What is the integral of x^2sqrt(7-x) ?

The integral of x^2sqrt(7-x) is -2/7 (7-x)^{7/2}+28/5 (7-x)^{5/2}-98/3 (7-x)^{3/2}+C

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integral of x^2sqrt(7-x)

\int x^{2} 7 - x d x

- \frac{2}{7} (7 - x)^{\frac{7}{2}} + \frac{28}{5} (7 - x)^{\frac{5}{2}} - \frac{98}{3} (7 - x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 7 - x d x

Apply u-substitution

= \int - (- u + 7)^{2} u d u

Take the constant out:

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

= - \int (- u + 7)^{2} u d u

Expand

(- u + 7)^{2} u : u^{\frac{5}{2}} - 14 u^{\frac{3}{2}} + 49 u

= - \int u^{\frac{5}{2}} - 14 u^{\frac{3}{2}} + 49 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 14 u^{\frac{3}{2}} d u + \int 49 u d u)

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

\int 14 u^{\frac{3}{2}} d u = \frac{28}{5} u^{\frac{5}{2}}

\int 49 u d u = \frac{98}{3} u^{\frac{3}{2}}

= - (\frac{2}{7} u^{\frac{7}{2}} - \frac{28}{5} u^{\frac{5}{2}} + \frac{98}{3} u^{\frac{3}{2}})

Substitute back

u = 7 - x

= - (\frac{2}{7} (7 - x)^{\frac{7}{2}} - \frac{28}{5} (7 - x)^{\frac{5}{2}} + \frac{98}{3} (7 - x)^{\frac{3}{2}})

Simplify

= - \frac{2}{7} (7 - x)^{\frac{7}{2}} + \frac{28}{5} (7 - x)^{\frac{5}{2}} - \frac{98}{3} (7 - x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{7} (7 - x)^{\frac{7}{2}} + \frac{28}{5} (7 - x)^{\frac{5}{2}} - \frac{98}{3} (7 - x)^{\frac{3}{2}} + C

What is the integral of x^2sqrt(7-x) ?
The integral of x^2sqrt(7-x) is -2/7 (7-x)^{7/2}+28/5 (7-x)^{5/2}-98/3 (7-x)^{3/2}+C