If x^5e^-x^2,dx=g(x)e^-x^2+c, where c is a constant of integration, then g(-1) is equal to…

Question

If $\displaystyle\int x^5e^{-x^2}\,dx=g(x)e^{-x^2}+c$, where c is a constant of integration, then $g(-1)$ is equal to :

Accepted Answer

Let $I=\displaystyle\int x^5e^{-x^2}dx=\int x^4\cdot xe^{-x^2}dx$. Put $-x^2=t\Rightarrow-2x\,dx=dt$, $x^4=t^2$:
$$I=\int\frac{t^2}{-2}e^tdt=-\frac12e^t(t^2-2t+2)+c$$

So $g(x)=-\dfrac12\left(x^4+2x^2+2\right)$ (using $t=-x^2$, $t^2-2t+2=x^4+2x^2+2$).

$$g(-1)=-\frac12(1+2+2)=-\frac52$$

Solutions · 1