What is the integral of (6-x)^{1/2} ?

The integral of (6-x)^{1/2} is -2/3 (6-x)^{3/2}+C

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integral of (6-x)^{1/2}

\int (6 - x)^{\frac{1}{2}} d x

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

Solution steps

\int (6 - x)^{\frac{1}{2}} d x

Apply u-substitution

: \int - u^{\frac{1}{2}} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 6 - x

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

Add a constant to the solution

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

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What is the integral of (6-x)^{1/2} ?
The integral of (6-x)^{1/2} is -2/3 (6-x)^{3/2}+C