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

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

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

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

\frac{128}{15}

Decimal Notation

8.53333 \dots

Solution steps

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

Expand

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

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

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

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

\int_{0}^{4} t^{\frac{3}{2}} d t = \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 (4-t)(sqrt(t)) ?
The integral from 0 to 4 of (4-t)(sqrt(t)) is 128/15