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

The integral from 1 to 4 of (e^{4/x})/(x^2) is (e^4-e)/4

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

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

\frac{e ^{4} - e}{4}

Decimal Notation

12.96996 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{1} - \frac{e ^{u}}{4} d u

= \int_{4}^{1} - \frac{e ^{u}}{4} d u

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

= - \int_{1}^{4} - \frac{e ^{u}}{4} d u

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{1}^{4} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{4} [e^{u}]_{1}^{4})

Simplify

= \frac{1}{4} [e^{u}]_{1}^{4}

Compute the boundaries:

e^{4} - e

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

Simplify

= \frac{e ^{4} - e}{4}

\frac{d}{d x} (a e^{2 x} + b e^{x})

\frac{\partial}{\partial z} (e^{x} sin (yz))

\int 10 sin^{3} (x) cos^{7} (x) d x

\frac{\partial}{\partial x} (y^{2} - e^{x y})

\frac{s}{s ^{4} + 1}

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