What is the integral of xsqrt(4-x) ?

The integral of xsqrt(4-x) is 2/5 (4-x)^{5/2}-8/3 (4-x)^{3/2}+C

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integral of xsqrt(4-x)

\int x 4 - x d x

\frac{2}{5} (4 - x)^{\frac{5}{2}} - \frac{8}{3} (4 - x)^{\frac{3}{2}} + C

Solution steps

\int x 4 - x d x

Apply u-substitution

: \int - (- u + 4) u d u

= \int - (- u + 4) u d u

Take the constant out:

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

= - \int (- u + 4) u d u

Expand

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

= - \int - u^{\frac{3}{2}} + 4 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 4 u d u)

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

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= - (- \frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = 4 - x

= - (- \frac{2}{5} (4 - x)^{\frac{5}{2}} + \frac{8}{3} (4 - x)^{\frac{3}{2}})

Simplify

= \frac{2}{5} (4 - x)^{\frac{5}{2}} - \frac{8}{3} (4 - x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (4 - x)^{\frac{5}{2}} - \frac{8}{3} (4 - x)^{\frac{3}{2}} + C

\int_{0}^{1} 2 y^{2} d y

ln (e^{- x} + x e^{- x})

\frac{d}{d x} (\frac{5 x}{x - 3})

\frac{y}{12000} + 0.058 y^{^{'}} = 0.0015

x \to \infty lim (- 18 x^{4})

What is the integral of xsqrt(4-x) ?
The integral of xsqrt(4-x) is 2/5 (4-x)^{5/2}-8/3 (4-x)^{3/2}+C