What is the integral from 0 to 1 of xsqrt(1+x) ?

The integral from 0 to 1 of xsqrt(1+x) is (4sqrt(2)+4)/(15)

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integral from 0 to 1 of xsqrt(1+x)

\int_{0}^{1} x 1 + x d x

\frac{4 2 + 4}{15}

Decimal

0.64379 \dots

Solution steps

\int_{0}^{1} x 1 + x d x

Apply u-substitution

= \int_{1}^{2} (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{1}^{2} u^{\frac{3}{2}} d u = \frac{8 2 - 2}{5}

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

= \frac{8 2 - 2}{5} - \frac{4 2 - 2}{3}

Simplify

\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3} : \frac{4 2 + 4}{15}

= \frac{4 2 + 4}{15}

What is the integral from 0 to 1 of xsqrt(1+x) ?
The integral from 0 to 1 of xsqrt(1+x) is (4sqrt(2)+4)/(15)