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

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

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

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

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

Decimal Notation

993.31932 \dots

Solution steps

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

\frac{d}{d x} (\frac{3 x}{4})

\frac{7}{3 x ^{2}}

\frac{\partial}{\partial x} (sin^{3} (2 x))

h (t) = (7 t^{2} + t)^{- 3}

y^{^{'}} = 2 y + x^{2} + 5

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