What is the integral of (x-1)sqrt(2-x) ?

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

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integral of (x-1)sqrt(2-x)

\int (x - 1) 2 - x d x

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

Solution steps

\int (x - 1) 2 - x d x

Apply u-substitution

: \int - 2 u^{2} (- u^{2} + 1) d u

= \int - 2 u^{2} (- u^{2} + 1) 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} (- u^{2} + 1) d u

Expand

u^{2} (- u^{2} + 1) : - u^{4} + u^{2}

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

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

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

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

Substitute back

u = 2 - x

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

Simplify

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

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

Add a constant to the solution

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

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