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

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

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integral from 0 to 2 of (x^2e^{x^3})

\int_{0}^{2} (x^{2} e^{x^{3}}) d x

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

Decimal Notation

993.31932 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{8} \frac{e ^{u}}{3} d u

= \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}

\int_{a}^{b} e^{t x} d x

\int_{0}^{2} \frac{4 x}{2 + x ^{2}} d x

\int_{0}^{2} 4 x + 1 d x

\int_{1}^{3} - x^{3} + 4 x^{2} - 3 x d x

\int_{2}^{4} (x^{2} - 2 x) d x

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