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

The integral of (4-t)sqrt(t) is 8/3 t^{3/2}-2/5 t^{5/2}+C

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integral of (4-t)sqrt(t)

\int (4 - t) t d t

\frac{8}{3} t^{\frac{3}{2}} - \frac{2}{5} t^{\frac{5}{2}} + C

Solution steps

\int (4 - t) t d t

Expand

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

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

\int 4 t d t = \frac{8}{3} t^{\frac{3}{2}}

\int t^{\frac{3}{2}} d t = \frac{2}{5} t^{\frac{5}{2}}

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

Add a constant to the solution

= \frac{8}{3} t^{\frac{3}{2}} - \frac{2}{5} t^{\frac{5}{2}} + C

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What is the integral of (4-t)sqrt(t) ?
The integral of (4-t)sqrt(t) is 8/3 t^{3/2}-2/5 t^{5/2}+C