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

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

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

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

e - 2

Solution steps

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

Take the constant out:

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

= 3 \cdot \int_{0}^{1} y^{2} (e^{y^{3}} - 1) d y

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot \frac{1}{3} (\int_{0}^{1} e^{u} d u - \int_{0}^{1} 1 d u)

\int_{0}^{1} e^{u} d u = e - 1

\int_{0}^{1} 1 d u = 1

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

Simplify

3 \cdot \frac{1}{3} (e - 1 - 1) : e - 2

= e - 2

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