Calculus / Unit 8: The Fundamental Theorem of Calculus
8.5 u-Substitution (Definite Integrals)
Notes
For a definite integral, either change the limits to $u$-values, or find the antiderivative in $x$ first.
- Changing limits: new limits are $u(a)$ and $u(b)$. Then you never go back to $x$.
- Example: $\displaystyle \int_0^1 2x(x^2 + 1)^3\,dx$ with $u = x^2 + 1$: limits go from $u = 1$ to $u = 2$. So $\displaystyle \int_1^2 u^3\,du = \left[\frac{u^4}{4}\right]_1^2 = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$.
- Mixing $x$-limits with a $u$ antiderivative is a common mistake.
Practice
Find $\displaystyle \int_0^1 2x(x^2 + 1)^3\,dx$.
With $u = x^2 + 1$ the limits become 1 to 2: $\displaystyle \frac{16}{4} - \frac{1}{4}$.
Find $\displaystyle \int_0^{\pi/2} \sin x \cos x\,dx$.
With $u = \sin x$ the limits go from 0 to 1, so $\displaystyle \int_0^1 u\,du = \frac{1}{2}$.
Find $\displaystyle \int_0^2 \frac{x}{x^2 + 1}\,dx$.
With $u = x^2 + 1$, limits are 1 to 5 and $\displaystyle x\,dx = \frac{du}{2}$. So $\displaystyle \frac{1}{2}[\ln u]_1^5 = \frac{1}{2}\ln 5$.
Ready for more? Take the Unit 8 test.