Substitution and integration by parts

What this concept is

Substitution reverses the chain rule; integration by parts reverses the product rule: ∫u v′ dx = uv−∫u′v dx.

Why it matters here

These two techniques handle many normalization, expectation and growth calculations without treating integration as a lookup trick.

Concrete example

For ∫ from 0 to 1 of 2x exp(x²)dx let u=x², du=2x dx and change bounds to 0,1; result e−1. For ∫ from 0 to 1 of x exp(x)dx choose u=x, v′=exp(x); result [x exp(x)]₀¹−[exp(x)]₀¹=1.

Common confusion / misconception

After substituting, do not mix old bounds with a new variable. Integration by parts can make the remaining integral harder; choose u so its derivative simplifies.

How to use it

First check whether an inner function and its derivative appear together. Otherwise try a product with a factor that simplifies under differentiation. Verify any proposed antiderivative by differentiating it. Repeated integration by parts, trigonometric substitution and partial fraction catalogs are optional extensions.

Check yourself

Why is ∫exp(x²)dx not handled by the same substitution as ∫2x exp(x²)dx?

Mini drill or application

Evaluate ∫ from 0 to 2 of x exp(−x)dx. By parts an antiderivative is −(x+1)exp(−x), so result is 1−3exp(−2). Differentiate it to verify all signs.

Read this only if stuck

OpenStax selective support. Use the reading guide for the section sequence and practice contract. Record your reasoning before consulting solutions.