chapter 13: integrals · chapter 13 overview: the integral calculus is essentially comprised of two...

Chapter 13:

Integrals

1001

Chapter 13 Overview: The Integral

Calculus is essentially comprised of two operations. Interspersed throughout the

chapters of this book has been the first of these operations––the derivative. The

surface of this field has barely been scratched and only those areas that are most

related to Analytic Geometry were viewed. College Calculus will begin with the

derivative and look much further into it, as well as looking at subjects other than

Analytic Geometry where the derivative plays a major role. But some basics of the

other operation––the integral––should be considered.

There are two kinds of integrals––the indefinite integral (or anti-derivative) and the

definite integral. The indefinite integral is referred to as the anti-derivative,

because, as an operation, it and the derivative are inverses (just as squares and

square roots, or exponential and logarithmic functions). As it pertains to Analytic

Geometry, the definite integral is an operation that gives the area between a curve

and the x-axis. It has nothing to do with traits or sketching a graph. This area is

purely Calculus and is an appropriate place to finish our introduction to the subject.

1002

13-1: Anti-Derivatives: The Power Rule

As seen in a previous chapter, certain traits of a function can be deduced if its

derivative is known. It would be valuable to have a formal process to determine

the original function from its derivative accurately. The process is called anti-

differentiation, or integration.

Symbol: “the integral of f of x dx”

The dx is called the differential. For now, just treat it as part of the integral

symbol. It tells the independent variable of the function (usually, but not always,

x) and, in a sense, is where the increase in the exponent comes from. It does have

meaning on its own which will be explored in a different course.

Looking at the integral as an anti-derivative, that is, as an operation that reverses

the derivative, a basic process for integration emerges.

Remember:

and

(or, multiply the power in front and subtract one from the power). In reversing the

process, the power must increase by one and divide by the new power. The

derivative does not allude to what constant, if any, may have been attached to the

original function.

f x( )( ) dx =⌠

⌡⎮

ddx

xn⎡⎣ ⎤⎦ = nxn−1 Dx constant[ ] = 0

1003

The Anti-Power Rule:

for

The “ ” is to account for any constant that might have been there before the

derivative was taken. NB. This rule will not work if , because it would

require division by zero. Recall from the derivative rules what yields

as the derivative–– . The complete the Anti-Power Rule is:

The Anti-Power Rule:

if

Since and

then

These allow integratation of a polynomial by integrating each term separately.

xn( )⌠⌡⎮

dx = xn+1

n+1+C n ≠ −1

+Cn = −1

x −1 or 1x⎛⎝⎜

⎞⎠⎟

ln x

xn( )⌠⌡⎮

dx = xn+1

n+1+C n ≠ −1

1x

⌠

⌡⎮⎮ dx = ln x +C

Dx f x( ) + g x( )⎡⎣ ⎤⎦ = Dx f x( )⎡⎣ ⎤⎦ + Dx g x( )⎡⎣ ⎤⎦ Dx cxn⎡⎣ ⎤⎦ = cDx xn⎡⎣ ⎤⎦

f x( )+ g x( )( )⌠

⌡⎮ dx = f x( ) dx+ g x( ) dx∫∫

c f x( )( ) ∫ dx = c f x( ) dx∫

1004

LEARNING OUTCOMES

Find the indefinite integral of a polynomial.

Use integration to solve rectilinear motion problems.

EX 1

EX 2

EX 3

3x2 + 4x+5( )⌠

⌡⎮ dx

3x2 + 4x+5( )⌠

⌡⎮ dx = 3 x2+1

2+1+ 4 x1+1

1+1+5 x0+1

0+1+C

= 3x3

3 + 4x2

2 + 5x1

1 +C

= x3 + 2x2 + 5x+C

x4 + 4x2 + 5+ 1x

⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx

x4 + 4x2 + 5+ 1x

⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx = x4+1

4 +1+4x2+1

2+1 + 5x0+1

0+1 + ln x+C

= 15 x5 + 43 x

3 + 5x+ ln x +C

x2 + x3 − 4x

⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx

x2 + x3 − 4x

⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx = x2 + x13 − 4

x⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx

= x2+1

2+1+x

13+1

13+1

− 4ln x +C

= 1

3x3 + 34 x

43 − 4ln x +C

1005

Integrals of products and quotients can be done easily IF they can be turned into a

polynomial.

EX 4

Example 5 is called an initial value problem. It has an ordered pair (or initial value

pair) that allows us to solve for C.

EX 5 . Find if .

x2 + x3( ) 2x+1( )⌠

⌡⎮ dx

x2 + x3( ) 2x+1( )⌠

⌡⎮⎮ dx = 2x3 + 2x

43 + x2 + x

13⎛

⎝⎜⎞⎠⎟

⌠

⌡

⎮⎮ dx

= 2x4

4 + 2x7

3

73

+ x3

3 + x4

3

43+C

= 1

2 x4 + 67 x

73 + 1

3x3 + 34 x

43 +C

f ′ x( ) = 4x3 − 6x+ 3 f x( ) f 0( ) = 13

f x( ) = 4x3 − 6x+ 3( )⌠

⌡⎮ dx

= x4 − 3x2 + 3x+C

f 0( ) = 04 − 3 0( )2 + 3 0( )+C =13∴ C =13

f x( ) = x4 − 3x2 + 3x+13

1006

EX 6 The acceleration of a particle is described by . Find the

distance equation for if and .

