What is the integral from-2 to 0 of x^2e^{x^3+8} ?

The integral from-2 to 0 of x^2e^{x^3+8} is (e^8-1)/3

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integral from-2 to 0 of x^2e^{x^3+8}

\int_{- 2}^{0} x^{2} e^{x^{3} + 8} d x

\frac{e ^{8} - 1}{3}

Decimal

993.31932 \dots

Solution steps

\int_{- 2}^{0} x^{2} e^{x^{3} + 8} d x

Apply u-substitution

= \int_{0}^{8} \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}^{8} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{8}

Compute the boundaries:

e^{8} - 1

= \frac{1}{3} (e^{8} - 1)

Simplify

= \frac{e ^{8} - 1}{3}

What is the integral from-2 to 0 of x^2e^{x^3+8} ?
The integral from-2 to 0 of x^2e^{x^3+8} is (e^8-1)/3