What is the integral of xsqrt(b-x) ?

The integral of xsqrt(b-x) is -2(1/3 x(b-x)^{3/2}+2/15 (b-x)^{5/2})+C

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integral of xsqrt(b-x)

\int x b - x d x

- 2 (\frac{1}{3} x (b - x)^{\frac{3}{2}} + \frac{2}{15} (b - x)^{\frac{5}{2}}) + C

Solution steps

\int x b - x d x

Apply u-substitution

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

Apply Integration By Parts

= - 2 (\frac{1}{3} u^{3} (- u^{2} + b) - \int - \frac{2 u ^{4}}{3} d u)

\int - \frac{2 u ^{4}}{3} d u = - \frac{2 u ^{5}}{15}

= - 2 (\frac{1}{3} u^{3} (- u^{2} + b) - (- \frac{2 u ^{5}}{15}))

Substitute back

u = b - x

= - 2 (\frac{1}{3} (b - x)^{3} (- (b - x)^{2} + b) - (- \frac{2 ( b - x ) ^{5}}{15}))

Simplify

- 2 (\frac{1}{3} (b - x)^{3} (- (b - x)^{2} + b) - (- \frac{2 ( b - x ) ^{5}}{15})) : - 2 (\frac{1}{3} x (b - x)^{\frac{3}{2}} + \frac{2}{15} (b - x)^{\frac{5}{2}})

= - 2 (\frac{1}{3} x (b - x)^{\frac{3}{2}} + \frac{2}{15} (b - x)^{\frac{5}{2}})

Add a constant to the solution

= - 2 (\frac{1}{3} x (b - x)^{\frac{3}{2}} + \frac{2}{15} (b - x)^{\frac{5}{2}}) + C

What is the integral of xsqrt(b-x) ?
The integral of xsqrt(b-x) is -2(1/3 x(b-x)^{3/2}+2/15 (b-x)^{5/2})+C