What is the integral of 15sqrt(9-2x) ?

The integral of 15sqrt(9-2x) is -5(-2x+9)^{3/2}+C

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integral of 15sqrt(9-2x)

\int 15 9 - 2 x d x

- 5 (- 2 x + 9)^{\frac{3}{2}} + C

Solution steps

\int 15 9 - 2 x d x

Take the constant out:

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

= 15 \cdot \int 9 - 2 x d x

Apply u-substitution

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

= 15 \cdot \int - \frac{u}{2} d u

Take the constant out:

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

= 15 (- \frac{1}{2} \cdot \int u d u)

Apply the Power Rule

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

= 15 (- \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 9 - 2 x

= 15 (- \frac{1}{2} \cdot \frac{2}{3} (9 - 2 x)^{\frac{3}{2}})

Simplify

15 (- \frac{1}{2} \cdot \frac{2}{3} (9 - 2 x)^{\frac{3}{2}}) : - 5 (- 2 x + 9)^{\frac{3}{2}}

= - 5 (- 2 x + 9)^{\frac{3}{2}}

Add a constant to the solution

= - 5 (- 2 x + 9)^{\frac{3}{2}} + C

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What is the integral of 15sqrt(9-2x) ?
The integral of 15sqrt(9-2x) is -5(-2x+9)^{3/2}+C