What is the integral of sqrt(8-4x) ?

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

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integral of sqrt(8-4x)

\int 8 - 4 x d x

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

Solution steps

\int 8 - 4 x d x

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= - \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 8 - 4 x

= - \frac{1}{4} \cdot \frac{2}{3} (8 - 4 x)^{\frac{3}{2}}

Simplify

- \frac{1}{4} \cdot \frac{2}{3} (8 - 4 x)^{\frac{3}{2}} : - \frac{4}{3} (- x + 2)^{\frac{3}{2}}

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

Add a constant to the solution

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

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