Question

express each definite integral in terms of u but don't evaluate

Answer 1

Recall that the change of variable theorem states that:

$\begin{gathered} \text{If }\varphi\colon\lbrack a,b\rbrack\rightarrow I\text{ is a differentiable function},\text{ with a continuous derivative, and }I \\ \text{is a interval, then:} \\ \int ^b_af(\varphi(x))\varphi^{\prime}(x)dx=\int ^{\varphi(b)}_{\varphi(a)}f(u)du\text{.} \\ \end{gathered}$

Now, notice that:

$\begin{gathered} u^{\prime}(x_{})=2x, \\ u\colon\lbrack-1,2\rbrack\rightarrow\lbrack0,3\rbrack. \end{gathered}$

Therefore, if we set f(x)=x², and use the change of variable theorem we get:

$\begin{gathered} \int ^2_{-1}6x(x^2-1)^2dx=3\int ^2_{-1}f(x^2-1)^{}\cdot2xdx \\ =\int ^{u(2)}_{u(-1)}f(u)^{}du=\int ^3_0u^2du\text{.} \end{gathered}$

Answer:

$\int ^3_0u^2du\text{.}$