What is the integral from 0 to 1 of x^7e^{x^4} ?

The integral from 0 to 1 of x^7e^{x^4} is 1/4

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integral from 0 to 1 of x^7e^{x^4}

\int_{0}^{1} x^{7} e^{x^{4}} d x

\frac{1}{4}

Decimal

0.25

Solution steps

\int_{0}^{1} x^{7} e^{x^{4}} d x

Apply u-substitution

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

Apply u-substitution

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

1

= \frac{1}{4} \cdot 1

Simplify

= \frac{1}{4}

What is the integral from 0 to 1 of x^7e^{x^4} ?
The integral from 0 to 1 of x^7e^{x^4} is 1/4