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

The integral from 0 to 1 of x^3(2+x^4) is 5/8

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

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

\frac{5}{8}

Decimal Notation

0.625

Solution steps

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

Apply Integration By Parts

: [\frac{1}{4} x^{4} (2 + x^{4}) - \int x^{7} d x]_{0}^{1}

= [\frac{1}{4} x^{4} (2 + x^{4}) - \int x^{7} d x]_{0}^{1}

\int x^{7} d x = \frac{x ^{8}}{8}

= [\frac{1}{4} x^{4} (2 + x^{4}) - \frac{x ^{8}}{8}]_{0}^{1}

Simplify

[\frac{1}{4} x^{4} (2 + x^{4}) - \frac{x ^{8}}{8}]_{0}^{1} : [\frac{1}{8} (4 x^{4} + x^{8})]_{0}^{1}

= [\frac{1}{8} (4 x^{4} + x^{8})]_{0}^{1}

Compute the boundaries:

\frac{5}{8}

= \frac{5}{8}

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\int_{0}^{2} 4 x^{3} - 12 x^{2} + 8 x d x

What is the integral from 0 to 1 of x^3(2+x^4) ?
The integral from 0 to 1 of x^3(2+x^4) is 5/8