1a 1b

§Appendix C Sigma Notation

Let a1 , a2 ,…,an ,… be a sequence of numbers.a2 is the second number in the sequencea i is the ith number in the sequence

define

∑i=1

n

ai=a1+a2+…+an

where i is the index of summation1 is the starting value of i in this examplen is the final value of i in this example

Example. Let a i=i

∑i=1

4

i=1+2+3+4=10■

Example. Let a i=i2

∑i=1

4

i2=12+22+32+42

¿1+4+9+16=30■

Four Basic Properties of Sums

Let a1 , a2 ,… and b1 , b2 ,…be sequences of numbers and let c be a constant.

1. ∑i=1

n

c=nc

2. ∑i=1

n

ca i=c∑i=1

n

ai

3. ∑i=1

n

(ai+bi )=∑i=1

n

a i+∑i=1

n

bi

4. ∑i=1

n

(ai−bi )=∑i=1

n

ai−∑i=1

n

bi

Why?

1. ∑i=1

3

c=c+c+c=3c

2. ∑i=1

3

ca i=c a1+c a2+c a3

¿c (a1+a2+a3)

2a 2b

¿c∑i=1

n

ai

3. ∑i=1

n

(ai+bi )=(a1+b1 )+(a2+b2 )

+(a3+b3)¿ (a1+a2+a3 )+(b1+b2+b3)

¿∑i=1

n

a i+∑i=1

n

bi

4. similar to 3. ■

Two Sums of Powers

A. ∑i=1

n

i=n (n+1)2

B. ∑i=1

n

i2=n (n+1 )(2n+1)6

Proof of (A)

1+2+…+(n−1 )+n

n+(n−1 )+…+2+1

(n+1 )+(n+1 )+…+(n+1 )+(n+1)

¿n(n+1)

This is equal to 2∑i=1

n

i, so we must divide by 2 to get

the final result.

Proof of (B)

recall ( x+ y )3=x3+3 x2 y+3 x y2+ y3

Consider the following telescoping sum

∑i=1

n

[ (1+ i)¿¿3−i3 ]=(23−13 )+(33−23 )+…¿

+(n3−(n−1)3 )+((n+1 )3−n3)

¿ (n+1 )3−1

¿n3+3n2+3n [1]

on the other hand

∑i=1

n

[ (1+ i)¿¿3−i3 ]=∑i=1

n

(1+3 i+3 i2)¿

¿∑i=1

n

1+∑i=1

n

3 i+∑i=1

n

3 i2

3a 3b

¿∑i=1

n

1+3∑i=1

n

i+3∑i=1

n

i2

¿n+3 n(n+1)2+3S [2]

where S is the sum we seek. Equating [1] with [2]

n+ 32

(n2+n )+3 S=n3+3n2+3n

3S=n3+ 32n2+1

2n

S=2n3+3n2+n6

¿ n(2n2+3n+1)6

=n (n+1 )(2n+1)

6

■

§5.1 Areas and Distances

The Area Problem

Find the area of the region S lying under the graph of y= f (x ) on the interval from x=a to x=b.

…a…b…x- , …, curve, f , S

Approx. the area under f by a sum of rectangles.

a…b…,…,f , four rectangles, x1¿ , x2

¿ , x3¿ , x4

¿

It’s easy to compute the area of a rectangle

rectangle, width Δ x of subinterval, height f (x i¿), x i

¿ sample point for iTH rectangle

4a 4b

Take the limit as the number of rectangles increases to ∞.

Example. Estimate the area A under the graph of f ( x )=x from x=0 to x=1.

0…1 ,0…1 , f ( x )=x

Divide the interval [0,1] into four subintervals. add

Estimate the area under the graph over each subinterval by a rectangle of height f (x i

¿), with sample point x i

¿ the right hand endpoint of each subinterval.

