What is the integral from 0 to 5 of xsqrt(4+x) ?

The integral from 0 to 5 of xsqrt(4+x) is 506/15

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integral from 0 to 5 of xsqrt(4+x)

\int_{0}^{5} x 4 + x d x

\frac{506}{15}

Solution steps

\int_{0}^{5} x 4 + x d x

Apply u-substitution

: \int_{4}^{9} (u - 4) u d u

= \int_{4}^{9} (u - 4) u d u

Expand

(u - 4) u : u^{\frac{3}{2}} - 4 u

= \int_{4}^{9} u^{\frac{3}{2}} - 4 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_{4}^{9} u^{\frac{3}{2}} d u - \int_{4}^{9} 4 u d u

\int_{4}^{9} u^{\frac{3}{2}} d u = \frac{422}{5}

\int_{4}^{9} 4 u d u = \frac{152}{3}

= \frac{422}{5} - \frac{152}{3}

Simplify

\frac{422}{5} - \frac{152}{3} : \frac{506}{15}

= \frac{506}{15}

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What is the integral from 0 to 5 of xsqrt(4+x) ?
The integral from 0 to 5 of xsqrt(4+x) is 506/15