What is the integral from 1 to 2 of (e^{(1/x)^3})/(x^4) ?

The integral from 1 to 2 of (e^{(1/x)^3})/(x^4) is (e-\sqrt[8]{e})/3

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integral from 1 to 2 of (e^{(1/x)^3})/(x^4)

\int_{1}^{2} \frac{e ^{(\frac{1}{x})^{3}}}{x ^{4}} d x

\frac{e - 8 e}{3}

Decimal

0.52837 \dots

Solution steps

\int_{1}^{2} \frac{e ^{(\frac{1}{x})^{3}}}{x ^{4}} d x

(\frac{1}{x})^{3} = \frac{1}{x ^{3}}

= \int_{1}^{2} \frac{e ^{\frac{1}{x ^{3}}}}{x ^{4}} d x

Apply u-substitution

: \int_{1}^{8} \frac{e ^{\frac{1}{u}}}{3 u ^{2}} d u

= \int_{1}^{8} \frac{e ^{\frac{1}{u}}}{3 u ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{1}^{8} \frac{e ^{\frac{1}{u}}}{u ^{2}} d u

Apply u-substitution

: \int_{1}^{\frac{1}{8}} - e^{v} d v

= \frac{1}{3} \cdot \int_{1}^{\frac{1}{8}} - e^{v} d v

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

= \frac{1}{3} (- \int_{\frac{1}{8}}^{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}{3} (- (- \int_{\frac{1}{8}}^{1} e^{v} d v))

Use the common integral:

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

= \frac{1}{3} (- (- [e^{v}]_{\frac{1}{8}}^{1}))

Simplify

= \frac{1}{3} [e^{v}]_{\frac{1}{8}}^{1}

Compute the boundaries:

e - 8 e

= \frac{1}{3} (e - 8 e)

Simplify

= \frac{e - 8 e}{3}

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What is the integral from 1 to 2 of (e^{(1/x)^3})/(x^4) ?
The integral from 1 to 2 of (e^{(1/x)^3})/(x^4) is (e-\sqrt[8]{e})/3