a t( ) = 3t2 + 8t +1x t( ) v 0( ) = 3 x 0( ) = 1

v t( ) = a t( )( )⌠

⌡⎮ dt = 3t2 + 8t +1( )⌠

⌡⎮ dt

= t 3 + 4t2 + t +C1

3= 0( )3 + 4 0( )2 + 0( )+C1

3=C1

v t( ) = t 3 + 4t2 + t + 3

x t( ) = v t( )( )⌠

⌡⎮ dt = t 3 + 4t2 + t + 3( )⌠

⌡⎮ dt

= 14 t

4 + 43 t

3 + 12 t

2 + 3t +C2

1= 14 0( )4 + 4

3 0( )3 + 12 0( )2 + 3 0( )+C2

1=C2

x t( ) = 14 t4 + 43 t

3 + 12 t2 + 3t +1

1007

EX 7 The acceleration of a particle is described by . Find the


a t( ) =12t2 − 6t + 4x t( ) v 1( ) = 0 x 1( ) = 3

v t( ) = a t( )( )⌠

⌡⎮ dt = 12t2 − 6t + 4( )⌠

⌡⎮ dt

= 4t 3 − 3t2 + 4t +C1

0 = 4 1( )3 − 3 1( )2 + 4 1( )+C1

−5 =C1

v t( ) = 4t 3 − 3t2 + 4t − 5

x t( ) = v t( )( ) ∫ dt = 4t 3 − 3t2 + 4t − 5( ) ∫ dt

= t 4 − t 3 + 2t2 − 5t +C2

3= 1( )4 − 1( )3 + 2 1( )2 − 5 1( )+C2

6 =C2

x t( ) = t 4 − t 3 + 2t2 − 5t + 6

1008

13-1 Free Response Homework

Perform the anti-differentiation.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

Solve the initial value problems.

13. . Find if .

14. . Find if .

15. . Find if .

16. The acceleration of a particle is described by . Find the


17. The acceleration of a particle is described by . Find the


6x2 − 2x+ 3( )⌠

⌡⎮ dx x3 + 3x2 − 2x+ 4( )⌠

⌡⎮ dx

2x3

⌠

⌡

⎮⎮

dx 8x4 − 4x3 + 9x2 + 2x+1( )⌠

⌡⎮ dx

x3 4x2 + 5( ) ∫ dx 4x−1( ) 3x+ 8( )⌠⌡⎮ dx

x − 6x

⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx x2 + x + 3x

⎛

⎝⎜

⎞

⎠⎟

⌠

⌡

⎮⎮⎮

dx

x+1( )3 ∫ dx 4x− 3( )2⌠⌡⎮

dx

x + 3 x32 − 6x

⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx 4x3 + x + 3x2

⎛

⎝⎜

⎞

⎠⎟

⌠

⌡

⎮⎮⎮

dx

f ′ x( ) = 3x2 − 6x+ 3 f x( ) f 0( ) = 2

f ′ x( ) = x3 + x2 − x+ 3 f x( ) f 1( ) = 0

f ′ x( ) = x − 2( ) 3 x +1( ) f x( ) f 4( ) = 1

a t( ) = 36t2 −12t + 8x t( ) v 1( ) = 1 x 1( ) = 3

a t( ) = t2 − 2t + 4x t( ) v 0( ) = 2 x 0( ) = 4

1009

13-1 Multiple Choice Homework

1.

a) b) c)

d) e)

2.

a) b) c)

d) e)

3.

a) b) c)

d) e)

1x2 dx⌠

⌡⎮=

ln x2 + C − ln x2 + C x−1 + C

−x−1 + C −2x−3 + C

x 10 + 8x4( ) dx∫ =

5x2 + 43x6 + C 5x2 + 8

5x5 + C 10x + 4

3x6 + C

5x2 + 8x6 + C 5x2 + 87x6 + C

x 3x dx∫ =

2 35

x52 + C

5 32

x52 + C

32x12 + C

2 3x + C5 32

x32 + C

1010

13-2: Integration by Substitution: The Chain Rule

The other three derivative rules––the Product, Quotient and Chain Rules––are a

little more complicated to reverse than the Power Rule. This is because they yield

a more complicated function as a derivative, one which usually has several

algebraic simplifications. The integral of a rational function is particularly difficult

to unravel because, as seen previously, a rational derivative can be obtained by

differentiating a composite function with a logarithm or a radical, or by

differentiating another rational function. Reversing the Product Rule is as

complicated, though for other reasons. Both these subjects can be left for a

traditional Calculus class. The Chain Rule is another matter.

Composite functions are among the most pervasive situations in math. Though not

as simple in reverse as the Power Rule, the overwhelming importance of this rule

makes it imperative that it be addressed here.

Remember:

The Chain Rule:

The derivative of a composite function turns into a product of a composite and a

non-composite. So if there is a product to integrate, it might be that the product

came from the Chain Rule. The integration is not done by a formula so much as a

process that might or might not work. One makes an educated guess and hopes it

works out. There are other processes in Calculus for when it does not work.

Integration by Substitution:

1. Identify the inside function of the composite and call it u.

2. Find du from u.

3. If necessary, multiply a constant inside the integral to create

du, and balance it by multiplying the reciprocal of that constant

outside the integral. (See EX 2)

4. Substitute u and du into the equation.

5. Perform the integration using the Anti-Power Rule (or transcendental

rules, in next section.)

6. Re-substitute for u.

ddx f g x( )( )⎡

⎣⎢⎤⎦⎥ = f ′ g x( )( ) i g′ x( )

1011

This is one of those mathematical processes that makes little sense when first seen.

But after seeing several examples, the meaning suddenly becomes clear.

BE PATIENT!

LEARNING OUTCOME

Use the Unchain Rule to integrate composite, product expressions.

EX 1

is the composite function. So

3x2 x3 + 5( )10⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx

x3 + 5( )10 u = x3 + 5du = 3x2 dx

3x2 x3 + 5( )10⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx = u10( )⌠

⌡⎮ du

= u11

11 +C

= 111 x3 + 5( )11 +C

1012

EX 2

is the composite function. So

EX 3

is the composite function.

