What is the integral from 1 to 2 of w^5e^{w^3} ?

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

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integral from 1 to 2 of w^5e^{w^3}

\int_{1}^{2} w^{5} e^{w^{3}} d w

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

Decimal Notation

6955.56863 \dots

Solution steps

\int_{1}^{2} w^{5} e^{w^{3}} d w

Apply u-substitution

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

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

Apply u-substitution

: \int_{1}^{8} \frac{5 e ^{v} v}{3} d v

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

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{1}^{8}

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

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

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

Simplify

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

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

Compute the boundaries:

7 e^{8}

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

Simplify

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

\frac{d y}{d x} - e^{2 x} = 3, y (0) = 4

x \to - 2 lim (\frac{x ^{2}}{x ^{2} + 2})

y = \frac{1}{x ^{6}}, (- 3, \frac{1}{729})

\int x e^{- 9 x^{2}} d x

y^{^{'}} + e^{x + y} = 0

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