = 4.

4. Answer: 1

Let g(x) = log f(x) = x log x. Then f(x)f(x) = g

(x) = 1 + log x. Therefore f(x) = f (x) when1 + log x = 1, that is, when x = 1.

5. Answer: 53 milesLet x be the amount of old road restored. Then the length of the new road is

9 + (5 x)2 using the

Pythagorean Theorem. Thus the total cost of the plan is C(x) = 200000x + 400000x2 10x+ 34.

The minimum cost occurs at one of the critical points which are x = 53. Clearly 5 +3 is not avalid answer and one can check 53 is indeed a minimum.

6. Answer: 2

The two graphs intersect at x2 2x2 + 8x2 = 28 or rather x = 2 with y = 4. At x = +2, m1 = 2and m2 = y(2). Using implicit differentiation on the second graph, we find y(x) = y2x4yx and pluggingin (2, 4) gives a slope of 0. If is the angle between the graphs then | tan()| = | m2m11+m1m2 |. Pluggingin the values yields the answer 2. x = 2 yields the same value.

7. Answer: pi/6

The mouse can wait some amount of time while the table rotates and then spend the remainder ofthe time moving along that ray at 1 m/s. He can reach any point between the starting point and thefurthest reachable point along the ray, (1 /pi) meters out. So the area is pi

0

(1/2)(1 /pi)2 d = (1/2)(1/pi)2 pi0

2 d = pi/6.

8. Answer: 27500 foot-pounds

Let x indicate the distance the cow has yet to travel. Then the work for a distance dx is (2x+ 20012 (100 x))dx. Thus the total work is

1000

( 52x+ 150)dx = 27500 foot-pounds.

1

9. Answer: 33

2

The base region is bounded on the left by x = y2 and on the right by 2y2 = 3 x. The intersectionpoints are (1, 1) and (1,1). Each cross-section, say x = a, is an equilateral triangle. The lengthof a side is 2y where y =

x for a 1 but it is y =

3x2 for a 1. The area of an equilateral

triangle is3s2

4 where s is the side length. Thus the volume is 10

3(2x)2

4 dx+ 31

34 (2

3x2 )

2dx =3 10xdx+

32

31(3 x)dx = 3

3

2 .

10. Answer: 1e

The ratio test tells us that the series converges if

limn

(n+1)!(c(n+1))n+1

n!(cn)n

=1c limn

(n

n+ 1

)nis less than one and diverges if it is greater than one. But

limn

(n

n+ 1

)n= lim

n

(1 +

1n

)n=

1e.

Then the limit above is just 1ce , so the series converges for c >1e and diverges for 0 < c

calculus from rice university

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