So

x x2 + 5( )3⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx

x2 + 5( )3 u = x2 + 5du = 2x dx

x x2 + 5( )3⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx = 12 x2 + 5( )3 2x dx( )⌠

⌡⎮⎮

= 12 u3( )⌠

⌡⎮ du

= 12 iu4

4 +C

= 18 x2 + 5( )4 +C

x3 + x( ) x4 + 2x2 − 55( )⌠

⌡⎮⎮ dx

x4 + 2x2 − 54

u = x4 + 2x2 − 5du = 4x3 + 4x( ) dx = 4 x3 + x( ) dx

x3 + x( ) x4 + 2x2 − 55( )⌠

⌡⎮⎮ dx = 1

4 x4 + 2x2 − 55( ) 4 x3 + x( )⌠

⌡⎮⎮ dx

= 14 u5( )⌠

⌡⎮ du

= 14 u

15⎛

⎝⎜⎞⎠⎟

⌠

⌡

⎮⎮ du

= 14u

65

65

+C

= 524 x4 + 2x2 − 5( )65 +C

1013

EX 4

3x2 + 4x − 5x3 + 2x2 − 5x + 2( )3

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

⌠

⌡

⎮⎮⎮⎮⎮

dx

u = x3 + 2x2 − 5x + 2du = 3x2 + 4x − 5( ) dx

3x2 + 4x− 5x3 + 2x2 − 5x+ 2( )3

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

⌠

⌡

⎮⎮⎮⎮⎮

dx = x3 + 2x2 − 5x+ 2( )−33x2 + 4x− 5( ) dx( ) ⌠

⌡⎮⎮

= u−3( )⌠

⌡⎮ du

= u−2

−2 +C

= −12 x3 + 2x2 − 5x + 2( )2

+C

1014

Sometimes, the other factor is not the du, or there is an extra x that must be

replaced with some form of u.

EX 5

x+1( ) x−1 ⌠⌡⎮

dx

u = x−1x = u +1du = dx

x+1( ) x−1⌠

⌡⎮ dx = u+1( )+1( ) u du⌠

⌡⎮⎮

= u+ 2( )u12⌠

⌡⎮⎮ du

= u3

2 + 2u1

2⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮ du

= u5

2

52

+ 2u3

2

32

+C

= 2

5 x−1( )52 + 43 x−1( )32 +C

1015

EX 6

x3 x2 + 4( )32⎛

⎝⎜⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx

u = x2 + 4x2 = u − 4du = 2x dx

x3 x2 + 4( )32⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx = 12 x2 x2 + 4( )32⎛

⎝⎜⎞⎠⎟

2x dx( )⌠

⌡

⎮⎮⎮

= 12 u − 4( )u32⎛

⎝⎜⎞⎠⎟ du

⌠

⌡

⎮⎮

= 12 u

52 − 4u32⎛

⎝⎜⎞⎠⎟

⌠

⌡

⎮⎮ du

= 12u

72

72

− 4u52

52

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟+C

= 17 x2 + 4( )72 − 45 x2 + 4( )52 +C

1016



1. 2.

3. 4.

5. 6.

7. 8.

9. 10.


1.

a) b) c)

d) e)

5x+ 3( )3⌠

⌡⎮ dx

x3 x4 + 5( )24⎛

⎝⎜⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx

1+ x3( )2⌠

⌡⎮⎮ dx 2− x( )2 3 ⌠

⌡⎮ dx

x 2x2 + 3( )⌠

⌡⎮⎮ dx dx

5x + 2( )3⌠

⌡

⎮⎮⎮

x3

1+ x4

⌠

⌡

⎮⎮⎮

dx

x +1x2 + 2x + 33

⌠

⌡

⎮⎮⎮

dx

x5 x2 + 4( )12⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx x+ 3 x+1( )2⌠⌡⎮

dx

xx2 − 4

dx⌠⌡⎮

=

−14 x2 − 4( )2

+C1

2 x2 − 4( ) +C12ln x2 − 4 + C

2 ln x2 − 4 + C12arctan x

2⎛⎝⎜

⎞⎠⎟+ C

1017

2.

a) b) c)

d) e)

3. When using the substitution , an anti-derivative of

is

a) b) c)

d) e)

4.

a) b) c)

d) e)

5.

a) b) c)

d) e)

e x

2 x dx⌠

⌡⎮=

ln x + C x + C ex + C

12e2 x + C e x + C

u = 1+ x60x 1+ x dx∫

20u3 − 60u + C 15u4 − 30u2 + C 30u4 − 60u2 + C

24u5 − 40u3 + C 12u6 − 20u4 + C

4x1+ x2 dx⌠

⌡⎮=

4arctan x +C 4xarctan x +C 1

2ln 1+ x2( ) + C

2 ln 1+ x2( ) + C 2x2 + 4 ln x +C

x x2 −1( )4 dx∫ =

110

x2 x2 −1( )5 + C 110

x2 −1( )5 + C 15x3 − x( )5 + C

15x2 −1( )5 + C 1

5x2 − x( )5 + C

1018

13-3: Anti-Derivatives: The Transcendental Rules

The proof of the transcendental integral rules can be left to a more formal Calculus

course. But since the integral is the inverse of the derivative, the discovery of the

rules should be obvious from looking at the comparable derivative rules.

