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

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

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

\int_{0}^{1} e^{x^{3}} \cdot 2 x^{2} d x

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

Decimal

1.14552 \dots

Solution steps

\int_{0}^{1} e^{x^{3}} \cdot 2 x^{2} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} e^{x^{3}} x^{2} d x

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \frac{1}{3} \cdot \int_{0}^{1} e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{3} [e^{u}]_{0}^{1}

Simplify

2 \cdot \frac{1}{3} [e^{u}]_{0}^{1} : \frac{2}{3} [e^{u}]_{0}^{1}

= \frac{2}{3} [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

= \frac{2}{3} (e - 1)

Simplify

= \frac{2 ( e - 1 )}{3}

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