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

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

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integral of sqrt(x)(6-x)

\int x (6 - x) d x

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

Solution steps

\int x (6 - x) d x

Expand

x (6 - 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

\frac{s}{( s ^{2} - k ^{2} )}

\frac{\partial}{\partial x} ((x^{3}) x^{2} + 9)

x \to 1 - lim (\frac{x ^{2} - 1}{x - 1})

\int \frac{x}{( x + 1 ) ( x + 2 ) ^{2}} d x

\frac{d y}{d x} = 2 y + 1

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