Integration

Tom Penick [email protected] www.teicontrols.com/notes 05/27/98

INTEGRATIONChapter 5

Definition of Integral Notation for Antiderivatives: 5.1p231

f x dx F x C( ) ( )= +∫where: F is an antiderivative of f

C is an arbitrary constant

Integration Formulas: 5.1 p232 To integrate, add one to theexponent, multiply the coefficient by the reciprocal ofthe new exponent, then add a constant C.

0 dx C=∫ k dx kx C k= + ≠∫ , 0

kf x dx k F x dx( ) ( )= ∫∫ dx x c= +∫[ ( ) ( )] ( ) ( )f x g x dx f x dx g x dx± = ± ∫∫∫x dx

x

nC nn

n

=+

+ ≠ −+

∫1

11,

Definition of Sigma Notation: 5.2 p234

a a a a ai n

i

n

= + + + +=

∑ 1 2 3

1

. . .

where: i is the index of summationai is the ith term of the sumn is the upper bound of summation1 is the lower bound of summation

Summation Formulas: 5.2 p240

ka k a ki i

i

n

i

n

===∑∑

11

, is a constant

[ ]a b a bi i i

i

n

i

i

n

i

n

± = ±= ==

∑ ∑∑1 11

c cni

n

==

∑1

in n

i

n

=+

=∑ ( )1

21

in n n

i

n2

1

1 2 1

6=

+ +

=∑ ( )( )

in n

i

n3

2 2

1

1

4=

+

=∑ ( )

Finding upper and lower sums of a region bounded by afunction, the x-axis, and two values of x: 5.2 p246

Upper sum: ( ) ( )

Sx x

nf x

i x x

ni

n

=−

+−

=∑ 2 1

12 1

1

Lower sum:

( ) ( )( )s

x x

nf x

i x x

ni

n

=−

+− −

=∑ 2 1

12 1

1

1

Where the function f(x) is increasing on the interval(x1, x2) and n is the number of divisions between x1

and x2. If the function is decreasing, reverse theformulas. ( )

n

xx 12 − represents the width of each

division and ( )f x1+.... represents each height

involved.

Definition of a Riemann Sum: 5.3 p252 If f is defined onthe interval [a, b] and ∆ is an arbitrary partition of[a, b], and

a x x x x x bn n= < < < < < =−0 1 2 1. . .

where ∆x1 is the width of the ith subinterval. If c i is anypoint in the ith subinterval, then the sum

∑−

− ≤≤∆n

iiiiii xcxxcf

11,)(

is called a Riemann sum of f for the partition ∆.

Definition of the Definite Integral: 5.3 p253 If f is definedon the interval [a, b] and the limit of a Riemann sumof f exists, then f is integrable on [a, b] and we denotethe limit by:

∑ ∫∑= =

∞→→∆=∆=∆

n

i

b

a

n

ii

nii dxxfxcfxcf

1 10

)()(lim)(lim

∆ is the width of the largest subinterval or norm of thepartition. If every subinterval is of equal width then:

n

abx

−=∆=∆ If not, then: n

ab≤

∆−

c a i xi = + ( )∆ = the x value where each verticalmeasurement is taken.


Area of a Region: 5.3 p254 If f is continuous and non-negative on the closed interval [a, b], then the area ofthe region bounded by the graph of f, the x-axis, andthe vertical lines x = a and x = b is given by:

x dxx

nC nn

n

=+

+ ≠−+

∫1

11,

Properties of Definite Integrals: 5.3 p256-8

If f is defined at x = a, then

f x dxa

a

( ) =∫ 0

If f is integrable on [a, b], then

f x dx f x dxa

b

b

a

( ) ( )= −∫∫Additive Interval Property: If f is integrable on the three

closed intervals determined by a, b, and c, then

f x dx f x dx f x dxa

c

a

b

c

b

( ) ( ) ( )= +∫∫ ∫If f and g are integrable on [a, b] and k is a constant,

then:

k f x dx k f x dxa

b

a

b

( ) ( )= ∫∫[ ]f x g x dx f x dx g x dx

a

b

a

b

a

b

( ) ( ) ( ) ( )± = ±∫∫ ∫Preservation of Inequality: If f is integrable and

nonnegative on the closed interval [a, b], then

0 ≤ ∫ f x dxa

b

( )

And if f and g are integrable on the closed interval[a, b], then

f x dx g x dxa

b

a

b

( ) ( )∫ ∫≤

Involving an absolute value: Find the zero of thefunction and rewrite as a sum.

The Fundamental Theorem of Calculus: 5.4 p260

If a function f is continuous on the closed interval [a, b],then:

[ ]f x dx F b F a F xa

b

a

b( ) ( ) ( ) ( )= − =∫

where F is any function that F'(x) = f(x) for all x in [a, b].In other words, F is the antiderivative of f. Note thatthe constant C has been dropped from theantiderivative because it cancels out in subtraction.

