Integration by Parts — Question 3

Question 3 – Solution

Recall the integration-by-parts formula $$\int u\,dv=uv-\int v\,du.$$ With the student’s choice $u=e^x$ and $dv=x^2\,dx$, we have $$du=e^x\,dx, \qquad v=\frac{x^3}{3}.$$ Therefore, $$\begin{align*} \int x^2e^x\,dx &=e^x\left(\frac{x^3}{3}\right) -\int \frac{x^3}{3}e^x\,dx \\ &=\frac{x^3e^x}{3}-\frac13\int x^3e^x\,dx. \end{align*}$$ The polynomial degree has increased from $2$ to $3$, so the remaining integral is more complicated. Repeating this choice would keep increasing the degree.

Instead, choose $$u=x^2,\quad dv=e^x\,dx, \qquad du=2x\,dx,\quad v=e^x.$$ Applying integration by parts gives $$\begin{align*} \int x^2e^x\,dx &=x^2e^x-\int 2xe^x\,dx \\ &=x^2e^x-2\int xe^x\,dx. \end{align*}$$

For the remaining integral, use integration by parts again: $$u=x,\quad dv=e^x\,dx, \qquad du=dx,\quad v=e^x.$$ Thus, $$\begin{align*} \int xe^x\,dx &=xe^x-\int e^x\,dx \\ &=xe^x-e^x. \end{align*}$$

Substituting this result into the original calculation gives $$\begin{align*} \int x^2e^x\,dx &=x^2e^x-2\left(xe^x-e^x\right)+C \\ &=x^2e^x-2xe^x+2e^x+C \\ &=e^x\left(x^2-2x+2\right)+C. \end{align*}$$ $$\boxed{\displaystyle \int x^2e^x\,dx =e^x\left(x^2-2x+2\right)+C}$$

Check by differentiating: $$\begin{align*} \frac{d}{dx}\left[e^x\left(x^2-2x+2\right)\right] &=e^x\left(x^2-2x+2\right)+e^x(2x-2) \\ &=x^2e^x, \end{align*}$$ which is the original integrand.