0…1 ,0…1 , f ( x )=x ,rectangles

5a 5b

The width of each rectangle is Δ x=1/4. Area of rectangles

R4=Δ x ⋅ 14+Δ x ⋅ 1

2+Δ x ⋅ 3

4+Δ x ⋅1

¿ Δ x ⋅( 14 + 12+ 34+1)

¿ Δ x ⋅ 1+2+3+44

¿ 14 ⋅104

=1016

=58

Clearly A<R4

Alternatively, let x i¿ be the left hand endpoint of each

subinterval.

0…1 ,0…1 , f ( x )=x ,rectangles

Area of rectangles

L4=Δ x ⋅ 04+Δ x ⋅ 1

4+Δ x ⋅ 1

2+Δ x ⋅ 3

4

¿ Δ x ⋅ 0+1+2+34

¿ 14 ⋅64= 616

=38

Clearly

L4<A<R4

38<A< 5

8

Improve estimate by dividing area into 8 strips

L8<A<R8

6a 6b

716

<A< 916

7a 7b

Example. Obtain the exact area under f ( x )=x from x=0 to x=1 by taking the limit as n increases.

Let the sample points be right hand endpoints of each subinterval.

0…1 ,…, f ( x )=x ,rectangles of heights 1n ,2n, 3n,…1

Rn=Δ x ⋅ 1n+Δ x ⋅ 2

n+…+ Δ x ⋅ n

n

¿ Δ x ⋅ 1n (1+2+…+n)

¿1n2

(1+2+…+n)

¿ 1n2

n(n+1)2

=n+12n

Exact area

A=limn→∞

Rn

¿ limn→∞

n+12n

¿ 12

Alternatively, let the sample points be left hand endpoints of each subinterval.

heights of rectangles ¿ 0n ,1n, 2n…, n−1

n

Ln=Δ x ⋅ 0n+Δ x ⋅ 1

n+…+Δ x ⋅ n−1

n

¿ Δ x ⋅ 1n (0+1+…+ (n−1 ))

¿1n2

(0+1+…+(n−1 ) )

¿ 1n2

(n−1)n2

=n−12n

Exact Area

8a 8b

A=limn→∞

Ln

¿ limn→∞

n−12n

=12 ■

Fact: We can take the height of the iTh rectangle as f (x i

¿) for any number x i¿ in the iTh subinterval.

A=limn→∞

[f (x1¿ ) Δ x+f (x2¿ )Δ x+…+ f ( xn¿) Δ x ]

A=limn→∞

∑i=1

n

f (x i¿) Δ x

This formula works for any continuous function f

…a…b…,0…,iTh rect., width Δ x , xi¿ ,height f (x i

¿)

The Distance Problem

Find the distance traveled by an object if its velocity is known

If the velocity is constantdistance=velocity×time

Example

120miles=60 mileshour

×2hours■

What if the velocity varies?…a…b…time t-, … velocity v-, curve

9a 9b

define time subintervals of length Δt add 4pick sample time t i

¿ in each subinterval addapproximate the velocity in each subinterval as v (ti

¿)

10a 10b

d=distance traveled

≈ v (t 1¿ ) Δt+v (t 2¿)Δ t+v (t 3¿ ) Δt+v (t 4¿ )Δt

¿∑i=1

4

v (t i¿) Δt

The exact distance is obtained in the limit as we use more and more subintervals.

d= limn→∞

∑i=1

n

v (ti¿ )Δt

§5.2 The Definite Integral…a…b…x-, …, curve f

define the definite integral

Steps

Divide [a ,b] into n subintervals of equal width

Δ x=(b−a)/n add subinterval i

x i=a+ i Δ x add x0 , x i−1 , x i , xn

Choose sample points in these subintervals

x1¿ , x2

¿ ,… xn¿ add x i

¿

Construct the Riemann Sum

∑i=1

n

f (x i¿) Δ x

Take the limit n→∞ to obtain the definite integral

∫a

b

f ( x )dx= limn→∞

∑i=1

n

f (x i¿) Δ x

Example. The distance problem

d=∫a

b

v (t )dt ■

Notation integral sign, lower limit of integration a, upper limit b, integrand, ghost of Δ x

11a 11b

Note that integral does not depend on “x”

∫a

b

f ( x )dx=∫a

b

f (t ) dt

x and t are dummy variables.

