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

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

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integral from 0 to 1 of x^4(2+2x^5)

\int_{0}^{1} x^{4} (2 + 2 x^{5}) d x

\frac{3}{5}

Decimal Notation

0.6

Solution steps

\int_{0}^{1} x^{4} (2 + 2 x^{5}) d x

Apply Integration By Parts

: [\frac{1}{5} x^{5} (2 + 2 x^{5}) - \int 2 x^{9} d x]_{0}^{1}

= [\frac{1}{5} x^{5} (2 + 2 x^{5}) - \int 2 x^{9} d x]_{0}^{1}

\int 2 x^{9} d x = \frac{x ^{10}}{5}

= [\frac{1}{5} x^{5} (2 + 2 x^{5}) - \frac{x ^{10}}{5}]_{0}^{1}

Simplify

[\frac{1}{5} x^{5} (2 + 2 x^{5}) - \frac{x ^{10}}{5}]_{0}^{1} : [\frac{1}{5} (2 x^{5} + x^{10})]_{0}^{1}

= [\frac{1}{5} (2 x^{5} + x^{10})]_{0}^{1}

Compute the boundaries:

\frac{3}{5}

= \frac{3}{5}

\int (e^{x} + x^{e} - e x) d x

\frac{d ^{2}}{d x ^{2}} (x 1 - x^{2})

\int_{0}^{r} x r^{2} - x^{2} d x

\int \frac{x}{( x ^{2} - 4 ) ( x + 3 )} d x

y = 1 - e^{x}

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