What is the integral from 0 to 1 of e^{y^3}y^2 ?

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

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integral from 0 to 1 of e^{y^3}y^2

\int_{0}^{1} e^{y^{3}} y^{2} d y

\frac{e - 1}{3}

Decimal Notation

0.57276 \dots

Solution steps

\int_{0}^{1} e^{y^{3}} y^{2} d y

Apply u-substitution

: \int_{0}^{1} \frac{e ^{u}}{3} d u

= \int_{0}^{1} \frac{e ^{u}}{3} 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_{0}^{1} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

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

Simplify

= \frac{e - 1}{3}

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