Example. Express

I= limn→∞

∑i=1

n 1

1+( in )2 ⋅1n

as an integral on [0,1].

Let Δ x=1n andf (x i )=1

1+(x i¿ )2 where x i¿= i

n .

Then I=∫0

1 11+x2

dx ■

?? Example. Determine a region whose area is equal to the give limit.

limn→∞

∑i=1

n

tan( π i4n ) π4 n

■

Interpretation of the Definite Integral

∫a

b

f ( x )dx= limn→∞

∑n=1

∞

f (x i¿) Δ x

12a 12b

For f positive and b>a, the Riemann sum approximates the area under the curve.

…a…b…x-, …, graph of f , area A, 4 rectangles

∫a

b

f ( x )dx=A=area under f ¿a¿b¿

Suppose f can be positive or negative…a…b…x-, …, f , A1 under f , A2 above f , rectangles

here f ( xi¿ )>0…∑ f ( xi¿ ) Δ x≈ A1, here f ( xi¿ )<0…∑ f (x i¿ )Δ x is a negative quantity whose magnitude ≈ A2

∫a

b

f ( x )dx=A1−A2=¿net area under or above f

from a to bEvaluating Integrals Recall

1. ∑i=1

n

c=nc

13a 13b

2. ∑i=1

n

ca i=c∑i=1

n

ai

3. ∑i=1

n

(ai+bi )=∑i=1

n

a i+∑i=1

n

bi

4. ∑i=1

n

(ai−bi )=∑i=1

n

ai−∑i=1

n

bi

where a1 , a2 ,… and b1 , b2 ,… are sequences of numbers and c is a constant (an expression that does not depend on the index of summation i).

Recall two sums of powers

1. ∑i=1

n

i=12n(n+1)

2. ∑i=1

n

i2=16n (n+1 )(2n+1)

Example. Prove that ∫a

b

x dx=12(b2−a2)

…a…b…x-, partition points, Δ x , Δ x=b−an

choose sample points in each subinterval

an easy choice is the right hand endpoints

x1¿=a+Δ x

x2¿=a+2Δ x

x i¿=a+i Δ x

Begin with the definition

∫a

b

x dx=limn→∞

∑i=1

n

f ( xi¿ )Δ x

¿ limn→∞

∑i=1

n

x i¿Δ x

¿ limn→∞

∑i=1

n

(a+i Δ x ) Δ x

14a 14b

¿ limn→∞

∑i=1

n

a Δ x+i Δ x2

¿ limn→∞

¿¿

∫a

b

x dx=¿ limn→∞

¿¿¿

¿ limn→∞

¿¿

¿ limn→∞ (a b−a

nn+( b−a

n )2 12n (n+1 ))

¿ limn→∞ (a (b−a )+(b−a )2 1

2n+1n )

¿a (b−a )+ 12

(b−a )2 limn→∞

n+1n

¿a (b−a )+ 12

(b−a )2

¿ (b−a )(a+ 12 (b−a ))¿ 12

(b−a ) (b+a )

¿ 12(b2−a2) we are done! ■

Example. Prove that ∫0

1

x2dx=13

0…1…x-, 0…1… y-, y=x2

width of subintervals Δ x=1n

sample points x i¿= i

n add

∫0

1

x2dx=limn→∞

∑i=1

n

(xi¿)2Δ x

¿ limn→∞

∑i=1

n

( in )2 1n

15a 15b

¿ limn→∞

1n3∑i=1

n

i2

¿ limn→∞

1n316n (n+1 ) (2n+1 )

¿ limn→∞

2n3+…6n3

¿ 13 ■

Properties of the definite integral

Reversing limits of integration

∫a

b

f ( x )dx=−∫b

a

f ( x )dx

Example. Suppose f >0 and b>a

…a…b…,…, f

∫a

b

f ( x )dx>0 and ∫b

a

f ( x )dx<0 ■

Corollary. Let b=a

∫a

a

f ( x )dx=0

Picture …a…, f

The “area” under a point is zero!

