What is the integral from 0 to 4 of x^2e^{x^3} ?

The integral from 0 to 4 of x^2e^{x^3} is (e^{64}-1)/3

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integral from 0 to 4 of x^2e^{x^3}

\int_{0}^{4} x^{2} e^{x^{3}} d x

\frac{e ^{64} - 1}{3}

Decimal

2.07838 E 27

Solution steps

\int_{0}^{4} x^{2} e^{x^{3}} d x

Apply u-substitution

: \int_{0}^{64} \frac{e ^{u}}{3} d u

= \int_{0}^{64} \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}^{64} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{64}

Compute the boundaries:

e^{64} - 1

= \frac{1}{3} (e^{64} - 1)

Simplify

= \frac{e ^{64} - 1}{3}

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What is the integral from 0 to 4 of x^2e^{x^3} ?
The integral from 0 to 4 of x^2e^{x^3} is (e^{64}-1)/3