Recipe

Integrate a definite integral by u-substitution

Use when: a definite integral \int_a^b f(g(x))g'(x)\,dx needs u-substitution, and the bounds a, b are given in terms of x

Goal: a single number, reached either by converting the bounds to u directly or by back-substituting to x before evaluating

Taught in U-substitution with definite integrals · extends Integrate by u-substitution

1. 1

   Same as the indefinite case: let u be the inner function, and compute du.

   Example

   For \int_0^2 2x(x^2+1)\,dx: let u=x^2+1, so du=2x\,dx.

   Why: nothing about having bounds changes how u and du are found. That part of the procedure is unchanged.

2. 2

   Pick one of two equally valid paths before continuing.

   • converting the bounds — find the u-value matching each x-bound, then never return to x at all.
   • back-substituting — treat this as an indefinite integral all the way through, substitute u back to x at the very end, and evaluate the ORIGINAL x-bounds on that result.

   Why: both paths are correct. They differ only in bookkeeping, not in the answer they produce.

3. 3

   Replace the lower and upper bounds with their corresponding u-values.

   Example

   x=0 gives u=0^2+1=1; x=2 gives u=2^2+1=5. The integral becomes \int_1^5 u\,du.

   Why: swapping the variable of integration means the bounds have to swap along with it. Leaving the old x-bounds on a u-integral would evaluate the wrong interval entirely.

4. 4

   Find the antiderivative in u, and plug the converted bounds straight in; no back-substitution is needed on this path.

   Example

   \int_1^5 u\,du=\left[\dfrac{u^2}{2}\right]_1^5=\dfrac{25}{2}-\dfrac12=12.

   Why: once the bounds are already in terms of u, the antiderivative never needs to return to x at all — that's the entire advantage of converting them.