What is the integral of 2xsqrt(4-x) ?

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

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integral of 2xsqrt(4-x)

\int 2 x 4 - x d x

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

Solution steps

\int 2 x 4 - x d x

Take the constant out:

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

= 2 \cdot \int x 4 - x d x

Apply u-substitution

= 2 \cdot \int - (- u + 4) u d u

Take the constant out:

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

= 2 (- \int (- u + 4) u d u)

Expand

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

= 2 (- \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

= 2 (- (- \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}}

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

Substitute back

u = 4 - x

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

Simplify

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

Add a constant to the solution

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

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