What is the integral from 0 to 1 of x^9e^{x^5} ?

The integral from 0 to 1 of x^9e^{x^5} is 1/5

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integral from 0 to 1 of x^9e^{x^5}

\int_{0}^{1} x^{9} e^{x^{5}} d x

\frac{1}{5}

Decimal Notation

0.2

Solution steps

\int_{0}^{1} x^{9} e^{x^{5}} d x

Apply u-substitution

: \int_{0}^{1} \frac{e ^{u^{\frac{5}{9}}} u ^{\frac{1}{9}}}{9} d u

= \int_{0}^{1} \frac{e ^{u^{\frac{5}{9}}} u ^{\frac{1}{9}}}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{0}^{1} e^{u^{\frac{5}{9}}} u^{\frac{1}{9}} d u

Apply u-substitution

: \int_{0}^{1} \frac{9 e ^{v} v}{5} d v

= \frac{1}{9} \cdot \int_{0}^{1} \frac{9 e ^{v} v}{5} d v

Take the constant out:

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

= \frac{1}{9} \cdot \frac{9}{5} \cdot \int_{0}^{1} e^{v} v d v

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{0}^{1}

= \frac{1}{9} \cdot \frac{9}{5} [e^{v} v - \int e^{v} d v]_{0}^{1}

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

= \frac{1}{9} \cdot \frac{9}{5} [e^{v} v - e^{v}]_{0}^{1}

Simplify

\frac{1}{9} \cdot \frac{9}{5} [e^{v} v - e^{v}]_{0}^{1} : \frac{1}{5} [e^{v} v - e^{v}]_{0}^{1}

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

Compute the boundaries:

1

= \frac{1}{5} \cdot 1

Simplify

= \frac{1}{5}

x \to 0 lim (\frac{( ln ( cos ( 2 x ) ) )}{x ^{2}})

x^{2} y^{^{''}} + 9 x y^{^{'}} - 20 y = \frac{5}{x ^{3}}

f (x) = csc (x) (x + cot (x))

x \to 0 lim (\frac{1}{x + 5})

\frac{\partial}{\partial x} ((5 x - y)^{7})

What is the integral from 0 to 1 of x^9e^{x^5} ?
The integral from 0 to 1 of x^9e^{x^5} is 1/5