Priscilla Johnston

Answered

2021-12-21

Evaluate the definite integral.

${\int}_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx$

Answer & Explanation

Anzante2m

Expert

2021-12-22Added 34 answers

Step 1

To evaluate the definite integral.

Step2

Given information:

Step 3

Calculation:

Integrate the function with respect to x.

Let

so, differentiate with respect to x.

[put in the given integral]

Step 4

So, the integral become,

limacarp4

Expert

2021-12-23Added 39 answers

Step 1

Let$u=\frac{1}{{x}^{4}}$ so that $du=-\frac{4}{{x}^{5}}dx$

By substitution,

$\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=\int {e}^{u}(-\frac{1}{4}du)$ (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,

$\int}_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx={[-\frac{1}{4}{e}^{\frac{1}{{x}^{4}}}]}_{1}^{2$

$=-\frac{1}{4}{e}^{\frac{1}{{2}^{4}}}-(-\frac{1}{4}{e}^{\frac{1}{{1}^{4}}})$

$=-\frac{1}{4}{e}^{\frac{1}{16}}+\frac{1}{4}e$

$=\frac{1}{4}(-{e}^{\frac{1}{16}}+e)$

$\approx 0.413$

Let

By substitution,

Recall:

Hence,

nick1337

Expert

2021-12-28Added 573 answers

Step 1

We have to calculate

Let us first calculate indefinite integral.

Given integral

Step 2

Substitute

Step 3

Substitute back

Step 4

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