What is the integral of xsqrt(10-x) ?

The integral of xsqrt(10-x) is 2/5 (10-x)^{5/2}-20/3 (10-x)^{3/2}+C

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integral of xsqrt(10-x)

\int x 10 - x d x

\frac{2}{5} (10 - x)^{\frac{5}{2}} - \frac{20}{3} (10 - x)^{\frac{3}{2}} + C

Solution steps

\int x 10 - x d x

Apply u-substitution

: \int - (- u + 10) u d u

= \int - (- u + 10) u d u

Take the constant out:

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

= - \int (- u + 10) u d u

Expand

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

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

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

\int 10 u d u = \frac{20}{3} u^{\frac{3}{2}}

= - (- \frac{2}{5} u^{\frac{5}{2}} + \frac{20}{3} u^{\frac{3}{2}})

Substitute back

u = 10 - x

= - (- \frac{2}{5} (10 - x)^{\frac{5}{2}} + \frac{20}{3} (10 - x)^{\frac{3}{2}})

Simplify

= \frac{2}{5} (10 - x)^{\frac{5}{2}} - \frac{20}{3} (10 - x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (10 - x)^{\frac{5}{2}} - \frac{20}{3} (10 - x)^{\frac{3}{2}} + C

(- ln (1 + 6 x))^{^{'}}

x \to - 5 lim (6 x - \frac{2}{x ^{3}})

x - 4

x \to 0 lim (\frac{1}{x} - \frac{1}{x ^{2}})

x \to - \infty lim (\frac{4}{x} - 5)

What is the integral of xsqrt(10-x) ?
The integral of xsqrt(10-x) is 2/5 (10-x)^{5/2}-20/3 (10-x)^{3/2}+C