What is the integral from 0 to 4 of (8-x)sqrt(x) ?

The integral from 0 to 4 of (8-x)sqrt(x) is 448/15

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integral from 0 to 4 of (8-x)sqrt(x)

\int_{0}^{4} (8 - x) x d x

\frac{448}{15}

Decimal

29.86666 \dots

Solution steps

\int_{0}^{4} (8 - x) x d x

Expand

(8 - x) x : 8 x - x^{\frac{3}{2}}

= \int_{0}^{4} 8 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} 8 x d x - \int_{0}^{4} x^{\frac{3}{2}} d x

\int_{0}^{4} 8 x d x = \frac{128}{3}

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

= \frac{128}{3} - \frac{64}{5}

Simplify

\frac{128}{3} - \frac{64}{5} : \frac{448}{15}

= \frac{448}{15}

\int_{0}^{1} x^{\frac{5}{2}} d x

\int_{0}^{1} 1 + 9 x^{\frac{1}{2}} d x

\int_{0}^{2} 3 + 6 x - x^{3} d x

\int_{1}^{2} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x

\int_{0}^{1} (x + 6 1 - x^{2}) d x

What is the integral from 0 to 4 of (8-x)sqrt(x) ?
The integral from 0 to 4 of (8-x)sqrt(x) is 448/15