petrusrexcs

2021-12-31

Evaluate the definite integral.
${\int }_{2}^{7}{e}^{2-x}dx$

Juan Spiller

Step 1
The given integral is as follows.
${\int }_{2}^{7}{e}^{2-x}dx$
Let u=2-x. Then, du=-dx.
The Lower limit will be u=2-2=0 and the upper limit will be u=2-7=-5.
Step 2
Apply the substitutions and evaluate the integral as follows.
${\int }_{2}^{7}{e}^{2-x}dx={\int }_{0}^{-5}{e}^{u}\left(-du\right)$
$={\int }_{0}^{-5}-{e}^{u}du$
$=-{\left[{e}^{u}\right]}_{0}^{-5}$
$=-\left[{e}^{-5}-{e}^{0}\right]$
$=1-{e}^{-5}$
$=1-\frac{1}{{e}^{5}}$
Therefore, the value of the given integral is,
$1-\frac{1}{{e}^{5}}$

yotaniwc

${\int }_{2}^{7}{e}^{2-x}dx$
Evaluate the indefinite integral
$\int {e}^{2-x}dx$
$\int -{e}^{t}dt$
$-\int {e}^{t}dt$
$-{e}^{t}$
$-{e}^{2-x}$
Return the limits
$-{e}^{2-x}{\mid }_{2}^{7}$
$-{e}^{2-7}-\left(-{e}^{2-2}\right)$
Simplify
$-\frac{1}{{e}^{5}}+1$

Vasquez

$\int {e}^{2-x}dx$
This is a tabular integral:
$\int {e}^{2-x}dx=-{e}^{2-x}+C$
Let's calculate a definite integral:
${\int }_{2}^{7}{e}^{2-x}dx=\left(-{e}^{2-x}\right){|}_{2}^{7}$
$F\left(7\right)=-{e}^{-5}$
$F\left(2\right)=-1$
$I=-{e}^{-5}-\left(-1\right)=1-{e}^{-5}$

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