What is the integral of 6(36-4x)^{1/2} ?

The integral of 6(36-4x)^{1/2} is -8(-x+9)^{3/2}+C

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integral of 6(36-4x)^{1/2}

\int 6 (36 - 4 x)^{\frac{1}{2}} d x

- 8 (- x + 9)^{\frac{3}{2}} + C

Solution steps

\int 6 (36 - 4 x)^{\frac{1}{2}} d x

Take the constant out:

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

= 6 \cdot \int (36 - 4 x)^{\frac{1}{2}} d x

Apply u-substitution

: \int - \frac{u ^{\frac{1}{2}}}{4} d u

= 6 \cdot \int - \frac{u ^{\frac{1}{2}}}{4} d u

Take the constant out:

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

= 6 (- \frac{1}{4} \cdot \int u^{\frac{1}{2}} d u)

Apply the Power Rule

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

= 6 (- \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 36 - 4 x

= 6 (- \frac{1}{4} \cdot \frac{2}{3} (36 - 4 x)^{\frac{3}{2}})

Simplify

6 (- \frac{1}{4} \cdot \frac{2}{3} (36 - 4 x)^{\frac{3}{2}}) : - 8 (- x + 9)^{\frac{3}{2}}

= - 8 (- x + 9)^{\frac{3}{2}}

Add a constant to the solution

= - 8 (- x + 9)^{\frac{3}{2}} + C

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What is the integral of 6(36-4x)^{1/2} ?
The integral of 6(36-4x)^{1/2} is -8(-x+9)^{3/2}+C