What is the integral of x^2sqrt(9-x) ?

The integral of x^2sqrt(9-x) is -2/7 (9-x)^{7/2}+36/5 (9-x)^{5/2}-54(9-x)^{3/2}+C

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integral of x^2sqrt(9-x)

\int x^{2} 9 - x d x

- \frac{2}{7} (9 - x)^{\frac{7}{2}} + \frac{36}{5} (9 - x)^{\frac{5}{2}} - 54 (9 - x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 9 - x d x

Apply u-substitution

= \int - (- u + 9)^{2} u d u

Take the constant out:

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

= - \int (- u + 9)^{2} u d u

Expand

(- u + 9)^{2} u : u^{\frac{5}{2}} - 18 u^{\frac{3}{2}} + 81 u

= - \int u^{\frac{5}{2}} - 18 u^{\frac{3}{2}} + 81 u 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

= - (\int u^{\frac{5}{2}} d u - \int 18 u^{\frac{3}{2}} d u + \int 81 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 18 u^{\frac{3}{2}} d u = \frac{36}{5} u^{\frac{5}{2}}

\int 81 u d u = 54 u^{\frac{3}{2}}

= - (\frac{2}{7} u^{\frac{7}{2}} - \frac{36}{5} u^{\frac{5}{2}} + 54 u^{\frac{3}{2}})

Substitute back

u = 9 - x

= - (\frac{2}{7} (9 - x)^{\frac{7}{2}} - \frac{36}{5} (9 - x)^{\frac{5}{2}} + 54 (9 - x)^{\frac{3}{2}})

Simplify

= - \frac{2}{7} (9 - x)^{\frac{7}{2}} + \frac{36}{5} (9 - x)^{\frac{5}{2}} - 54 (9 - x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{7} (9 - x)^{\frac{7}{2}} + \frac{36}{5} (9 - x)^{\frac{5}{2}} - 54 (9 - x)^{\frac{3}{2}} + C

What is the integral of x^2sqrt(9-x) ?
The integral of x^2sqrt(9-x) is -2/7 (9-x)^{7/2}+36/5 (9-x)^{5/2}-54(9-x)^{3/2}+C