Derivative Rules:

Integral Rules:

LEARNING OUTCOME

Integrate functions involving transcendental operations.

ddx sinu

⎡⎣ ⎤⎦ = cosu( )dudxddx cosu⎡⎣ ⎤⎦ = −sinu( )dudxddx tanu⎡⎣ ⎤⎦ = sec2u( )dudxddx eu⎡⎣ ⎤⎦ = eu( )dudxddx au⎡⎣ ⎤⎦ = au i lna( )dudx

ddx cscu⎡⎣ ⎤⎦ = −cscucotu( )dudxddx secu⎡⎣ ⎤⎦ = secu tanu( )dudxddx cotu⎡⎣ ⎤⎦ = −csc2u( )dudxddx lnu

⎡⎣ ⎤⎦ =1u

⎛⎝⎜

⎞⎠⎟dudx

ddx loga u

⎡⎣ ⎤⎦ =1

u i lna⎛⎝⎜

⎞⎠⎟dudx

cosu( )⌠⌡⎮ du = sinu +C

sinu( )⌠⌡⎮

du = −cosu +C

sec2u( )⌠

⌡⎮ du = tanu +C

eu( )⌠⌡⎮

du = eu +C

au( )⌠⌡⎮

du = aulna +C

cscucotu( )⌠⌡⎮ du = −cscu +C

sec u tan u( )⌠⌡⎮ du = secu +C

csc2u( )⌠

⌡⎮ du = −cotu +C

1u

⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

du = ln u +C

1019

EX 1

EX 2

EX 3

The Unchain Rule will apply to the transcendental functions quite well.

EX 4

sin x+ 3cosx( )⌠⌡⎮

dx

sin x+ 3cosx( )⌠⌡⎮

dx = sin x( )⌠⌡⎮

dx+ 3 cosx( )⌠⌡⎮ dx

= −cosx+ 3sin x+C

ex + 4 + 3csc2 x( )⌠

⌡⎮ dx

ex + 4 + 3csc2 x( ) ⌠

⌡⎮ dx = ex( )⌠

⌡⎮ dx+ 4 dx+ 3 csc2 x( )⌠

⌡⎮∫ dx

= ex + 4x− 3cot x+C

sec x sec x + tan x( )( )⌠

⌡⎮ dx

secx secx+ tan x( )( )⌠

⌡⎮ dx = sec2 x( )⌠

⌡⎮ dx + secx tan x( )⌠

⌡⎮ dx

= tan x + sec x +C

sin 5x( ) ⌠⌡⎮

dx

u = 5xdu = 5 dx

sin5x( )⌠⌡⎮

dx = 15 sin5x( )5⌠

⌡⎮ dx

= 15 sinu( )⌠

⌡⎮ du

= 15 −cosu( )+C

= − 15 cos5x+C

1020

EX 5

EX 6

sin6 xcosx( )⌠

⌡⎮ dx

u = sin xdu = cosx dx

sin6 xcosx( )⌠

⌡⎮ dx = u6( )⌠

⌡⎮ du

= 17u

7 +C

= 17 sin7 x +C

x5 sin x6( )⌠

⌡⎮ dx

u = x6

du = 6x5 dx

x5 sin x6( )⌠

⌡⎮ dx = 1

6 sin x6( ) 6x5 dx( )⌠

⌡⎮

= 16 sinu( )⌠

⌡⎮ du

= − 16 cosu +C

= − 16 cosx6 +C

1021

EX 7

EX 8

cot3 xcsc2 x( )⌠

⌡⎮ dx

u = cot xdu = −csc2 x dx

cot3 xcsc2 x( )⌠

⌡⎮ dx = cot3 x( ) −csc2 x dx( )⌠

⌡⎮

= − u3( )⌠

⌡⎮ du

= − 14 u

4 +C

= − 14 cot4 x +C

cos xx

⎛

⎝⎜

⎞

⎠⎟

⌠

⌡

⎮⎮⎮

dx

u = x = x12

du = 12 x

−12 dx = 12x

12 dx

cos xx

⎛

⎝⎜

⎞

⎠⎟

⌠

⌡

⎮⎮⎮

dx = 2 cos x( ) 12 x

dx⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

= 2 cosu( )⌠⌡⎮ du

= 2sinu +C

= 2sin x +C

1022

EX 9

EX 10

xex2 +1( )⌠

⌡⎮⎮ dx

u = x2 +1du = 2x dx

xex2+1( )⌠

⌡⎮⎮ dx = 1

2 ex2+1( ) 2x dx( )⌠

⌡⎮⎮

= 12 eu( )⌠

⌡⎮ du

= 12 e

u +C

= 12 ex2 +1 +C

ln3 xx

⎛

⎝⎜⎞

⎠⎟⌠

⌡

⎮⎮⎮

dx

u = lnx

du = 1x dx

ln3 xx

⎛

⎝⎜⎞

⎠⎟⌠

⌡

⎮⎮⎮

dx = u3 du∫

= 14 u

4 +C

= 14 ln4 x+C

1023



1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.


1.

a) b)

c) d)

e)

x4 cos x5( )⌠

⌡⎮ dx sin 7x+1( )( )⌠

⌡⎮ dx

sec2 3x −1( )( )⌠

⌡⎮ dx sin x

x⎛

⎝⎜

⎞

⎠⎟

⌠

⌡

⎮⎮⎮

dx

tan4 xsec2 x( )⌠

⌡⎮ dx ln x

x⌠

⌡⎮⎮

dx

e6x( )⌠

⌡⎮⎮ dx

cos2xsin3 2x

⌠

⌡

⎮⎮ dx

x ln x2 +1( )x2 +1

⌠

⌡

⎮⎮⎮

dx e x

x

⌠

⌡

⎮⎮⎮

dx

cot x csc2 x( )⌠

⌡⎮ dx

1x2

sin 1x

⎛⎝⎜

⎞⎠⎟cos 1

x⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx

x3 + 2 + 1x2 +1

⎛⎝⎜

⎞⎠⎟ dx⌠

⌡⎮=

x4

4+ 2x + tan−1 x + C x4 + 2 + tan−1 x + C

x4

4+ 2x + 3

x3 + 3+ C

x4

4+ 2x + tan−1 2x2 + C

4 + 2x + tan−1 x + C

1024

2.

