1

MATH HL - OPTION CALCULUS REVISION

(one example for each case) by Christos Nikolaidis

Example Remarks

1. L’ Hôpital

Solution

It works only for fractions

00 or

∞

∞

2. Continuity - Differentiability

Solution

• f is continuous at x=2, so 2x

lim→

(x2+4)=2x

lim→

ax = f(2)

Thus, 8=2a=8 ⇒ a=4 • f is continuous at x=3, so

3xlim

→ax=

3xlim

→(b-cx2) =f(3)

Thus 12=b-9c=12 ⇒ b-9c=12 (*)

• f must be differentiable at x=2. Indeed, for x=2 =′− (2)f (x2+4)΄ =2x=4, =′+ (2)f (ax)΄=a=4

• f must be differentiable at x=3. That is, for x=3 =′− (3)f (ax)΄=a=4, =′+ (3)f (b-cx2)΄=-2cx=-6c so that 4=-6c (**)

(*) and (**) give c=-2/3 and b=6

Therefore a=4, b=6, c=-2/3

Continuity at x=a means

−→axlim f =

+→axlim f = f(a)

Differentiability at x=a means (a)f(a)f +− ′=′

i.e. the derivatives before and after point a are equal for x=a

Find 3

2

x0.5xx1xe

0xlim

−−−

→

00

Let

>−

≤<

≤+

=

3x,cxb3x2ax,

2x4,xf(x)

2

2

Find the values of a, b, c given that the function is continuous and differentiable.

2

3. Rolle Theorem – Mean Value Theorem (MVT)

Solution

(a) sinxf(x) = is continuous and differentiable in [0,π] and )f(π0f(0) ==

Thus there is some value c∈]0,π[ s.t. 0(c)f =′ Indeed, xc(x)f os=′ and cosc=0 for c=π/2

(b) f(x) is still continuous/differentiable in [0.1,π-0.1] )0.1-f(πf(0.1) =

Again c=π/2

(c) f(x) is continuous and differentiable in [0,π/2] Thus there is some value c∈]0,π/2[ such that

0π/2f(0)-)/2f(π(c)f

−=′ , i.e. cosc=

π2

Indeed the GDC gives c=0.881

The minimum requirement is: - Continuous in [a,b] - Differentiable in ]a,b[ - f(a)=f(b)

Then 0(c)f =′ for some c∈]a,b[ The minimum requirement is: - Continuous in [a,b] - Differentiable in ]a,b[

Then abf(a)-f(b)(c)f

−=′

for some c∈[a,b]

Solution

-1f(1) = <0 and f(2)=11>0 Thus there is at least one root between 1 and 2 Suppose that there are two roots a and b

Then - f(x) is continuous and differentiable in [a,b] - )f(b0f(a) ==

Thus, by Rolle, there is a value c∈[a,b] such that 0(c)f =′

But, 53x(x)f 2 +=′ which has no roots which contradicts the existence of c

Therefore there is only one root.

A polynomial is continuous and differentiable everywhere In general, if f(x) has n distinct roots, then (x)f ′ must have at least n-1 roots (apply Rolle between any two consecutive roots)

Let sinxf(x) = (a) Explain why Rolle’s Thm holds in the interval

[0,π] and find the corresponding value of c (b) Does it hold in the interval [0.1 , π-0.1]? (c) Apply the Mean Value Theorem in [0 , π/2]

and find the corresponding value of c

Let 75xxf(x) 3 −+= Use Rolle’s Thm to show that f(x) has only one root

3

Solution

Consider f(x)=ex

It is continuous and differentiable everywhere. We apply the theorem in the interval [0, x] where x>0

0xe-e(c)f

0x

−=′ for come c∈[0,x]

That is 1ex

1-e cx

>= ⇒ ex-1 > x ⇒ ex > x +1

We also apply in the interval [x, 0] where x<0

0xe-e(c)f

0x

−=′ for come c∈[x, 0]

