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

The integral of sqrt(4-2x) is -1/3 (4-2x)^{3/2}+C

integral of sqrt(4-2x)

\int 4 - 2 x d x

- \frac{1}{3} (4 - 2 x)^{\frac{3}{2}} + C

Solution steps

\int 4 - 2 x d x

Apply u-substitution

= \int - \frac{u}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int u d u

Apply the Power Rule

= - \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 - 2 x

= - \frac{1}{2} \cdot \frac{2}{3} (4 - 2 x)^{\frac{3}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (4 - 2 x)^{\frac{3}{2}} : - \frac{1}{3} (4 - 2 x)^{\frac{3}{2}}

= - \frac{1}{3} (4 - 2 x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (4 - 2 x)^{\frac{3}{2}} + C

2 x - y \frac{d y}{d x} = 3

\int \frac{x ^{2} + 1}{x ^{4}} d x

t a y l or 2 + x, - 1

\int \frac{11}{x ^{3} - 27} d x

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

What is the integral of sqrt(4-2x) ?
The integral of sqrt(4-2x) is -1/3 (4-2x)^{3/2}+C