a) b)

c) d)

e)

3.

a) b)

c) d)

e)

4.

a) b) c)

d) e)

cos 3− 2x( ) dx∫ =

sin 3− 2x( ) + C − sin 3− 2x( ) + C

12sin 3− 2x( ) + C −

12sin 3− 2x( ) + C

−15sin 3− 2x( ) + C

x − 2x −1

dx⌠⌡⎮

=

− ln x −1 + C x + ln x −1 + C

x − ln x −1 + C x + x −1 + C

x − x −1 + C

ex2− 2xex

2 dx⌠

⌡⎮=

x − ex2

+C x − e− x2

+ C x + e− x2

+ C

−ex2

+ C −e− x2

+ C

1025

5.

a) b) c)

d) e)

6sin xcos2 x dx∫ =

2sin3 x + C −2sin3 x + C 2cos3 x + C

−2cos3 x + C 3sin2 xcos2 x +C

1026

13-4: Definite Integrals

As noted in the overview, anti-derivatives are known as indefinite integrals

because the answer is a function, not a definite number. But there is a time when

the integral represents a number. That is when the integral is used in an Analytic

Geometry context of area. Though it is not necessary to know the theory behind

this to be able to do it, the theory is a major subject of Integral Calculus, so it will

be explored briefly here.

Geometry presented how to find the exact area of various polygons, but it never

considered figures where one side is not made of a line segment. Here, consider a

figure where one side is the curve and the other sides are the x-axis and

the lines and .

As seen above, the area can be approximated by rectangles whose height is the y

value of the equation and whose width is . The more rectangles made, the

better the approximation. The area of each rectangle would be and the

total area of n rectangles would be . If an infinite number of

rectangles (which would be infinitely thin) could be made, the exact area would be

the resulting sum. The rectangles can be drawn several ways––with the left side at

the height of the curve (as drawn above), with the right side at the curve, with the

rectangle straddling the curve, or even with rectangles of different widths. But

y = f x( )x = a x = b

y=f(x)

a b

Δx

f x( ) iΔx

f xi( ) iΔx

i=1

n

∑

1027

once they become infinitely thin, it will not matter how they were drawn––they

will have no width and a height equal to the y-value of the curve.

An infinite number of rectangles can be made mathematically by taking the limit as

n approaches infinity, or

.

This limit is rewritten as the Definite Integral:

b is the “upper bound” and a is the “lower bound,” and would not mean much if it

were not for the following rule:

The Fundamental Theorem of Calculus:

1.

2. If , then .

The First Fundamental Theorem of Calculus simply says what is already known––

that an integral is an anti-derivative. The Second Fundamental Theorem says the

answer to a definite integral is the difference between the anti-derivative at the

upper bound and the anti-derivative at the lower bound.

This idea of the integral meaning the area may not make sense initially, mainly

because Geometry was used, where area is always measured in square units. But

that is only because the length and width are always measured in the same kind of

units, so multiplying length and width must yield square units. Consider a graph

limn→∞ f xi( ) iΔx

i=1

n

∑

f x( ) dxa

b

∫

d dx f t( ) dt

c

x

∫ = f x( )

′F x( ) = f x( ) f x( ) dxa

b

∫ = F b( )− F a( )

1028

where the x-axis is time in seconds and the y-axis is velocity in feet per second.

The area under the curve would be measured as seconds multiplied by feet/sec––

that is, feet. So the area under the curve equals the distance traveled in feet. In

other words, the integral of velocity is distance.

LEARNING OUTCOMES

Evaluate definite integrals.

Interpret definite integrals as area under a curve.

EX 1 Find

Notice that the arbitrary constant C does not effect the definite integral. It cancels

itself out when we substitute a and b and subtract. So, with a definite integral,

the + C can be ignored.

EX 2 Find

x2 dx1

4

∫

x2 dx1

4

∫ = x3

3 +C⎡

⎣⎢⎢

⎤

⎦⎥⎥1

4

= 43

3 +C⎛

⎝⎜

⎞

⎠⎟ −

13

3 +C⎛

⎝⎜

⎞

⎠⎟

= 21

sinx dxπ4

π2∫

sin x dxπ4

π2∫ = −cosx⎡⎣ ⎤⎦π 4

π2

= −cosπ2⎡⎣⎢

⎤⎦⎥− −cosπ4⎡⎣⎢

⎤⎦⎥

= 0− − 12

⎛⎝⎜

⎞⎠⎟

1029

EX 3

EX 4

If a composite function requires integration by substitution, substitute for the

boundaries as well. a and b are normally what x would equal. If switching to u,

the boundaries must equal what u would equal when and .

= 12

or 2 2

3x2 + 4x + 5( ) dx0

3

∫

3x2 + 4x + 5( ) dx0

3

∫ = x3 + 2x2 + 5x⎡⎣ ⎤⎦03

= 33 + 2 3( )2 + 5 3( )⎡⎣⎢

⎤⎦⎥− 03 + 2 0( )2 + 5 0( )⎡⎣⎢

⎤⎦⎥

= 60 − 0= 60

3x

⎛⎝⎜

⎞⎠⎟

dx1

3

∫

3x

⎛⎝⎜

⎞⎠⎟

dx1

3

∫ = 3 ln x⎡⎣ ⎤⎦13

= 3 ln3⎡⎣ ⎤⎦ − 3 ln1⎡⎣ ⎤⎦

= 3ln3= 3.295

x = a x = b

1030

EX 5

EX 6

x x2 + 5( )3⎛⎝⎜

⎞⎠⎟

dx0

2

∫

u = x2 + 5, du = 2x, u 0( ) = 5, u 2( ) = 9

x x2 + 5( )3⎛⎝⎜

⎞⎠⎟

dx0

2

∫ = 12 x2 + 5( )3 2x dx( )

