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.
Examples Integrate
Solution We use integration by parts. Notice that we need to use substitution to find the integral of e^{x}.
Hence we have
Exercise Evaluate x ln x dx
Integration By Parts Twice Example Evaluate x^{2} e^{x} dx We use integration by parts
We have x^{2}e^{x}  2xe^{x} dx = x^{2}e^{x}  2xe^{x} dx Have we gone nowhere? Now we now use integration by parts a second time to find this integral
We get x^{2}e^{x}  2xe^{x} + 2e^{x} dx = x^{2}e^{x}  2xe^{x} + 2e^{x} + 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
We get x lnx  dx = x lnx  x + C
When to Use Integration By Parts
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
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
For our example, we have c(t) = 2000 t r = 0.05 t_{1} = 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
This gives us
We have already found the antiderivative for this last integral. We have
You should ask for $211,393.
Back to the Math 116 Home Page Back to the Math Department Home email Questions and Suggestions
