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

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

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

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

\frac{e ^{2} - e}{3}

Decimal

1.55692 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} \frac{e ^{u}}{3} d u

= \int_{1}^{2} \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_{1}^{2} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

e^{2} - e

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

Simplify

= \frac{e ^{2} - e}{3}

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