0

2

∫

= 12 u3du

5

9

∫

= 12 i u4

4⎡

⎣⎢⎢

⎤

⎦⎥⎥ 5

9

= 12 i 94

4 − 54

4⎡

⎣⎢⎢

⎤

⎦⎥⎥

= 742

sin6 xcosx( ) dx 0

π2∫

u = sin x, du = cosx dx, u π2

⎛⎝⎜

⎞⎠⎟=1, u 0( ) = 0

sin6 xcosx( ) dx 0

π2∫ = u6( ) du0

1

∫= 1

7u7⎡

⎣⎢⎤⎦⎥0

1

= 17

1031


Evaluate the definite integrals.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.


1.

a) b) c)

d) e)

x2 + 5( ) dx0

3

∫ 12x− 3

dx2

6

∫

x 1− x2( ) dx−1

1

∫ x+ 2( )3 dx0

1

∫

x+ 5( ) x2 − 3( ) dx−2

2

∫ 1− xx

dx4

9⌠

⌡

⎮⎮⎮

cos5 xsin x dxπ6

π2∫ x

2 +1x2 dx

1

3⌠

⌡

⎮⎮

dx2x+ 5−1

2⌠

⌡⎮⎮

sin x2− cos x0

π⌠

⌡⎮⎮ dx

dxx lnx2

4⌠

⌡⎮⎮

ex

1+ e2x dx0

ln 2⌠

⌡

⎮⎮

1x+ 2x⎛

⎝⎜⎞⎠⎟ dx

2

6⌠⌡⎮

=

ln 4 + 32 ln 3+ 40 ln 3+ 32

ln 4 + 40 ln12 + 32

1032

2.

a) b) c) 0 d) e)

3.

a) b) c)

d) e)

4.

a) 1 b) –1 c) –2 d) 2 e) None of these

5. If , then

a) 3 b) 6 c) 9 d) 12 e) 15

sinπ x dx0

1

∫ =

2π

1π

−2π

−1π

sec2 xtan x

dxπ

4

π3⌠

⌡⎮=

ln 3 − ln 3 ln 2

3 −1 ln π3− ln π

4

x2 − 4x + 4 dx0

2

∫ =

f x( ) dx2

4

∫ = 6 f x( ) + 3( ) dx2

4

∫ =

1033

6. Let f be the function defined by . Then

a) b) c)

d) e)

f x( ) = x +1 for x < 01+ sinπ x for x ≥ 0

⎧⎨⎩

f x( )dx−1

1

∫ =

32

32−2π

12−2π

32+2π

12+2π

1034

13-5: Definite Integrals and Area

Since the definite integrals were originally defined in terms of “area under a

curve,” what this context of “area” really means needs to be considered in relation

to the definite integral.

Consider this function, on . The graph looks like this:

In the last section, . But it can be seen there is area under the

curve, so how can the integral equal the area and equal 0? Remember that the

integral was created from rectangles with width dx and height . So the area

below the x-axis would be negative, because the values are negative.

EX 1 What is the area under on ?

It is already known that , so this integral cannot

represent the area. The question is to find the positive number that

represents the area (total distance), not the difference between the positive

and negative “areas” (displacement). The commonly accepted context for

area is a positive value. So,

y = xcos x2( )( )

x∈ 0, π⎡⎣⎢

⎤⎦⎥

xcos x2( )( )

0

π⌠⌡⎮ dx = 0

f x( )f x( )

y = xcos x2( )( )

x∈ 0, π⎡⎣⎢

⎤⎦⎥

xcos x2( )( )

0

π⌠⌡⎮ dx = 0

Area = xcos x2( )0

π⌠⌡ dx

= xcos x2( )0

1.244∫ dx− xcos x2( )1.244

π∫ dx

=1

1035

The calculator could find this answer:

When the phrase “area under the curve” is used, what is really meant is the area between the curve and the x-axis. CONTEXT IS EVERYTHING. The area under the curve is only equal to the definite integral when the curve is completely above the x-axis. When the curve goes below the x-axis, the definite integral is negative, but the area, by definition, is positive.

LEARNING OUTCOMES

Relate definite integrals to area under a curve.

Understand the difference between displacement and total distance.

Extend that idea to understanding the difference between the two concepts in

other contexts.

Properties of Definite Integrals:

1.

2.

3. , where

f x( ) dx = −

a

b

∫ f x( ) dxb

a

∫

f x( ) dx

a

a

∫ = 0

f x( )a

b

∫ dx = f x( )a

c

∫ dx+ f x( ) c

b

∫ dx a < c < b

1036

EX 2 Find the area under on .

A quick look at the graph reveals that the curve crosses the x-axis at .

Integrate y on to get the difference between the areas, not the

sum. To get the total area, set up two integrals:

y = x3 − 2x2 − 5x − 6 x∈ −1, 2⎡⎣ ⎤⎦

x = 2

x∈ −1, 2⎡⎣ ⎤⎦

Area = x3 − 2x2 −5x−6( )−1

1⌠⌡ dx+ − x3 − 2x2 −5x−6( )

1

2⌠⌡ dx⎛

⎝⎜⎞⎠⎟

= x4

4 − 2x3

3 − 5x2

2 −6x⎤

⎦⎥⎥−1

1

+ x4

4 − 2x3

3 − 5x2

2 −6x⎡

⎣⎢⎢

⎤

⎦⎥⎥1

2

= 323 + 29

12

= 15712

1037

EX 3 Find the area between and the x-axis on .

