Calculus Notes

Section 4.5 Definite Integrals with U-Substitution

There are two different strategies to perform u-substitution:

• focus on u-substitution, temporarily ignore the limits of integration, then convert back to x

  Example 4.5.1.

  Find \(\displaystyle \int_0^2 x (x^2 + 1)^3 dx\text{:}\)

  1. Focus on u-substitution:

     \begin{align*} u \amp = x^2 + 1\\ du \amp = 2x dx \end{align*}

  2. Temporarily ignore limits of integration:

     \begin{align*} \int^*_* \underbrace{(x^2 + 1)^3}_{u^3} \cdot \underbrace{xdx}_{\frac{1}{2}du} \amp = \int^*_* u^3 \cdot \frac{1}{2} du \\ \amp = \frac{1}{2} \left( \dfrac{u^4}{4} \right) \Big|_*^* \end{align*}

  3. Limits reappear:

     \begin{align*} \frac{1}{8} \left( x^2 + 1 \right)^4 \Big|_0^2 \amp = \frac{1}{8} (2^2 + 1)^4 - \frac{1}{8}(0^2 + 1^4)\\ \amp = 78 \end{align*}

• or, change the limits after performing u-substitution.

  Example 4.5.2.

  Find \(\displaystyle \int_0^2 x (x^2 + 1)^3 dx\text{:}\)

  1. Change limits:

     \begin{align*} u \amp = x^2 + 1 \\ du \amp = 2xdx \end{align*}
     \begin{equation*} \begin{cases} x = 0, \amp u = 0^2 + 1 \amp = 1\\ x = 2, \amp u = 2^2 + 1 \amp = 5 \end{cases} \end{equation*}

  2. Rewrite integral:

     \begin{align*} \int_{x = 0}^{x = 2} (x^2 + 1)^3 x dx \amp = \int_{u = 5}^{u = 1} u^3 \cdot \frac{1}{2} du \\ \amp = \frac{1}{2} \left( \dfrac{u^4}{4} \right) \Big|_1^5\\ \amp = \frac{1}{8} (5^4) - \frac{1}{8} (1^4) \\ \amp = \frac{625}{8} - \frac{1}{8}\\ \amp = 78 \end{align*}