Question

Evaluate the integral\(\int {{t^3}{e^{ - {t^2}}}dt} \)`

Short answer

We will use the substitute method and integral by parts method to solve the integral.

Step-by-step solution

Given Data

Given =\({\rm I} = \int {{t^3}{e^{ - {t^2}}}dt} \)

Let \(x = {t^2}\)

\(\begin{aligned}{l}\frac{{dx}}{{dt}} = 2tdt\\dt = \frac{{dx}}{{2t}}\end{aligned}\)

Substitute

\(\)\(\begin{aligned}{l}{\rm I} = \int {{t^3}{e^{ - {t^2}}}dt} \\{\rm I} = \int {x{e^{ - x}}\frac{{dx}}{2}} \\{\rm I} = \frac{1}{2}\int {x{e^{ - x}}dx} \end{aligned}\)

Using Integration by Part Method

\({\rm I} = \mu v - \int {vd\mu } \)

Let,\(\begin{aligned}{l}\mu = x\\d\mu = dx\end{aligned}\) \(\begin{aligned}{l}dv = {e^{ - x}}\\v = - {e^{ - x}}\end{aligned}\)

\(\begin{aligned}{l}{\rm I} = \frac{1}{2}\left( { - x{e^{ - x}} + \int {{e^{ - x}}dx} } \right)\\{\rm I} = \frac{1}{2}\left( { - x{e^{ - x}} - \frac{1}{2}{e^{ - x}} + {c_1}} \right)\\{\rm I} = \frac{1}{2}.x{e^{ - x}} - \frac{1}{2}{e^{ - x}} + \frac{{{c_1}}}{2}\\{\rm I} = - \frac{1}{2}{t^2}{e^{ - {t^2}}} - \frac{1}{2}{e^{ - {t^2}}} + c\\\therefore {\rm I} = - \frac{1}{2}{e^{ - {t^2}}}\left( {{t^2} + 1} \right) + c\end{aligned}\)

Hence, \(\int {{t^3}{e^{ - {t^2}}}dt} = - \frac{1}{2}{e^{ - {t^2}}}\left( {{t^2} + 1} \right) + c\)