assignment submission\nfor this assignment, you submit answers by question parts. the number \nassignment…

Explanation:

Step1: Use integration - by - parts formula $\int_{a}^{b}u\mathrm{d}v=uv|{a}^{b}-\int{a}^{b}v\mathrm{d}u$

Let $u = x^{2}+4$, $\mathrm{d}v=e^{-x}\mathrm{d}x$. Then $\mathrm{d}u = 2x\mathrm{d}x$, $v=-e^{-x}$. [ \begin{align*} \int_{0}^{1}(x^{2}+4)e^{-x}\mathrm{d}x&=-(x^{2}+4)e^{-x}\big|{0}^{1}+\int{0}^{1}2xe^{-x}\mathrm{d}x\ &=-(1 + 4)e^{-1}+(0 + 4)e^{0}+2\int_{0}^{1}xe^{-x}\mathrm{d}x\ &=- \frac{5}{e}+4+2\int_{0}^{1}xe^{-x}\mathrm{d}x \end{align*} ]

Step2: Apply integration - by - parts again on $\int_{0}^{1}xe^{-x}\mathrm{d}x$

Let $u = x$, $\mathrm{d}v=e^{-x}\mathrm{d}x$. Then $\mathrm{d}u=\mathrm{d}x$, $v=-e^{-x}$. [ \begin{align*} \int_{0}^{1}xe^{-x}\mathrm{d}x&=-xe^{-x}\big|{0}^{1}+\int{0}^{1}e^{-x}\mathrm{d}x\ &=-e^{-1}+0-\left(e^{-x}\big|_{0}^{1}\right)\ &=-\frac{1}{e}-(e^{-1}-e^{0})\ &=-\frac{1}{e}-\frac{1}{e}+1\ &=1-\frac{2}{e} \end{align*} ]

Step3: Substitute the result of $\int_{0}^{1}xe^{-x}\mathrm{d}x$ back

[ \begin{align*} \int_{0}^{1}(x^{2}+4)e^{-x}\mathrm{d}x&=-\frac{5}{e}+4+2\left(1-\frac{2}{e}\right)\ &=-\frac{5}{e}+4 + 2-\frac{4}{e}\ &=6-\frac{9}{e} \end{align*} ]

Answer:

$6-\frac{9}{e}$