Four basic properties of integrals

1. ∫a

b

cdx=c(b−a)

16a 16b

2. ∫a

b

c f ( x ) dx=c∫a

b

f ( x ) dx

3. ∫a

b

f ( x )+g ( x )dx=∫a

b

f ( x )dx+∫a

b

g ( x )dx

4. ∫a

b

f ( x )−g (x ) dx=∫a

b

f ( x )dx−∫a

b

g ( x ) dx

Notes

2. constant multiple rule3. integral of sum is sum of integrals4. integral of difference is difference of integrals.Property 1 has a simple interpretation

…a…b…,…c…,function c, area ¿c (b−a)

Example. Evaluate I=∫0

3

(5−2x )dx

use the difference rule

I=∫0

3

5dx−∫0

3

2 xdx

property #1 and constant multiple rule

¿5 (3−0 )−2∫0

3

xdx

rule that ∫a

b

x dx=12(b2−a2)

¿15−2(12 )(9−0 )

¿6 ■

Addition Property wrt Interval of IntegrationFor any real nos. a ,b and c

5. ∫a

c

f ( x )dx+∫c

b

f ( x )dx=∫a

b

f ( x )dx

Picture for a<c<b and f >0…a…c…b…x-, …, f , A , A1 , A2

17a 17b

Example: Evaluate

I=∫3

4

f ( x )dx+∫1

3

f ( x )dx+∫4

1

f ( x )dx

by addition property reverse limits

so I=0 ■

Comparison Property of Integrals

6. If f ( x )≥0 and b>a then

∫a

b

f ( x )dx ≥0

7. If f ( x )≥g (x) and b>a then

∫a

b

f ( x )dx ≥∫a

b

g ( x )dx

8. If m≤ f (x )≤M then

m (b−a )≤∫a

b

f (x )dx≤ M (b−a)

Picture for M >m>0:…a…b… , …m…M… function m, function M , f

Example. Show that

∫2

5

√x2−1dx≤10.5

without evaluating the integral.

Solution. For x>1:

√ x2−1<√x2=x

Then by property 7:

∫2

5

√1−x2dx≤∫2

5

xdx

¿ 12 (52−22) ¿

12

(25−4 )=212

18a 18b

¿10.5■

Example. Show that

π6

<∫0

π3

cos ( x )dx≤ π3

without evaluating the integral.

0… π3… π2…x-, 0…1… y-, cos (x)

on the interval [0 , π3 ]

12≤ cos ( x )≤1

by property 8

12 ( π3−0)<∫

0

π3

cos ( x )dx<1( π3−0)

which gives the result we are trying to show ■

§5.3 The Evaluation TheoremIf f is continuous on the closed interval [a ,b], then

∫a

b

f ( x )dx=F (b )−F (a)

where F is any antiderivative of f , that is F '=f .

Example. ∫a

b

x dx=¿ F (b )−F (a)¿

where F ( x )=12x2+C.

F (b )=12b2+C ,F (a )=1

2a2+C

∫a

b

x dx=12(b2−a2)

Notice that the constant of integration C cancels. We may as well set C=0. ■

Example. ∫0

1

x2dx=F (1 )−F (0)

where F ( x )=13x3

19a 19b

F (1 )=13 , F (0 )=0

∫0

1

x2dx=13 ■

Example. The distance problem.

v velocitys positiond distance traveleda starting timeb ending time

we have seen previously

d=∫a

b

v (t )dt

recall s' (t )=v (t)

By the Evaluation Theorem

∫a

b

v (t )dt=s (b )−s (a )=¿¿distance traveled! ■

Recall the Mean Value Theorem. Let f be continuous on [a ,b] and differentiable on (a ,b). Then there is a no. c in (a ,b) such that

f ' (c )=f (b )−f (a)

b−a

Proof of the Evaluation Theorem.