Evaluating the integral :

Obviously, the area cannot be zero. Look at the graph:

What has happened with our integral here is that on the curve is

above the x-axis, so the “area” represented by the integral is positive. But on

, the “area” is negative. Since these areas are the same size, the

integrals added to zero.

There are several ways to account for this to find the area directly. The most

common is to split the integral into two integrals where the boundaries are

the zeros, and multiply the “negative area” by another negative to make it

positive.

y = sin x x∈ 0, 2π⎡⎣ ⎤⎦

sin x dx0

2π

∫

sin x dx0

2π

∫ = −cosx⎡⎣ ⎤⎦02π

= − −1( )( )−1= 0

x∈ 0, π⎡⎣ ⎤⎦

x∈ π , 2π⎡⎣ ⎤⎦

Area = sin x dx0

π

∫ − sin x dxπ

2π

∫= −cos x⎡⎣ ⎤⎦0

π − −cos x⎡⎣ ⎤⎦π2π

= − −1( )( )− −1( )⎡⎣⎢

⎤⎦⎥ − −1( )− − −1( )( )⎡

⎣⎢⎤⎦⎥

= 4

1038

Here’s a fun scenario––imagine leaving your house to go to school, and that school

is 6 miles away. You leave your house and halfway to school you realize you have

forgotten your Calculus homework (gasp!). You head back home, pick up your

assignment, and then head to school.

There are two different questions that can be asked here. How far are you from

where you started? And how far have you actually traveled? You are six miles

from where you started but you have traveled 12 miles. These are the two different

ideas behind displacement and total distance.

Vocabulary: 1. Displacement – how far apart the starting position and ending position of an

object are (it can be positive or negative)

2. Total Distance – how far an object travels in total (this can only be positive)

Displacement

Total Distance

EX 4 A particle moves along a line so that its velocity at any time t is

(measured in meters per second).

a) Find the displacement of the particle during the time period .

b) Find the distance traveled during the time period .

= v dta

b

∫

= v dta

b

∫

v t( ) = t2 − t − 6

1≤ t ≤ 4 1≤ t ≤ 4

a) v dta

b

∫ = t 2 − t − 6( ) dt1

4

∫

= t3

3− t

2

2− 6t

1

4

= −4.5

1039

Note that the properties of integrals are used to split the integral into two integrals

that represent the separate positive and negative distance traveled. Putting a “–” in

front of the first integral turns the negative value from the integral into a positive

value. Split the integral at because that would be where .

b) v dta

b

∫ = t 2 − t − 6 dt1

4

∫ = − t 2 − t − 6( ) dt

1

2

∫ + t 2 − t − 6( ) dt2

4

∫

= − t3

3+ t

2

2+ 6t

1

2

+ t3

3− t

2

2− 6t

2

4

= 10 16

t = 2 v t( ) = 0

1040


1. Find the area between and the x-axis on .





6. Find the area between and the x-axis on



9. The velocity function (in meters per second) for a particle moving along a

line is for . Find a) the displacement and b) the distance

traveled by the particle during the given time interval.

10. The velocity function (in meters per second) for a particle moving along a

line is for . Find a) the displacement and b) the distance

traveled by the particle during the given time interval.

11. Find the distance traveled by a particle in rectilinear motion whose velocity,

in feet/sec, is described from second to seconds.

12. Find the distance traveled by a particle in rectilinear motion whose velocity,

in feet/sec, is described from second to seconds.

[Be Careful!!!]

y = x2 +1 x∈ −1, 1⎡⎣ ⎤⎦

y = x 9 − 4x2 x∈ 0, 32

⎡⎣⎢

⎤⎦⎥

y = cos x x∈ 0, 2π⎡⎣ ⎤⎦

y = xe−x2

x∈ −2, 2⎡⎣ ⎤⎦

y = x 9 − 4x2 x∈ − 32 , 3

2⎡⎣⎢

⎤⎦⎥

y = x 2x2 −18 x∈ −2, 1⎡⎣ ⎤⎦

y = 3sin x 1− cos x x∈ −π , 0⎡⎣ ⎤⎦

y = x2ex3 x∈ 0, 1.5⎡⎣ ⎤⎦

v t( ) = 3t −5 0 ≤ t ≤ 3

v t( ) = t2 − 2t −8 1≤ t ≤ 6

v t( ) = 4t +1 t = 1 t = 5

v t( ) = 3t2 − 4t +1 t = 0 t = 4

1041


1. The area under the graph of on the interval is

a) b) c)

d) e)

2. A particle starts at when and moves along the x-axis in such a

way that at time its velocity is given by . Determine the position

of the particle at .

a) b) c) d) e)

3. At , a particle starts at the origin with a velocity of 6 feet per second

and moves along the x-axis in such a way that at time t its acceleration is feet

per second per second. Through how many feet does the particle move during the

first 2 seconds?

a) 16 ft b) 20 ft c) 24 ft d) 28 ft e) 32 ft

4.

a) b) c) d) 0 e)

y = 4x3 + 6x − 1x

1≤ x ≤ 2

32 − ln 2 units2 30 − ln 2 units2 24 − ln 2 units2

994

units2 21 units2

5, 0( ) t = 0

t > 0 v t( ) = 11+ t

t = 3

9716

9516

7916

5 + ln 4 1+ ln 4

t = 012t 2

x2 − xx3 dx

1

2⌠⌡⎮

=

ln 2 − 12

ln 2 + 12

12

14

1042

5.

a) b) 1 c) 2 d) e)

x1+ x2

dx0

3⌠⌡⎮

=

12

ln 2 arctan2 − π4

1043

Integrals Practice Test

Part 1: CALCULATOR REQUIRED

Round to 3 decimal places. Show all work.

Multiple Choice (3 pts. each)

1. A particle moves along the x-axis with the velocity given by

for time . If the particle is at position at time , what is the

position of the particle at time ?

