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

The integral from 0 to 4 of sqrt(x)(4-x) is 128/15

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integral from 0 to 4 of sqrt(x)(4-x)

\int_{0}^{4} x (4 - x) d x

\frac{128}{15}

Decimal

8.53333 \dots

Solution steps

\int_{0}^{4} x (4 - x) d x

Expand

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

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

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

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

= \frac{64}{3} - \frac{64}{5}

Simplify

\frac{64}{3} - \frac{64}{5} : \frac{128}{15}

= \frac{128}{15}

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What is the integral from 0 to 4 of sqrt(x)(4-x) ?
The integral from 0 to 4 of sqrt(x)(4-x) is 128/15