Partition [a ,b] into n subintervals of equal width.

…a…b…x-, x0 , x1 , x2 ,…,xn−1 , xn

width of each subinterval Δx=(b−a) /n

Let F be any antiderivative of f . Write F (b )−F (a) as a telescoping sum

F (b )−F (a )=F (xn )−F(x0)

¿ (F (xn)−F (xn−1 ))+(F (xn−1 )−F (xn−2 ))+¿

20a 20b

…+(F (x2 )−F ( x1 ))+ (F (x1 )−F (x0 ))

Apply the Mean Value Theorem to F on an arbitrary subinterval i : [ xi−1 , x i ]

F (x i )−F (xi−1 )=F' (x i¿) (x i−x i−1 )=f ( xi¿ ) Δx,

where x i¿ is some no. on the interval (x i−1 , x i).

Then

F (b )−F (a )= f ( xn¿ ) Δx+f (xn−1¿ )Δx+…f ( x1¿ ) Δx

¿∑i=1

n

f ( xi¿ )Δx

the last expression is a Riemann sum!

Let n→∞

limn→∞

F (b )−F (a )=limn→∞

∑i=1

n

f (x i¿ ) Δx

F (b )−F (a) does not depend on n and the limit on the right is the integral. Therefore we get

F (b )−F (a )=∫a

b

f (x )dx

■

Notation F (b )−F (a )=F ( x )|ab

Then the Evaluation Theorem may be written

∫a

b

f ( x )dx=F (x )|ab

where F is any antiderivative of F.

Example. Evaluate ∫0

1

x37 dx

For f ( x )=xn, an antiderivative of f is

F ( x )= 1n+1

xn+1

For n=3/7

F ( x )= x107

( 107 )=710 x

107

Thus

∫0

1

x37 dx= 7

10x107 |0

1

¿ 710 (1 )107 −

710

(0 )107

21a 21b

¿ 710−0

¿ 710 ■

Example. Evaluate I=∫0

π /4

sec2 (θ )dθ

Recall ddθ tan (θ )=sec2(θ)

Then I=tan (θ )|0π4=tan ( π4 )−tan (0 )=1■

The Evaluation Theorem in Applications

Interpret derivatives as rates of change

s position v velocity

Q electric charge I=dQdt electric current

y=h(x) height of trail dydx

=h' (x) slope of trail

x miles from start

rewrite the evaluation theorem in terms of rate of change

Net Change TheoremThe integral of a rate of change is a net change

∫a

b

F ' ( x )dx=F (b )−F (a)

where F '(x ) is the rate of change of F wrt x and F (b )−F (a) is the net change in F.

Note that the word “net” connotes that there can be positive and negative contributions to the rate of change

Examples.

The integral of velocity gives the net change in position

∫a

b

v (t )dt=s (b )−s (a)

The integral of current gives the net charge passing through a wire

22a 22b

∫a

b

I ( t ) dt=Q (b )−Q(a)

The integral of the slope of trail gives the net change in height of trail

∫a

b

h' ( x )dx=h (b )−h(a) ■

Particle Motion (back and forth) along a straight line

s(t ) position at time t

v (t) velocity at time t

¿ v (t )∨¿ speed at time t

Consider the following particle path

0…4…s-, t 1 at 0, t 2 at 4, t 3 at 3

the particle’s displacement (distance traveled)

s (t 3 )−s (t 1)=3=∫t1

t3

v ( t ) dt

the total distance the particle travels

5=4+1=∫t 1

t 3

|v ( t )|dt

¿∫t 1

t 2

v ( t )dt+∫t 2

t 3

(−v ( t ) )dt

Example. The velocity function for a particle moving along a line is

v (t )=4−2t meters/sec

Find (a) the displacement and (b) the total distance traveled over the time interval

