What is the integral from 0 to 4 of sqrt(x)(x+1) ?

The integral from 0 to 4 of sqrt(x)(x+1) is 272/15

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integral from 0 to 4 of sqrt(x)(x+1)

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

\frac{272}{15}

Solution steps

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

Expand

x (x + 1) : x^{\frac{3}{2}} + x

= \int_{0}^{4} x^{\frac{3}{2}} + x d x

Apply the Sum Rule:

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

= \int_{0}^{4} x^{\frac{3}{2}} d x + \int_{0}^{4} x d x

\int_{0}^{4} x^{\frac{3}{2}} d x = \frac{64}{5}

\int_{0}^{4} x d x = \frac{16}{3}

= \frac{64}{5} + \frac{16}{3}

Simplify

\frac{64}{5} + \frac{16}{3} : \frac{272}{15}

= \frac{272}{15}

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What is the integral from 0 to 4 of sqrt(x)(x+1) ?
The integral from 0 to 4 of sqrt(x)(x+1) is 272/15