That is 1ex1-e c

x<= ⇒ ex-1 > x ⇒ ex > x +1

Finally, for x=0, ex = x +1

Therefore, in any case ex ≥ x +1

since x is positive

since x is negative

4. Fundamental Theorem of Calculus

Solution

(a) ∫′

++

−+1

02

2dx

1xx2x3x =

1

01xx2x3x

2

2

++

−+=

32)(

32 8

=−−

(b) ∫x

2

3dt(lnt)dxd = 3(lnx)

(c) ∫2x

2

3dt(lnt)dxd = 32 )2x(lnx

(d) ∫2x

xdt(lnt)

dxd 3 =

dxd

− ∫∫ x

0

3

0

3 dt(lnt)dt(lnt)2x

= 32 )2x(lnx - 3(lnx)

• [ ] f(a)f(b)f(x)(x)dxf ba

a

b

−==′∫

• ∫x

af(t)dt

dxd = f(x)

• ∫g(x)

af(t)dt

dxd = f(g(x)) (x)g′

use =∫g(x)

h(x)f(t)dt ∫∫ −

h(x)g(x)

aaf(t)dtf(t)dt

Use the MVT for f(x)=ex to show that 1xex +≥

Find

(a) ∫′

++

−+1

02

2dx

1xx2x3x (b) ∫

x

2

3dt(lnt)dxd

(c) ∫2x

2

3dt(lnt)dxd (d) ∫

2x

xdt(lnt)

dxd 3

4

5. Riemann Sum

Solution

(a) Subintervals [0,4π ] [

4π ,

2π ] [

2π ,

43π ] [

43π ,π]

(i) R.S. = f(0)4π + f(

4π )

4π + f(

43π )

4π +f(π)

4π = 1.11

(ii) R.S. = f(4π )

4π + f(

2π )

4π + f(

2π )

4π +f(

43π )

4π = 2.68

(b) Subintervals [0,2π ] [

2π ,π]

R.S. = f(4π )

2π + f(

43π )

2π = 2.22

(c) 2sinxdxπ

0=∫

So the approximation in (b) is much better despite the larger subintervals

Riemann sum = ∑i

ii ∆x)f(x

6. Improper Integrals

Solution

(a) dxx1lim

1b ∫∞→

b

[ ]b1blnxlim

∞→= [ ]0lnblim

b−=

∞→+∞=

(b) dxx1lim

12b ∫∞→

b b

1b x1lim

−=∞→

+−=∞→

1b1lim

b1=

(c) Find the indefinite integral first

dxexx∫ c+=+== ∫∫ xx

x-x-x-

e1-

ex-dxe-xedxxe

Thus,

dxex

0x∫

+∞ b

0xb e1x-lim

+=

∞→

+

+=

∞→1

e1b-lim bb

1=

It diverges

It converges to 1

(in the last limit use l’Hôpital)

It converges to 1

Consider ∫π

0sinxdx . Find the Riemman sum

(a) if you consider 4 subintervals of equal length and as representative values xi (i) the minima (ii) the maxima

(b) if you consider 2 subintervals of equal length and the midpoints as representative values

(c) Compare with the actual value of the integral

Find (a) dxx1

1∫+∞

, (b) dxx1

12∫

+∞

, (c) dxex

0x∫

+∞

5

7. Differential Equations (D.E.)

Solution

xxdy2y

d=+1

∫∫ =+

⇒ xx2y

dyd

1

c2xarctany2

+=⇒

For x=0, y=1 c4π

=⇒

Hence ,

4π

2xarctany2

+=

Notice: If they ask to express y in terms of x:

General solution: c)2xtany2

+= ( ,

Particcular solution: )4π

2xtany2

+= (

Separable

as we can separate x’s from y’s and get

∫∫ = g(x)dxf(y)dy

This is the general solution Using the initial condition

This is the particular solution

Solution

Homogeneous

as it can take the form

)xy

F(dxdy

=

[this line is always the same!] Now it is a D.E. of separable variables [this is always the case!] General solution Using the initial condition Particular solution

Solve 2

22

xy13xy36x

dxdy ++

= , if y=1 when x=1.

