What is the integral from 0 to 2 of (12-t)sqrt(t) ?

The integral from 0 to 2 of (12-t)sqrt(t) is 16sqrt(2)-(8sqrt(2))/5

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integral from 0 to 2 of (12-t)sqrt(t)

\int_{0}^{2} (12 - t) t d t

162 - \frac{8 2}{5}

Decimal Notation

20.36467 \dots

Solution steps

\int_{0}^{2} (12 - t) t d t

Expand

(12 - t) t : 12 t - t^{\frac{3}{2}}

= \int_{0}^{2} 12 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} 12 t d t - \int_{0}^{2} t^{\frac{3}{2}} d t

\int_{0}^{2} 12 t d t = 162

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

= 162 - \frac{8 2}{5}

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What is the integral from 0 to 2 of (12-t)sqrt(t) ?
The integral from 0 to 2 of (12-t)sqrt(t) is 16sqrt(2)-(8sqrt(2))/5