# Evaluate the definite integral. \int_{1}^{2}\frac{e^{\frac{1}{x^{4}}}}{x^{5}}dx

Evaluate the definite integral.
${\int }_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx$
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Anzante2m

Step 1
To evaluate the definite integral.
Step2
Given information:
${\int }_{1}^{2}\left(\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}\right)dx$
Step 3
Calculation:
Integrate the function with respect to x.
${\int }_{1}^{2}\left(\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}\right)dx$
Let $\frac{1}{{x}^{4}}=u$
so, differentiate with respect to x.
$\frac{d}{dx}\left(\frac{1}{{x}^{4}}\right)=\frac{d}{dx}\left(u\right)$
$\frac{-4}{{x}^{5}}=\frac{du}{dx}$
$\left(\frac{1}{{x}^{5}}\right)dx=\frac{1}{-4}du$
[put in the given integral]
Step 4
So, the integral become,
${\int }_{x=1}^{2}\left(\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}\right)dx={\int }_{x=1}^{2}\left(\frac{{e}^{u}}{-4}\right)du$
$\left[put\left(\frac{1}{{x}^{5}}\right)dx=\left(\frac{1}{{x}^{-4}}\right)du,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}u=\frac{1}{{x}^{4}}\right]$
$=-\frac{1}{4}{\left[{e}^{u}\right]}_{x=1}^{2}$
$\left[\because \int {e}^{x}dx={e}^{x}\right]$
$=-\frac{1}{4}{\left[{e}^{\frac{1}{{x}^{4}}}\right]}_{x=1}^{2}$

$=-\frac{1}{4}\left[{e}^{\frac{1}{{2}^{\left(4\right)}}}-{e}^{\frac{1}{{1}^{\left(4\right)}}}\right]$

###### Not exactly what you’re looking for?
limacarp4
Step 1
Let $u=\frac{1}{{x}^{4}}$ so that $du=-\frac{4}{{x}^{5}}dx$
By substitution,
$\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=\int {e}^{u}\left(-\frac{1}{4}du\right)$ (1)
$\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\frac{1}{4}\int {e}^{u}du$
Recall: $\int {e}^{u}du={e}^{u}+C$ so:
$\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\frac{1}{4}{e}^{u}+C$
$\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\frac{1}{4}{e}^{\frac{1}{{x}^{4}}}+C$
Hence,
${\int }_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx={\left[-\frac{1}{4}{e}^{\frac{1}{{x}^{4}}}\right]}_{1}^{2}$
$=-\frac{1}{4}{e}^{\frac{1}{{2}^{4}}}-\left(-\frac{1}{4}{e}^{\frac{1}{{1}^{4}}}\right)$
$=-\frac{1}{4}{e}^{\frac{1}{16}}+\frac{1}{4}e$
$=\frac{1}{4}\left(-{e}^{\frac{1}{16}}+e\right)$
$\approx 0.413$
###### Not exactly what you’re looking for?
nick1337

Step 1
We have to calculate
${\int }_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx$
Let us first calculate indefinite integral.
Given integral
Step 2
$\int \frac{{e}^{\frac{1}{{4}^{4}}}}{{x}^{5}}dx=-\frac{1}{4}\int {e}^{u}du$
Substitute $u=\frac{1}{{x}^{4}}$ this gives us
$dx=-\frac{{x}^{5}}{4}du$
Step 3
$-\frac{1}{4}\int {e}^{u}du=-\frac{{e}^{u}}{4}=-\frac{{e}^{\frac{1}{{x}^{4}}}}{4}$
Substitute back $u=\frac{1}{{x}^{4}}$
Step 4
$\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\frac{{e}^{\frac{1}{{4}^{5}}}}{4}+C$
${\int }_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\left[\frac{{e}^{\frac{1}{{4}^{5}}}}{4}{\right]}_{1}^{2}$
$=\frac{e}{4}-\frac{\sqrt[16]{e}}{4}$