Single Variable Calculus Questions and Answers

Start Your Free Trial

Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 34

Expert Answers info

eNotes eNotes educator | Certified Educator

calendarEducator since 2007

write13,548 answers

starTop subjects are Math, Literature, and Science

Evaluate $\displaystyle \int t^3 e^{-t^2} dt$ by making a substitution first, then by using Integration by parts.

If we use $ z = -t^2$, $t^2 = -z$, then $dz = -2t dt$

so,

$ \begin{equation} \begin{aligned} \int t^3 e^{-t^2} dt = \int t^2 \cdot t e^{-t^2} dt &= \int - z \cdot e^z \cdot \left( \frac{dz}{-z} \right)\\ \\ &= \int \frac{1}{2} ze^z dz\\ \\ &= \frac{1}{2} \int z e^z dz \end{aligned} \end{equation} $

By using integration by parts, if we let $u =z $ and $dv = e^z dz$, then

$du = dz$ and $v = e^z$

$ \begin{equation} \begin{aligned} \frac{1}{2} \int ze^z dz = uv - v\int du &= \frac{1}{2} \left[ ze^z - \int e^z dz \right]\\ \\ &= \frac{1}{2} \left[ ze^z - e^z \right]\\ \\ &= \frac{e^z}{z} [z -1] \end{aligned} \end{equation} $

but, $z = -t^2$

So, $\displaystyle \frac{e^z}{z} [z-1] = \frac{e^{-t^2}}{2} \left[ -t^2 -1 \right] + c$