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

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

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integral from 0 to y^2 of e^{x/(y^2)}

\int_{0}^{y^{2}} e^{\frac{x}{y ^{2}}} d x

y^{2} (e - 1)

Solution steps

\int_{0}^{y^{2}} e^{\frac{x}{y ^{2}}} d x

Apply u-substitution

: \int_{0}^{1} e^{u} y^{2} d u

= \int_{0}^{1} e^{u} y^{2} d u

Take the constant out:

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

= y^{2} \cdot \int_{0}^{1} e^{u} d u

Use the common integral:

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

= y^{2} [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

= y^{2} (e - 1)

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\int_{- π}^{π} x^{2} \cdot cos (n x) d x

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