What is the integral of x(10-x)^{1/2} ?

The integral of x(10-x)^{1/2} is -2(-1/5 (10-x)^{5/2}+10/3 (10-x)^{3/2})+C

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integral of x(10-x)^{1/2}

\int x (10 - x)^{\frac{1}{2}} d x

- 2 (- \frac{1}{5} (10 - x)^{\frac{5}{2}} + \frac{10}{3} (10 - x)^{\frac{3}{2}}) + C

Solution steps

\int x (10 - x)^{\frac{1}{2}} d x

Apply u-substitution

: \int - 2 (- u^{2} + 10) u^{2} d u

= \int - 2 (- u^{2} + 10) u^{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \int (- u^{2} + 10) u^{2} d u

Expand

(- u^{2} + 10) u^{2} : - u^{4} + 10 u^{2}

= - 2 \cdot \int - u^{4} + 10 u^{2} d u

Apply the Sum Rule:

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

= - 2 (- \int u^{4} d u + \int 10 u^{2} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int 10 u^{2} d u = \frac{10 u ^{3}}{3}

= - 2 (- \frac{u ^{5}}{5} + \frac{10 u ^{3}}{3})

Substitute back

u = (10 - x)^{\frac{1}{2}}

= - 2 - \frac{( ( 10 - x ) ^{\frac{1}{2}} ) ^{5}}{5} + \frac{10 ( ( 10 - x ) ^{\frac{1}{2}} ) ^{3}}{3}

Simplify

- 2 - \frac{( ( 10 - x ) ^{\frac{1}{2}} ) ^{5}}{5} + \frac{10 ( ( 10 - x ) ^{\frac{1}{2}} ) ^{3}}{3} : - 2 (- \frac{1}{5} (10 - x)^{\frac{5}{2}} + \frac{10}{3} (10 - x)^{\frac{3}{2}})

= - 2 (- \frac{1}{5} (10 - x)^{\frac{5}{2}} + \frac{10}{3} (10 - x)^{\frac{3}{2}})

Add a constant to the solution

= - 2 (- \frac{1}{5} (10 - x)^{\frac{5}{2}} + \frac{10}{3} (10 - x)^{\frac{3}{2}}) + C

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What is the integral of x(10-x)^{1/2} ?
The integral of x(10-x)^{1/2} is -2(-1/5 (10-x)^{5/2}+10/3 (10-x)^{3/2})+C