The Mean Value Theorem for Integrals: 5.4 p263 If f iscontinuous on the closed interval [a, b], then thereexists a number c in the closed interval [a, b] suchthat:

))(()( abcfdxxfb

a−=∫

a cb

y

bc

f(c)

x

f(x)

Definition of the Average Value of a function on aninterval: 5.4 p264 If f is integrable on [a, b], then theaverage value of f on this interval is given by:

1

b af x dx

a

b

− ∫ ( )

The Second Fundamental Theorem of Calculus: 5.4

p269 If f is continuous on an open interval I containinga, then for every x in the interval,

[ ]d

dxf t dt f x

a

x

( ) ( )∫ =

Antidifferentiation of a Composite Function: 5.5 p269 Let fand g be functions such that f go and g' are

continuous on an interval I. If F is an antiderivative off on I, then

f g x g x dx F g x C( ( )) ( ) ( ( ))′ = +∫Change of Variables: 5.5 p272 If we let u = g(x), then

du = g' (x) dx, and the integral above takes the form:

f g x g x dx f u du F u C( ( )) ( ) ( ) ( )′ = = +∫∫And for definite integrals: If the function u = g(x) has a

continuous derivative on the closed interval [a, b] andf has an antiderivative over the range of g, then

f g x g x dx f u dua

b

g a

g b

( ( )) ( ) ( )( )

( )

′ =∫ ∫

The above illustrates in equation form how thesubstitution process below works. The purpose of allthis is to allow us to reduce a complex integral to aform that fits a rather limited number of formulas forevaluating integrals.


Guidelines for Integration by Substitution: 5.5 p274

1. Choose a substitution u = g(x). Usually, it is best tochoose the inner part of a composite function, such asa quantity raised to a power (or root). Sometimes itmay be better to let u equal the entire root rather thanthe value under the root.

2. Take the derivative of u and write in terms ofdu = g' (x) dx. In order to get the terms on the rightside of this equation to be identical to terms present inthe integral, it may be necessary to perform somealgebraic manipulation.

3. Rewrite the integral in terms of u and du. It may benecessary here to solve for x in terms of u to completethe substitution.

4. Evaluate the resulting integral.

5. Replace u by g(x) to obtain the antiderivative in termsof x.

The substitution technique needs to be thoroughlyunderstood because it will be used repeatedly in futurechapters.

The General Power Rule for Integration: 5.5 p274 If g isa differentiable function of x, then

[ ( )] ( )[ ( )]

,g x g x dxg x

nC nn

n

′ =+

+ ≠ −+

∫1

11

Equivalently, if u = g(x), then

u duu

nC nn

n

=+

+ ≠ −+

∫1

11,

The Trapezoidal Rule: 5.6 p280 Let f be continuous on[a, b].

f x dxb a

nf x f x f x f x f xn na

b

( ) [ ( ) ( ) ( ) . . . ( ) ( )]≈−

+ + + + +−∫ 22 2 20 1 2 1

where n is the number of subintervals from a to b. Thismakes the values of x:

x a0 =

x a b an1 = + −

x x b an2 1= + −

. . .x bn =

where b an− is the width of the subinterval.

Moreover, as n → ∞ , the right-hand side approaches

f x dxa

b

( )∫

Error using the Trapezoidal Rule: 5.6 p284 If f has acontinuous second derivative on [a, b], then the error

E in approximating f x dxa

b

( )∫ by the Trapezoidal

Rule is:

bxaxfn

abE ≤≤′′−

≤ |],)(|[max12

)(2

3

Simpson's Rule (n is even): 5.6 p282 Let f be continuouson [a, b].

f x dxb a

nf x f x f x f x f x f xn na

b( ) [ ( ) ( ) ( ) ( ) . . . ( ) ( )]≈

−+ + + + + +−∫ 3

4 2 4 40 1 2 3 1

Moreover, as n → ∞ , the right-hand side approaches

f x dxa

b

( )∫ .

Error in Simpson's rule: 5.6 p284 If f has a continuousfourth derivative on [a, b], then the error E in

approximating f x dxa

b

( )∫ by Simpson's Rule is:

Eb a

nf x a x b≤

−≤ ≤

( )[max| ( )| ],( )

5

4

4

180

Theorem for integrals of 2nd Degree Polymonials: 5.6 p282

If p x Ax Bx C( ) = + +2 , then

p x dxb a

p a pa b

p ba

b

( ) ( ) ( )=−

+

+

+

∫ 6

42

Even and Odd Functions:

y

x

(Even powers of x)

EVEN FUNCTION

y

x

ODD FUNCTION

(Odd powers of x)

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