MAT 213Brief Calculus

Section 5.6Integration by Substitution or Algebraic

Manipulation

• See if you can figure out what functions would give the following derivatives

2

2 5

( 1)

3 4

( ) 6(2 4 ) (4 4)

( ) 2

( ) 4 1

t

g x x x x

h t te

k x x x

• Recall the chain rule

• Now imagine taking the antiderivative of both sides

• Therefore any we can find the derivative of any function of this form using the method of substitution

)('))(('))'((

)('))(('))((

xgxgfxgf

xgxgfxgfdx

d

dxxgxgfxgf

dxxgxgfdxxgfdx

d

)('))(('))((

)('))(('))((

The Method of Substitution

• When integrating something of the form

we let u = g(x) or the “inside function”

• Then and we get an integral that

can be done in terms of u

• Let’s look at our previous examples using this method which I refer to as u substitution

dxxgxgf )('))(('

)(' xgdx

du

2

2 5

( 1)

3 4

6(2 4 ) (4 4)

2

4 1

t

x x x dx

te dt

x x dx

• Now we can always adjust our substitution if we are off by a constant

• For example, let’s find the following antiderivative

dxxx 543

• Substitution can also be used for definite integrals

22

20 3?

1

xdx

x

WARNING!!!

We may not be able to use substitution if anything other than a constant multiple is missing. (We cannot

just “add in” a variable)

Example

“x2dx” is not a constant multiple of ”dw = 4x3dx”

dxxx342 1

• Let’s try a few from the book– Each group choose any 2 from the book and put

your final solution up on the board