Evaluate the integral:

${\int}_{0}^{1}({x}^{2}+1){e}^{-x}dx=$

${\int}_{0}^{1}({x}^{2}+1){e}^{-x}dx=$

Alex Baird
2022-07-28
Answered

Evaluate the integral:

${\int}_{0}^{1}({x}^{2}+1){e}^{-x}dx=$

${\int}_{0}^{1}({x}^{2}+1){e}^{-x}dx=$

You can still ask an expert for help

Ali Harper

Answered 2022-07-29
Author has **16** answers

${\int}_{0}^{1}({x}^{2}+1){e}^{-x}dx={\int}_{0}^{1}{x}^{2}{e}^{-x}dx+{\int}_{0}^{1}{e}^{-x}dx$

${\int}_{0}^{1}{e}^{-x}dx=(-{e}^{-x}{)}_{0}^{1}=-{e}^{-1}+1$

${\int}_{0}^{1}{x}^{2}{e}^{-x}dx=[u={x}^{2}\to {u}^{\prime}=2x,{v}^{\prime}={e}^{-x}\to v=-{e}^{-x}]=uv-\int uv=-{x}^{2}{e}^{-x}+\int 2x{e}^{-x}dx$

$\int 2x{e}^{-x}dx=[u=2x\to u=2,{v}^{\prime}={e}^{-x}\to v=-{e}^{-x}]=-2x{e}^{-x}+\int 2{e}^{-x}dx$

$\to {\int}_{0}^{1}{x}^{2}{e}^{-x}dx=(-{x}^{2}{e}^{-x}{)}_{0}^{1}-(2x{e}^{-x}{)}_{0}^{1}+{\int}_{0}^{1}2{e}^{-x}dx=-{e}^{1}-2{e}^{-1}+2(-{e}^{-1}+1)=-5{e}^{-1}+1$

${\int}_{0}^{1}({x}^{2}+1){e}^{-x}dx={\int}_{0}^{1}{x}^{2}{e}^{-x}dx+{\int}_{0}^{1}{e}^{-x}dx=-5{e}^{-1}+1-{e}^{-1}+1=-6{e}^{-1}+2$

${\int}_{0}^{1}{e}^{-x}dx=(-{e}^{-x}{)}_{0}^{1}=-{e}^{-1}+1$

${\int}_{0}^{1}{x}^{2}{e}^{-x}dx=[u={x}^{2}\to {u}^{\prime}=2x,{v}^{\prime}={e}^{-x}\to v=-{e}^{-x}]=uv-\int uv=-{x}^{2}{e}^{-x}+\int 2x{e}^{-x}dx$

$\int 2x{e}^{-x}dx=[u=2x\to u=2,{v}^{\prime}={e}^{-x}\to v=-{e}^{-x}]=-2x{e}^{-x}+\int 2{e}^{-x}dx$

$\to {\int}_{0}^{1}{x}^{2}{e}^{-x}dx=(-{x}^{2}{e}^{-x}{)}_{0}^{1}-(2x{e}^{-x}{)}_{0}^{1}+{\int}_{0}^{1}2{e}^{-x}dx=-{e}^{1}-2{e}^{-1}+2(-{e}^{-1}+1)=-5{e}^{-1}+1$

${\int}_{0}^{1}({x}^{2}+1){e}^{-x}dx={\int}_{0}^{1}{x}^{2}{e}^{-x}dx+{\int}_{0}^{1}{e}^{-x}dx=-5{e}^{-1}+1-{e}^{-1}+1=-6{e}^{-1}+2$

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