What is the integral of (6-x)sqrt(x) ?

The integral of (6-x)sqrt(x) is 4x^{3/2}-2/5 x^{5/2}+C

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integral of (6-x)sqrt(x)

\int (6 - x) x d x

4 x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} + C

Solution steps

\int (6 - x) x d x

Expand

(6 - x) x : 6 x - x^{\frac{3}{2}}

= \int 6 x - x^{\frac{3}{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 6 x d x - \int x^{\frac{3}{2}} d x

\int 6 x d x = 4 x^{\frac{3}{2}}

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

= 4 x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}}

Add a constant to the solution

= 4 x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} + C

What is the integral of (6-x)sqrt(x) ?
The integral of (6-x)sqrt(x) is 4x^{3/2}-2/5 x^{5/2}+C