Integration by Parts

Derivation of Integration by Parts

Recall the product rule:

(uv)' = u' v + uv'

or

uv' = (uv)' - u' v

Integrating both sides, we have that

uv' dx     =     (uv)' dx  -   u' v dx

=    uv   -   u' v dx.

 Theorem:  Integration by Parts Let u and v be differentiable functions, then

Examples

Integrate

Solution

We use integration by parts.  Notice that we need to use substitution to find the integral of ex.

 u = x dv= e3x dx du = dx v = 1/3 e3x

Hence we have

Exercise

Evaluate

x ln x dx

Integration By Parts Twice

Example

Evaluate

x2 ex dx

We use integration by parts

 u = x2 dv = ex dx du = 2x dx v = ex

We have

x2ex - 2xex dx     =     x2ex - 2xex dx

Have we gone nowhere?  Now we now use integration by parts a second time to find this integral

 u = x dv = ex dx du = dx v = ex

We get

x2ex   -   2xex  +  2ex dx

=     x2ex   -   2xex  +   2ex + C

Other By Parts

Occasionally there is not an obvious pair of u and dv.  This is where we get creative.

Example:

Find

ln x dx

What should we let u and dv be?  Try

 u = ln x dv = dx du =  1/x dx v =  x

We get

x lnx   -   dx      =    x lnx  -  x  +  C

When to Use Integration By Parts

1. When u-substitution does not work

2. When there is a mix of two types of functions such as an exponential and polynomial, polynomial and log, etc.

3. With  ln x.

4. When all else fails.

Evaluating a Definite Integral

We can use integration by parts to evaluate definite integrals.  We just have to remember that all terms receive the limits.

Example

Evaluate

Solution

Use integration by parts

 u = ln x dv = x2 dx du =  1/x dx v =  1/3 x3

We get

Application:  Present Value

Your patent brings you a annual income of 3,000 t dollars where t is the number of years since the the patent begins.  The patent will expire in 20 years.  A business has offered to purchase the patent from you.  How much should you ask for it?  Assume an inflation rate of 5%.

This question is a present value problem.  Since there is inflation, your later earnings will be worth less than this year's earnings.  The formula to determine this is given by

 Present Value Formula If c(t) is the continuous annual income over t1 years with an inflation rate r, then the present value can by found by

For our example, we have

c(t)  =  2000 t        r  =  0.05        t1  =  20

We integrate

Use integration by parts and note that with the substitution

u  =  -0.05t        du  =  -0.05dt

or

-20du  =  dt

we get

so that

 u = 2000t dv = e-0.05t dt du =  2000 dt v =  -20e-0.05t

This gives us

We have already found the antiderivative for this last integral.  We have