What is the integral from 1 to 2 of (e^{1/(x^5)})/(x^6) ?

The integral from 1 to 2 of (e^{1/(x^5)})/(x^6) is (e-\sqrt[32]{e})/5

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integral from 1 to 2 of (e^{1/(x^5)})/(x^6)

\int_{1}^{2} \frac{e ^{\frac{1}{x ^{5}}}}{x ^{6}} d x

\frac{e - 32 e}{5}

Decimal

0.33730 \dots

Solution steps

\int_{1}^{2} \frac{e ^{\frac{1}{x ^{5}}}}{x ^{6}} d x

Apply u-substitution

: \int_{1}^{32} \frac{e ^{\frac{1}{u}}}{5 u ^{2}} d u

= \int_{1}^{32} \frac{e ^{\frac{1}{u}}}{5 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int_{1}^{32} \frac{e ^{\frac{1}{u}}}{u ^{2}} d u

Apply u-substitution

: \int_{1}^{\frac{1}{32}} - e^{v} d v

= \frac{1}{5} \cdot \int_{1}^{\frac{1}{32}} - e^{v} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{5} (- \int_{\frac{1}{32}}^{1} - e^{v} d v)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} (- (- \int_{\frac{1}{32}}^{1} e^{v} d v))

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{5} (- (- [e^{v}]_{\frac{1}{32}}^{1}))

Simplify

= \frac{1}{5} [e^{v}]_{\frac{1}{32}}^{1}

Compute the boundaries:

e - 32 e

= \frac{1}{5} (e - 32 e)

Simplify

= \frac{e - 32 e}{5}

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What is the integral from 1 to 2 of (e^{1/(x^5)})/(x^6) ?
The integral from 1 to 2 of (e^{1/(x^5)})/(x^6) is (e-\sqrt[32]{e})/5