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}

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}

f (x) = x - \frac{4}{x}, at 0

2 y^{^{''}} + 4 y^{^{'}} = 0

y^{^{'}} + 2 y = 2

5 x^{3} lo g_{3} (x)

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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