(a) 4

(b) 6

(c) 9

(d) 11

(e) 12

2. If and , what is the value of ?

(a) –21

(b) –13

(c) 0

(d) 13

(e) 21

3.

(a)

(b)

(c)

(d)

(e)

v t( ) = 3t 2 + 6tt ≥ 0 x = 2 t = 0

t = 1

f x( ) dx−5

2

∫ = −17 f x( ) dx5

2

∫ = −4 f x( ) dx−5

5

∫

x2

ex3 dx⌠

⌡⎮=

−13ln ex

3

+ C

−ex

3

3+ C

−13ex

3 + C

13ln ex

3

+ C

x3

3ex3 + C

1044

4.

(a)

(b)

(c)

(d)

(e)

5.

(a)

(b)

(c)

(d)

(e)

6.

(a)

(b)

(c)

(d)

sin t dt0

x

∫ =

sin x− cos xcos xcos x −11− cos x

x2 +1x

dx1

e⌠⌡⎮

=

e2 −12

e2 +12

e2 + 22

e2 −1e2

2e2 − 8e + 63e

x 4 − x2 dx∫ =

4 − x2( )323

+ C

− 4 − x2( )32 + Cx2 4 − x2( )32

3+ C

−x2 4 − x2( )32

3+ C

1045

(e)

7. If , then

(a) –12

(b) 12

(c) –4

(d) 4

(e) 0

8.

(a)

(b)

(c)

(d)

(e)

−4 − x2( )323

+ C

x7 + k( ) dx−2

2

∫ = 16 k =

sin 2x + 3( )∫ dx =

12cos 2x + 3( ) + C

cos 2x + 3( ) + C− cos 2x + 3( ) + C−12cos 2x + 3( ) + C

−15cos 2x + 3( ) + C

1046

Integrals Practice Test

Part 2: NO CALCULATOR ALLOWED

Round to 3 decimal places. Show all work.

Free Response (10 pts. each)

1. Find the area between and the x-axis on the interval .

2. Find the area between and the x-axis from to .

3. A particle moves along a line with velocity function , where v is

measured in meters per second. Set up, but do not solve, and integral expression

for a) the displacement and b) the distance traveled by the particle during the time

interval .

y =1x2

x ∈ 1, 2[ ]

y =x

1+ x2x = −2 x = 2

v t( ) = t 2 − t

0, 5[ ]

1047

Integrals Homework Answer Key


1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17.


1. D 2. A 3. A


1. 2.

2x3 − x2 + 3x +C 14 x

4 + x3 − x2 + 4x +C

3x23 +C 8

5 x5 − x4 + 3x3 + x2 + x +C

23 x

6 + 54 x4 +C 4x3 + 292 x2 − 8x +C

23 x

32 −12x12 +C 1

2 x2 + 2x

12 + 3ln x +C

14 x

4 + x3 + 32 x2 + x +C 16

3 x3 −12x2 + 9x +C

23 x

32 + 65 x52 −12x

12 +C 2x2 − 2x−12 − 1x +C

f x( ) = x3 − 3x2 + 3x + 2 f x( ) = 14 x4 + 13 x

3 − 12 x2 + 3x − 3712

f x( ) = 32 x2 − 103 x

32 − 2x + 323x t( ) = 3t 4 − 2t 3 + 4t2 −13t +11

x t( ) = 112 t

4 − 13 t3 + 2t2 + 2t + 4

120 5x + 3( )4 +C 1

100 x4 + 5( )25 +C

1048

3. 4.

5. 6.

7. 8.

9.

10.


1. C 2. E 3. D 4. D 5. B


1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

17 x

7 + 12 x4 + x +C − 35 2 − x( )53 +C

16 2x

2 + 3( )32 +C − 110 5x + 2( )-2 +C

12 1+ x

4( )12 +C 34 x2 + 2x + 3( )23 +C

110 x2 + 4( )5 − x2 + 4( )4 + 83 x2 + 4( )3 +C

27 x + 3( )72 − 85 x + 3( )52 + 83 x + 3( )32 +C

15 sin x5 +C − 1

7 cos 7x +1( ) +C

13 tan 3x −1( ) +C −2cos x +C

15 tan

5 x+C 12 ln x( )2 +C

16 e

6x +C − 14 sin−2 2x+C = − 1

4 csc2 2x+C

14 ln x

2 +1( )( )2 +C 2e x +C

− 23cot32 x+C − 1

2 sin 1x

⎛⎝⎜

⎞⎠⎟

2

+C, or 12 cos 1x⎛⎝⎜

⎞⎠⎟

2

+C

1049


1. A 2. D 3. C 4. C 5. D


1. 24 2. 2 3. 0 4. 16.25

5. 6. 7. 0.070 8.

9. 10. 11. 12.


1. C 2. A 3. A 4. D 5. D 6. D


1. 2. 2.25 3. 4 4. 0.982

5. 4.5 6. 7.

8. 9a. b.

10a. b. 11. 52 feet 12. 36.296 feet


1. C 2. E 3. D 4. A 5. B

−3313 −3 83

0.549 0.693 0.693 0.322

83

2cos π + 2 − 32

13 e

3.375 −1( ) −32m

416 m

−103 m

983 m

1050

Integrals Practice Test Answer Key

Multiple Choice

1. B 2. B 3. C 4. E

5. B 6. E 7. D 8. D

Free Response

1.

2.

3a.

b.

Area = 12 Area = 2 ln5

Displacement = t2 − t( ) dt0

5

∫Total Distance Traveled = − t2 − t( ) dt0

1

∫ + t2 − t( ) dt1

5

∫

chapter 13: integrals · chapter 13 overview: the integral calculus is essentially comprised of two...

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