What is the integral from 0 to 3 of x^7e^{x^4} ?

The integral from 0 to 3 of x^7e^{x^4} is (80e^{81}+1)/4

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integral from 0 to 3 of x^7e^{x^4}

\int_{0}^{3} x^{7} e^{x^{4}} d x

\frac{80 e ^{81} + 1}{4}

Decimal

3.01219 E 36

Solution steps

\int_{0}^{3} x^{7} e^{x^{4}} d x

Apply u-substitution

: \int_{0}^{2187} \frac{e ^{u^{\frac{4}{7}}} u ^{\frac{1}{7}}}{7} d u

= \int_{0}^{2187} \frac{e ^{u^{\frac{4}{7}}} u ^{\frac{1}{7}}}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int_{0}^{2187} e^{u^{\frac{4}{7}}} u^{\frac{1}{7}} d u

Apply u-substitution

: \int_{0}^{81} \frac{7 e ^{v} v}{4} d v

= \frac{1}{7} \cdot \int_{0}^{81} \frac{7 e ^{v} v}{4} d v

Take the constant out:

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

= \frac{1}{7} \cdot \frac{7}{4} \cdot \int_{0}^{81} e^{v} v d v

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{0}^{81}

= \frac{1}{7} \cdot \frac{7}{4} [e^{v} v - \int e^{v} d v]_{0}^{81}

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

= \frac{1}{7} \cdot \frac{7}{4} [e^{v} v - e^{v}]_{0}^{81}

Simplify

\frac{1}{7} \cdot \frac{7}{4} [e^{v} v - e^{v}]_{0}^{81} : \frac{1}{4} [e^{v} v - e^{v}]_{0}^{81}

= \frac{1}{4} [e^{v} v - e^{v}]_{0}^{81}

Compute the boundaries:

80 e^{81} + 1

= \frac{1}{4} (80 e^{81} + 1)

Simplify

= \frac{80 e ^{81} + 1}{4}

\int_{4}^{9} \frac{x}{x - 36} d x

f^{^{'}} (x) = x^{2} + 1

\int x^{- 5} (x + 1) d x

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

x e^{x - 1}

What is the integral from 0 to 3 of x^7e^{x^4} ?
The integral from 0 to 3 of x^7e^{x^4} is (80e^{81}+1)/4