What is the integral from 0 to 2 of x^5e^{x^3} ?

The integral from 0 to 2 of x^5e^{x^3} is (7e^8+1)/3

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integral from 0 to 2 of x^5e^{x^3}

\int_{0}^{2} x^{5} e^{x^{3}} d x

\frac{7 e ^{8} + 1}{3}

Decimal Notation

6955.90196 \dots

Solution steps

\int_{0}^{2} x^{5} e^{x^{3}} d x

Apply u-substitution

: \int_{0}^{32} \frac{e ^{u^{\frac{3}{5}}} u ^{\frac{1}{5}}}{5} d u

= \int_{0}^{32} \frac{e ^{u^{\frac{3}{5}}} u ^{\frac{1}{5}}}{5} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int_{0}^{32} e^{u^{\frac{3}{5}}} u^{\frac{1}{5}} d u

Apply u-substitution

: \int_{0}^{8} \frac{5 e ^{v} v}{3} d v

= \frac{1}{5} \cdot \int_{0}^{8} \frac{5 e ^{v} v}{3} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \frac{5}{3} \cdot \int_{0}^{8} e^{v} v d v

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{0}^{8}

= \frac{1}{5} \cdot \frac{5}{3} [e^{v} v - \int e^{v} d v]_{0}^{8}

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

= \frac{1}{5} \cdot \frac{5}{3} [e^{v} v - e^{v}]_{0}^{8}

Simplify

\frac{1}{5} \cdot \frac{5}{3} [e^{v} v - e^{v}]_{0}^{8} : \frac{1}{3} [e^{v} v - e^{v}]_{0}^{8}

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

Compute the boundaries:

7 e^{8} + 1

= \frac{1}{3} (7 e^{8} + 1)

Simplify

= \frac{7 e ^{8} + 1}{3}

4 x^{2} - 3, - x^{2} + 9

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 4 x + 4 y)

f (x) = ln (x), at x = 2

\frac{d}{d x} (x^{2} + 2 x + 3)

\frac{d}{d x} (x^{\frac{3}{2}} - 4)

What is the integral from 0 to 2 of x^5e^{x^3} ?
The integral from 0 to 2 of x^5e^{x^3} is (7e^8+1)/3