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

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

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integral from 1 to 2 of (xe^{x/y})/(y^2)

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

- e^{\frac{x}{2}} + e^{x}

Solution steps

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

Take the constant out:

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

= x \cdot \int_{1}^{2} \frac{e ^{\frac{x}{y}}}{y ^{2}} d y

Apply u-substitution

: \int_{x}^{\frac{x}{2}} - \frac{e ^{u}}{x} d u

= x \cdot \int_{x}^{\frac{x}{2}} - \frac{e ^{u}}{x} d u

Take the constant out:

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

= x (- \frac{1}{x} \cdot \int_{x}^{\frac{x}{2}} e^{u} d u)

Use the common integral:

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

= x (- \frac{1}{x} [e^{u}]_{x}^{\frac{x}{2}})

Simplify

x (- \frac{1}{x} [e^{u}]_{x}^{\frac{x}{2}}) : - [e^{u}]_{x}^{\frac{x}{2}}

= - [e^{u}]_{x}^{\frac{x}{2}}

Compute the boundaries:

e^{\frac{x}{2}} - e^{x}

= - (e^{\frac{x}{2}} - e^{x})

Simplify

= - e^{\frac{x}{2}} + e^{x}

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