Solve 1)x(ydxdy 2 += , if y=1 when x=0.

6

Solution

With Integrating Factor as it can take the form

Q(x)P(x)ydxdy

=+ [see booklet]

Spot P(x), Q(x)

[see booklet for I= ∫P(x)dxe ] Don’t put (+c) here You have to remember that

Iy= ∫ IQdx

Solution

ytanxcosxedxdy 3x −=

0.2xx n1n +=+

0.2y n1n +=+ y )tanxycosx(e nnn3xn −

Euler

as it says so! ☺

Always in the form y)F(x,dxdy

=

[see booklet for formulas]

The GDC does it all!

Recursion 0.2aa n1n +=+

0.2bb n1n +=+ nnn

3a tanabcosa(e n − )

Set 0a0 = , 1b0 =

Notice: the smaller step h, the better approximation!

Solve cosxtanxyedxdy

e 3x3x =+ −− , if y=1 when x=0.

Let cosxtanxyedxdy

e 3x3x =+ −− , if 1y = when 0x =

Use Euler with step h=0.2 to estimate y for 1x =

7

8. Isoclines – Slope Fields

Solution (a) The isoclines are the curves cyx =+ .

That is cxy +−=

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

x

y

(b) We sketch the slope on each point

(c) We draw a curve passing through (0,0)

The grid is usually given!

If y)f(x,dxdy

=

the isoclines are cy)f(x, =

← that is all lines of slope -1

Slope 0 means Slope 1 means Slope -1 means

For example, at (0,1) the slope is

1yxdxdy

=+= , that is

Notice that on each isocline the slopes are equal. E.g. on y=-x+1 all slopes are 1. Just look at the point (0,0) and “follow the stream”. Indeed, if you solve the d.e. (using integrating factor) you will find the particular solution y=ex-x-1 and the graph looks like that!

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

x

y

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

x

y

Consider yxdxdy

+= . In the following grid

(a) Sketch the isoclines (b) Draw the Slope Field (c) Sketch the partic. solution passing through (0,0)

8

9. Series

Solution

It diverges since 032

43n12nlim

n≠=

++

∞→

Divergence Test

[rarely used] if the limit is not 0, we stop there! The series diverges! If it is 0, we use another test.

Solution

We check the corresponding integral dxx(lnx)

1I2

2∫+∞

=

=∫ dxx(lnx)

12 =∫ dx

x(lnx)-2

c+−lnx1

Thus ∞→

=blimI

b

lnx1

2

−∞→

=blim

+−ln21

lnb1

ln21

=

The integral converges so the series converges.

Integral Test

2x(lnx)1

f(x) = must be positive,

continuous and decreasing Usually

• for p-series ∑∞

=1npn

1 → ∫∞

1px

dx

• if you see lnn • if you recognize a known

integral, e.g.∑∞

=1nnen → ∫

∞−

1

xdxxe

Solution

Since 22 n

143n

sinn≤

+ and ∑

∞

=1n2n

1 converges

our series converges as well.

Comparison Test

[rarely used! they usually mention it in the question] larger converges ⇒ smaller too smaller diverges ⇒ larger too

Solution

na looks like 2n n

1b = and ∑∞

=1n2n

1 converges

n

nn b

alim∞→ 1

n43n52nlim

2

3n +

+=

∞→0const

32

43n5n2nlim 3

23

n>==

+

+=

∞→

Our series converges as well.

Limit Comparison Test

[Popular!] We usually compare with a

p-series ∑∞

=1npn

1

1pifdiv1pifconv

≤

>

←any positive constant will do

[we know the result beforehand]

Solution

n

1nn a

alim +

∞→

( )( )

( )( )2n

21n

n !n2!2n

!22n!1)(n2lim

+

+=

+

∞→

( ) 2)(2n12n

1)n2lim2

n ++

+=

∞→

( 121

<=

Our series converges.

