What is the integral of xsqrt(6-x) ?

The integral of xsqrt(6-x) is 2/5 (6-x)^{5/2}-4(6-x)^{3/2}+C

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integral of xsqrt(6-x)

\int x 6 - x d x

\frac{2}{5} (6 - x)^{\frac{5}{2}} - 4 (6 - x)^{\frac{3}{2}} + C

Solution steps

\int x 6 - x d x

Apply u-substitution

: \int - (- u + 6) u d u

= \int - (- u + 6) u d u

Take the constant out:

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

= - \int (- u + 6) u d u

Expand

(- u + 6) u : - u^{\frac{3}{2}} + 6 u

= - \int - u^{\frac{3}{2}} + 6 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 6 u d u)

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

\int 6 u d u = 4 u^{\frac{3}{2}}

= - (- \frac{2}{5} u^{\frac{5}{2}} + 4 u^{\frac{3}{2}})

Substitute back

u = 6 - x

= - (- \frac{2}{5} (6 - x)^{\frac{5}{2}} + 4 (6 - x)^{\frac{3}{2}})

Simplify

= \frac{2}{5} (6 - x)^{\frac{5}{2}} - 4 (6 - x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (6 - x)^{\frac{5}{2}} - 4 (6 - x)^{\frac{3}{2}} + C

x \to 3 lim (\frac{x - 1}{x ^{2} - 1})

e^{- 2 s} \frac{1}{s - 9}

\int \frac{1}{1 + x} d x

(0, - 6), (3, - 5)

What is the integral of xsqrt(6-x) ?
The integral of xsqrt(6-x) is 2/5 (6-x)^{5/2}-4(6-x)^{3/2}+C