0≤ t ≤3 seconds

(a) ∫0

3

v (t ) dt=∫0

3

(4−2 t )dt

¿ 4 t−t2|03

¿ (12−9 )−(0−0)

¿3meters

23a 23b

(b) ∫0

3

|v ( t )|dt=∫0

2

v ( t ) dt+∫2

3

(−v (t ))dt

¿∫0

2

(4−2 t ) dt+∫2

3

(2 t−4 )dt

¿¿

¿ ( (8−4 )−(0−0 ) )+ ( (9−12 )−(4−8 ) )

¿ (4−0 )+( (−3 )−(−4 ) )

¿4+1 ¿5 meters ■

Indefinite Integrals

The symbol

∫ f ( x )dx

is called the indefinite integral and means the general antiderivative of f (x).

If F is any antiderivative of f then

∫ f ( x )dx=F (x )+C

C is the constant of integration.

Antiderivative Formulas using indefinite integral notation

Let k be a constant and let f and g be functions

∫ k dx=kx+C

∫ k f ( x )dx=k∫ f ( x )dx+C

∫ f ( x )+g ( x )dx= ∫ f ( x )dx+ ∫ g ( x )dx

∫ f ( x )−g (x )dx= ∫ f ( x )dx− ∫ g ( x )dx

we also have formulas for specific functions

∫ xndx= 1n+1

xn+1+C valid for n≠−1

∫ 1x dx=ln¿ x∨¿+C ¿

∫ exdx=ex+C

∫ ax dx= ax

ln (a )+C

∫ sin ( x )dx=−cos (x )+C

∫ cos (x )dx=sin (x)+C

∫ sec2 ( x )dx=tan (x )+C

24a 24b

∫ sec (x ) tan ( x )dx=sec ( x )+C

∫ 11+x2

dx=tan−1 (x )+C

∫ 1√1−x2

dx=sin−1 ( x )+C

Example. Find the indefinite integralI= ∫ ¿

solution

I=∫cos ( x )dx−∫2sin ( x )dx

¿∫cos ( x )dx−2∫ sin ( x ) dx

¿ sin ( x )+2cos ( x )+C ■

Example. Find the indefinite integral

I= ∫ ¿

solution

I= ∫ 15 x

dx+ ∫ 3 sec ( x ) tan(x)¿dx ¿

¿ 15∫1xdx+3 ∫ sec ( x ) tan ( x )dx

¿ 15 ln ¿ x∨+3 sec(x)+C ■

§5.4 The Fundamental Theorem of CalculusThe “Area so far” function…a…x…b…t-, 0…, f (t), area g(x ) from a to x

25a 25b

g ( x )=∫a

x

f ( t ) dt

If f >0, g is the “area so far” under f

Example. Let a=1 and f ( t )=1t .

g ( x )=∫1

x 1tdt

¿ ln|t||1x

¿ ln ( x )− ln (1) where x>0 ¿ ln (x)

Notice that

g' (x )=1x=f (x) This is not a coincidence! ■

26a 26b

The Fundamental Theorem of Calculus, part 1 (FTC 1)

If f is continuous on [a ,b], then the function g defined by

g ( x )=∫a

x

f (t )dt a≤ x≤b

is an antiderivative of f , that is

g' (x )=f (x ) for a≤ x≤b.

Plausibility Argument. Why is g' (x )=f (x )?…a…x… x+h…b…,… ,area g(x ) from a to x

g' (x )=limh→ 0

g (x+h )−g(x )h

from the graph

g ( x+h )−g ( x )≈ f ( x )h

this becomes more accurate as h→0

g' (x )=limh→ 0

f ( x )h+small correctionh

¿ f (x) ■

Our text gives a rigorous proof.

Using Leibniz notation, the FTC 1 becomes

ddx∫a

x

f ( t ) dt=f (x )

Example. Find the derivative of

g ( x )=∫−1

x

√ t 3+1dt

Solution f (t )=√t 3+1 is continuous on ¿.

g' (x )=√x3+1 ■

27a 27b

Example. Find the derivative of

y=∫x2

π sin (t)t

dt

Solution. f (t )= sin (t)t is continuous for t>0.