Ratio Test

[Popular!]

Use if you see n!, an or similar expressions e.g. 1×3×…………×(2n+1)

← <1, converges >1, diverges =1, don’t know (use another test)

∑∞

= +

+

1n 43n12n . Determine if it converges or diverges

∑∞

=2n2n(lnn)

1 . Determine if it converges or diverges

∑∞

= +1n 43nsinn2

. Determine if it converges or diverges

∑∞

= +

+

1n 43n52n

3 . Determine if it converges or diverges

( )( )∑

∞

=1n

2n

!2n!n2 . Determine if it converges or diverges

9

10. Alternating Series

Solution

043n

1limn

=+∞→

.

The sequence 43n

1+

decreases (see the graph in GDC)

The series converges

Alternating series Test

For ∑ − nn a1)(

or ∑ +− n1n a1)(

• 0alim nn

=∞→

.

• The sequence na decreases

Solution =1S 0.142, =3S 0.120, =5S 0.110

=2S 0.042, =4S 0.057, =6S 0.064

The result is always between two consecutive values so it is above 0.064, and thus above 0.06

Approximation

← upper bounds

← lower bounds (or vice-versa)

Solution (a) It is simply less that a7= 0.04

251

=

(b) It is simply less that a333= 0.0011000

11003

1=<

So the approximation is correct to 3 sf.

Error of approximation

When you accept Sn as an approximation then |error| < an+1

Solution By the alternating series test both series converge

We check the series of the absolute values

∑∞

=1n2n

1 converges, so the original converges absolutely

∑∞

=1n n1 diverges, so the original converges conditionally

Absolute convergence vs

Conditional convergence

If the series of the absolute values converges the original also converges (absolutely)

If the series of the absolute values diverges, check original by the alternating series test (it may converge conditionally)

Check ∑∞

=

−

1n2

n

n1)( and ∑

∞

=

−

1n

n

n1)( for absolute

or conditional convergence.

Let ∑∞

= +

−

1n

n

43n1)( . Find an upper bound of the error

(a) if you accept S6=0.064 as an approximation (b) If you accept S332 as an approximation; show

that the approximation is correct to 3sf.

Show that the sum ∑∞

=

+

+

−

1n

1n

43n1)( is > 0.06

∑∞

=

+

+

−

1n

1n

43n1)( ,Determine if it converges or diverges

10

11. Power Series

Solution

n

n

1n

1n

nn

1nn x

n31)3(n

xlima

alim+

+

∞→

+

∞→ += x

33nnlim

n +=

∞→=

3x

We solve the equation 13R

=

So the radius of convergence is 3R =

For x=3 the series becomes ∑∞

=1n n1 , so it diverges

For x=-3 the series becomes ∑∞

=1n

n

n(-1) , so it converges

Hence, the interval of convergence is x∈[-3,3[

Always Ratio Test

[the result can be something like that i.e. a|x| or 0, or ∝; see details in the last box]

← This means that the series converges for x∈]-3,3[ diverges for x outside we don’t know yet for x=±3

← Check the endpoints

Solution

n

n

1n

1n

nn

1nn 5)-(x

n31)3(n5)-(xlim

aalim

+

+

∞→

+

∞→ += 5-x

33nnlim

n +=

∞→

=35-x

We solve the equation 13R

=

So the radius of convergence is 3R =

For x=8 the series becomes ∑∞

=1n n1 , so it diverges

For x=2 the series becomes ∑∞

=1n

n

n(-1) , so it converges

Hence, the interval of convergence is x∈[2,8[

Always Ratio Test

Exactly the same procedure but center of convergence = 5 Interval = ]5-R,5+R[ (and check the endpoints)

← This means that the series converges for x∈]2,8[ diverges for x outside we don’t know yet for x=2,8 ← Check the endpoints

