What is the integral from 0 to 4 of (5-t)sqrt(t) ?

The integral from 0 to 4 of (5-t)sqrt(t) is 208/15

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integral from 0 to 4 of (5-t)sqrt(t)

\int_{0}^{4} (5 - t) t d t

\frac{208}{15}

Decimal

13.86666 \dots

Solution steps

\int_{0}^{4} (5 - t) t d t

Expand

(5 - t) t : 5 t - t^{\frac{3}{2}}

= \int_{0}^{4} 5 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} 5 t d t - \int_{0}^{4} t^{\frac{3}{2}} d t

\int_{0}^{4} 5 t d t = \frac{80}{3}

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

= \frac{80}{3} - \frac{64}{5}

Simplify

\frac{80}{3} - \frac{64}{5} : \frac{208}{15}

= \frac{208}{15}

\frac{d}{d x} ((sin (2 x))^{2})

\int_{- 5}^{0} ∣ x + 3 ∣ d x

\int_{1}^{4} (\frac{6}{x ^{2}}) d x

y^{^{'}} = \frac{t}{y}

x + 6, x^{2}

What is the integral from 0 to 4 of (5-t)sqrt(t) ?
The integral from 0 to 4 of (5-t)sqrt(t) is 208/15