Reverse the limits of integration.

y=−∫π

x2 sin (t )t

dt

¿−∫π

u sin (t)t

dt

where u=x2. By the FTC 1

dydu

=−sin(u)

u

By the chain rule

dydx

=dydu

dudx

¿ −sin (u)u

dudx

¿ −sin (x2)x2

⋅2x

¿−2x sin (x2) ■

The Fundamental Theorem of Calculus (FTC)

Suppose f is continuous on [a ,b]

1. ddx∫a

x

f ( t ) dt=f (x )

2. ∫a

b

f ( x )dx=F (b )−F (a)

where F is any antiderivative of f . This is the Evaluation Theorem!

Part 2 can be rewritten

∫a

b

F ' ( x )dx=F (b )−F (a)

Roughly speaking, the FTC says that differentiation and integration are inverse processes!

Recall the Mean Value Theorem:

28a 28b

If f is continuous on [a ,b] and differentiable on (a ,b), then there is a no. c in (a ,b) such that

f ' (c )=f ( b )− f (a)

b−a

Mean Value Theorem for Integrals

If f is continuous on [a ,b], then there is a no. c in (a ,b) such that

f ( c )=∫a

b

f ( x )dx

b−a

Proof. Let F be the “Area so far” function

F ( x )=∫a

x

f (t )dt

F ' ( x )=f (x) by FTC 1.

F is continuous on [a ,b] and differentiable on (a ,b).

By the Mean Value Theorem, there is a no. c in [a ,b] such that

F ' (c )=F (b )−F (a)

b−a

By FTC 2 (the Evaluation Theorem), this becomes

f ( c )=∫a

b

f ( x )dx

b−a

which is what we want to prove! ■

Geometrical Interpretation…a…b…x-, …, f add c , f (c), rectangle

∫a

b

f ( x )dx=f (c )(b−a)

area under curve ¿ area of rectangle

29a 29b

Note:

f ( c )=f ave=1

b−a∫ab

f ( x )dx

is the average value of f on [a ,b]

Examplea) Find the average value of f ( x )=x3 on [0,1]

f ave=11−0∫0

1

x3dx=14x4|0

1= 14

b) Find c such that f ave=f (c)

f ( c )=c3=14

c=( 14 )13≈0.63

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STOP §5.5 The Substitution RuleDifferential Notation revisited

Consider y=f (x ). By the definition of derivative:

dydx=lim

h→0

f ( x+h )−f (x )h

this is not exactly a ratio…

Let dx be any real no. and define

dy=f ' ( x) dx

another real no. This allows us to interpret the left hand side of

dydx

=f ' (x )

30a 30b

as a true ratio.

Practice. Let’s write u=g (x). Then du=g' ( x )dx

Let u=x2. Then du=2 xdx

?? Let u=√x. Then du=¿

?? Let u=ln ¿ x∨¿¿. Then du=¿

?? Let u=tan (θ). Then du=¿

Integration by Substitution

Recall the indefinite integral

∫ F ' ( x )dx=F ( x )+C

Recall the chain rule

let F ( x )=f (g (x ) )

F ' ( x )= f ' (g ( x ) )g '(x )

Thus

∫ f ' (g ( x ) )g ' ( x )dx=f (g ( x ) )+C

Example. Let F ( x )=sin (x2)

F ' ( x )=cos (x2 )2x

so ∫ cos (x2 )2 xdx=sin ( x2 )+C ■

This is taking the chain rule backwards!

How to recognize f ' (g ( x ) )g '(x ) in an integrand?

Trick – “Integration by Substitution”

Suppose you have

I= ∫ f ' (g ( x ) )g ' ( x )dx

but you might not recognize the form.