Ratio test Radius Interval of convergence

|x|a R=a1 x∈]-R,R[ and check at x=±R

0 R=∞ x∈]-∞,∞ [ ∞ R=0 x=0

The limit

n

1nn a

alim +

∞→

results in

a|x| or 0 or ∞

∑∞

=1nn

n

n3x , Find radius and interval of convergence

∑∞

= ⋅

−

1nn

n

3n5)(x , Find radius & interval of convergence

11

12. Maclaurin Series – Taylor Series

Solution

x)ln(1f(x) −−= 0f(0) =

x11(x)f−

=′ 1(0)f =′

2x)(11(x)f−

=′′ 1(0)f =′′

3x)(12(x)f−

=′′′ 2(0)f =′′′

Thus

L+′′′

+′′

+′

+= 32 x3!(0)fx

2!(0)fx

1!(0)ff(0)f(x)

L++++=⇒3x

2xx0f(x)

32

Just use the formula in the

booklet!

We find the first three derivatives and values at x=0

Notice: the result is a power

series , namely ∑∞

=1n

n

nx with

R=1, interval of conv x∈[-1,1[ Now we know that this series converges to

x11lnf(x)−

=

Solution

0.6663

0.52

0.50.5f(0.5)32

=++≡

4(4)

3 0.54!(c)f(0.5)R = where 0 < c < 0.5

We need 4

(4)

x)(16(x)f−

= . Thus

0.2541

0.5)-4(11(0.5)(0.5)

c)-24(16(0.5)R 4

44

43 ==≤=

The actual value of f(0.5) is ln2 = 0.693

So the actual error is ln2-0.666 = 0.026 (by GDC)

Lagrange form of error

Approximation

[see formula booklet]

Max value for c=0.5

Thus the estimated error was no so good.

Solution

ln2f(0.5) = , 2(0.5)f =′ , 4(0.5)f =′′ , 61(0.5)f =′′′

Thus

L+′′

+′+= 20.5)-(x2!(0.5)f0.5)-(0.5)(xff(0.5)f(x)

L++++=⇒ 32 0.5)-(x380.5)-2(x0.5)-2(xln2f(x)

Just use the formula in the

booklet!

We find (as above) the first three derivatives and the values at x=0.5

Find the Maclaurin series for x1

1lnf(x)

−= , up to x3

Find the approximation for f (0.5) given by the Maclaurin series above. Estimate the error using the Lagrange form. Compare with the actual error.

Taylor series for x1

1lnf(x)

−= up to (x- a)3, a=0.5

12

Solution

We know already 1f(0) = and 4dxdy(0)f

1y0x

==′

==

We need dxdy8y

dxdy5x5y2x

dxyd2

2

+++=

Thus

3732050dx

yd(0)f1y0x

2

2

=+++==′′

==

L+′′

+′

+= 2x2!(0)fx

1!(0)ff(0)f(x) L+++= 2x

2374x1

Maclaurin found by a D.E.

Because f(x)=y and dxdy(x)f =′

Implicit differentiation

Solution

We know that

L++++= 3!x

2!xx1xe

32

Thus

L+−+−= 3!x

2!x1x-e

6422

x

Maclaurin found by another

known Maclaurin series.

(see formula booklet)

Solution

L++−= 5!x

3!xxsinx

53

(a) LL ++−=++−=′=

′

4!x

2!x15!

5x3!

3xx)(sinxcosx42

(b) x

04!

4x2!

2xdx5!

5x3!

3xxsinxdxx

0

x

0

∫∫ +−=

++−= LL

L+−=+−⇒4!

4x2!

2x1cosx

L++−=⇒ 4!x

2!x1cosx

42

Differentiating and integrating

a series, term by term (see formula booklet) Differentiate both sides. Integrate both sides; definite integrals from 0 to x.

Let 22 4y5xyxdxdy

++= , and 1y = when 0x = .

Find the Maclaurin series of y = f(x) up to x2.

Find the Maclaurin series for 2-xe f(x) = , up to x6

Based on the maclaurin series of sinx find the Maclaurin series for cosx by using term by term (a) Differentiation (b) Integration