What is the integral from 0 to x^3 of e^{y/x} ?

The integral from 0 to x^3 of e^{y/x} is x(e^{x^2}-1)

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integral from 0 to x^3 of e^{y/x}

\int_{0}^{x^{3}} e^{\frac{y}{x}} d y

x (e^{x^{2}} - 1)

Solution steps

\int_{0}^{x^{3}} e^{\frac{y}{x}} d y

Apply u-substitution

: \int_{0}^{x^{2}} e^{u} x d u

= \int_{0}^{x^{2}} e^{u} x d u

Take the constant out:

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

= x \cdot \int_{0}^{x^{2}} e^{u} d u

Use the common integral:

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

= x [e^{u}]_{0}^{x^{2}}

Compute the boundaries:

e^{x^{2}} - 1

= x (e^{x^{2}} - 1)

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What is the integral from 0 to x^3 of e^{y/x} ?
The integral from 0 to x^3 of e^{y/x} is x(e^{x^2}-1)