Guess the inner function of the composition

u=g (x)

form the differential

du=g' ( x )dx

substitute into I

I= ∫ f ' (u )du

¿ f (u )+C

31a 31b

¿ f (g ( x ) )+C

Example. Evaluate

I=∫2 xcos (x2 )dx

guess the inner function

u=x2

form the differential

du=2 xdx

substitute into I

I= ∫ cos (u )du

¿ sin (u )+C

¿ sin (x2 )+C ■

Example. EvaluateI= ∫ √3 x+4 dx

guess the inner functionu=3x+4

form the differentialdu=3dx

solve for dx

dx=13du

substitute into I

I=∫u12 13du

¿ 13⋅ 23u32+C

¿ 29 (3 x+4 )32+C

can check by differentiation

32a 32b

ddx ( 29 (3x+4 )

32+C)= 29 ⋅ 32 (3 x+4 )

12 (3 )+0=(3 x+4 )

12

■

33a 33b

Example. Evaluate

I= ∫ tan2 (x ) sec2 ( x )dx

guess the inner function

u=tan(x )

form the differential

du=sec2 ( x )dx

substitute into I

I= ∫ u2du

¿ 13 u3+C

¿ 13 tan3 ( x )+C

■

Example. Evaluate

I= ∫ x1+x4

dx

guess the inner function

u=x2

form the differential

du=2 xdx

solve for x dx

x dx=12du

substitute into I

I= ∫ 11+u2

12du

¿ 12 tan−1 (u )+C

¿ 12 tan−1 (x2 )+C

■

34a 34b

Evaluating Definite Integrals By Substitution

Recall the Fundamental Theorem of Calculus, part 2

∫a

b

F ' ( x )dx=F (b )−F (a)

Method I Find the indefinite integral by substitution and then apply the FTC, part 2.

Example. Evaluate

∫−1

0

√3 x+4dx

let

u=3x+4 , du=3dx , 13du=dx

then

∫ √3 x+3dx= ∫ u12 ⋅ 13du=1

3⋅ 23u32+C

¿ 29 (3 x+4 )32+C

Now let F ( x )=29(3 x+4). Then by FTC 1

I=F (0 )−F (−1 )

¿ 29⋅ 4

32−29⋅1

32

¿ 29 (8−1 )

¿ 149 ■

35a 35b

36a 36b

Method II. Transform limits of integration while substituting and apply FTC 2.

This method is better once you get used to it because it involves less writing

Example. Evaluate

I=∫−1

0

√3 x+4 dx

Let

u=3x+4 , du=3dx , 13du=dx

Also note that if x=−1 then u=1 if x=0 then u=4then

I=∫1

4

u12 13du

¿ 13⋅ 23u32|14

¿ 29

(432−1

32)

¿ 29 (8−1 )=149 ■

Example. Evaluate

I=∫−π4

π4

tan2 (x ) sec2 ( x )dx

Let

u=tan(x )

du=sec2(x )

Also note that if x=−π4 then u=−1

if x=π4 then u=1

then

I=∫−1

1

u2du

¿ 13 u3|−11

¿ 13 (13− (−1 )3 )

¿ 23

37a 37b

■

Example. Evaluate

I=∫1

2 cos( πx )x2

dx

Let u=πx=π x−1.

Then du=π (−x−2 )dx and −1π du=x−2dx.

If x=1 then u=π

x=2 u=π2

Then

I=∫π

π2

cos (u )(−1π du)

¿1π∫π

2

π

cos (u )du

¿1πsin (u )|π

2

π

¿ 1π (sin (π )−sin( π2 )) ¿ 1π (0−1 )

¿−1/π ■

Example. Evaluate

I=∫2

4 1x ln( x)

dx

Let u=ln ( x ) .

Then du=1x dx

If x=2 then u=ln 2 x=4 u=ln 4=ln22=2 ln2

Then

I=∫ln2

ln 4 1udu

¿ ln u|ln2ln 4

¿ ln (ln 4¿)−ln ¿¿¿

¿ ln (2 ln2¿)− ln¿¿¿

38a 38b

¿ ln 2+ ln¿¿¿

¿ ln 2

■