If int {x^5 e^{-x^2} dx} = g(x) e^{-x^2} + c, | vedclass

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(C) Let $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$.We can rewrite the integral as $\int x^4 \cdot e^{-x^2} \cdot x dx = \int t^

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$-\frac{5}{2}$

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(C) Let $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$.We can rewrite the integral as $\int x^4 \cdot e^{-x^2} \cdot x dx = \int t^2 \cdot e^{-t} \cdot \frac{1}{2} dt = \frac{1}{2} \int t^2 e^{-t} dt$.Using integration by parts $\int u dv = uv - \int v du$,let $u = t^2$ and $dv = e^{-t} dt$,so $du = 2t dt$ and $v = -e^{-t}$.$\frac{1}{2} \int t^2 e^{-t} dt = \frac{1}{2} [ -t^2 e^{-t} - \int (-e^{-t}) 2t dt ] = \frac{1}{2} [ -t^2 e^{-t} + 2 \int t e^{-t} dt ]$.Applying integration by parts again for $\int t e^{-t} dt$ with $u = t$ and $dv = e^{-t} dt$:$\int t e^{-t} dt = -t e^{-t} - \int (-e^{-t}) dt = -t e^{-t} - e^{-t}$.Substituting back: $\frac{1}{2} [ -t^2 e^{-t} + 2(-t e^{-t} - e^{-t}) ] = (-\frac{1}{2} t^2 - t - 1) e^{-t} + c$.Substituting $t = x^2$: $g(x) = -\frac{x^4}{2} - x^2 - 1$.Thus,$g(-1) = -\frac{(-1)^4}{2} - (-1)^2 - 1 = -\frac{1}{2} - 1 - 1 = -\frac{5}{2}$.

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

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