What is the integral from 0 to 2 of (2-t)sqrt(t) ?

The integral from 0 to 2 of (2-t)sqrt(t) is (16sqrt(2))/(15)

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integral from 0 to 2 of (2-t)sqrt(t)

\int_{0}^{2} (2 - t) t d t

\frac{16 2}{15}

Decimal Notation

1.50849 \dots

Solution steps

\int_{0}^{2} (2 - t) t d t

Expand

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

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

\int_{0}^{2} 2 t d t = \frac{8 2}{3}

\int_{0}^{2} t^{\frac{3}{2}} d t = \frac{8 2}{5}

= \frac{8 2}{3} - \frac{8 2}{5}

Simplify

\frac{8 2}{3} - \frac{8 2}{5} : \frac{16 2}{15}

= \frac{16 2}{15}

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What is the integral from 0 to 2 of (2-t)sqrt(t) ?
The integral from 0 to 2 of (2-t)sqrt(t) is (16sqrt(2))/(15)