Julia White

2021-12-28

Evaluate the integral.
${\int }_{0}^{1}\left({x}^{2}+1\right){e}^{-x}dx$

xandir307dc

Step 1: Given that:
Evaluate the integral.
${\int }_{0}^{1}\left({x}^{2}+1\right){e}^{-x}dx$
Step 2: Formula Used:
Integration by Parts:
$\int f\left(x\right)\cdot g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int \left[\frac{d}{dx}\left(f\left(x\right)\right)\int g\left(x\right)dx\right]dx$
Where f(x) is the first function and g(x) is second function.
Step 3: Evaluate:
To find the given definite integral first we find the value of the indefinite integral
$\int \left({x}^{2}+1\right){e}^{-x}dx=\left({x}^{2}+1\right)\int {e}^{-x}dx-\int \left[\frac{d}{dx}\left({x}^{2}+1\right)\int {e}^{-x}dx\right]dx$
$=-\left({x}^{2}+1\right){e}^{-x}-\int 2x\left(-{e}^{-x}\right)dx$
$=-\left({x}^{2}+1\right){e}^{-x}+\int 2x{e}^{-x}dx$
$=-\left({x}^{2}+1\right){e}^{-x}+2\left(x\int {e}^{-x}dx-\int \left(\frac{d}{dx}\left(x\right)\int {e}^{-x}dx\right)dx\right)$
$=-\left({x}^{2}+1\right){e}^{-x}+2\left(-x{e}^{-x}-{e}^{-x}\right)$
$=-\left({x}^{2}+1\right){e}^{-x}-2x{e}^{-x}-2{e}^{-x}+C$
$=-\left({x}^{2}+2x+3\right){e}^{-x}+C$
Now, Finding the definite integral for the given integral we have,
${\int }_{0}^{1}\left({x}^{2}+1\right){e}^{-x}dx={\left[-\left({x}^{2}+2x+3\right){e}^{-x}\right]}_{0}^{1}$

Paineow

$\int \left({x}^{2}+1\right)\cdot {e}^{-x}dx$
Integration by parts formula:
$\int U\left(x\right)\cdot dV\left(x\right)=U\left(x\right)\cdot V\left(x\right)-\int V\left(x\right)\cdot dU\left(x\right)$
Put
$U={x}^{2}+1$
$dV={e}^{-x}dx$
Then:
dU=2x dx
$V=-{e}^{-x}$
Therefore
$\int \left({x}^{2}+1\right)\cdot {e}^{-x}\cdot dx=-\left({x}^{2}+1\right)\cdot {e}^{-x}-\int -2\cdot x\cdot {e}^{-x}\cdot dx=-\left({x}^{2}+1\right)\cdot {e}^{-x}+\int 2\cdot x\cdot {e}^{-x}\cdot dx$
Find the integral
$\int 2\cdot x\cdot {e}^{-x}dx$
U=x
dU=dx
$dV=2{e}^{-x}dx$
$V=-2{e}^{-x}$
$\int 2\cdot x\cdot {e}^{-x}dx=-2\cdot x\cdot {e}^{-x}-\int -2\cdot {e}^{-x}dx=-2\cdot x\cdot {e}^{-x}+\int 2\cdot {e}^{-x}dx$
Find the integral
$\int 2\cdot {e}^{-x}dx=-2\cdot {e}^{-x}$
$\int \left({x}^{2}+1\right)\cdot {e}^{-x}=-2\cdot x\cdot {e}^{-x}+\left(-{x}^{2}-1\right)\cdot {e}^{-x}-2\cdot {e}^{-x}+C$
or
$\int \left({x}^{2}+1\right)\cdot {e}^{-x}=-\left({x}^{2}+2\cdot x+3\right)\cdot {e}^{-x}+C$
Calculate the definite integral:
${\int }_{0}^{1}\left({x}^{2}+1\right)\cdot {e}^{-x}dx=\left(-\left({x}^{2}+2\cdot x+3\right)\cdot {e}^{-x}\right){\mid }_{0}^{1}$

karton

$\begin{array}{}{\int }_{0}^{1}\left({x}^{2}+1\right){e}^{-x}dx\\ \int \left({x}^{2}+1\right){e}^{-x}dx\\ \int \left({x}^{2}+1\right)×\frac{1}{{e}^{x}}dx\\ \int \frac{{x}^{2}+1}{{e}^{x}}dx\\ \int \frac{{x}^{2}}{{e}^{x}}+\frac{1}{{e}^{x}}dx\\ \int \frac{{x}^{2}}{{e}^{x}}dx+\int \frac{1}{{e}^{x}}dx\\ -{x}^{2}{e}^{-x}-2x{e}^{-x}-2{e}^{-x}-\frac{1}{{e}^{x}}\\ \left(-{x}^{2}{e}^{-x}-2x{e}^{-x}-2{e}^{-x}-\frac{1}{{e}^{x}}\right){|}_{0}^{1}\\ -{1}^{2}{e}^{-1}-2×1{e}^{-1}-2{e}^{-1}-\frac{1}{{e}^{1}}-\left(-{0}^{2}{e}^{-0}-2×0{e}^{-0}-2{e}^{-0}-\frac{1}{{e}^{0}}\right)\\ -\frac{6}{e}+3\end{array}$