What is the integral from 0 to 1 of x^3(2+x^4)^3 ?

The integral from 0 to 1 of x^3(2+x^4)^3 is 65/16

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integral from 0 to 1 of x^3(2+x^4)^3

\int_{0}^{1} x^{3} (2 + x^{4})^{3} d x

\frac{65}{16}

Decimal

4.0625

Solution steps

\int_{0}^{1} x^{3} (2 + x^{4})^{3} d x

Apply u-substitution

: \int_{2}^{3} \frac{u ^{3}}{4} d u

= \int_{2}^{3} \frac{u ^{3}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{2}^{3} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2}^{3}

= \frac{1}{4} [\frac{u ^{4}}{4}]_{2}^{3}

Compute the boundaries:

\frac{65}{4}

= \frac{1}{4} \cdot \frac{65}{4}

Simplify

= \frac{65}{16}

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What is the integral from 0 to 1 of x^3(2+x^4)^3 ?
The integral from 0 to 1 of x^3(2+x^4)^3 is 65/16