What is the integral of 6rsqrt(4-r^2) ?

The integral of 6rsqrt(4-r^2) is -2(-r^2+4)^{3/2}+C

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integral of 6rsqrt(4-r^2)

\int 6 r 4 - r^{2} d r

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

Solution steps

\int 6 r 4 - r^{2} d r

Take the constant out:

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

= 6 \cdot \int r 4 - r^{2} d r

Apply u-substitution

= 6 \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

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

Apply the Power Rule

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

Substitute back

u = 4 - r^{2}

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

Simplify

6 (- \frac{1}{2} \cdot \frac{2}{3} (4 - r^{2})^{\frac{3}{2}}) : - 2 (- r^{2} + 4)^{\frac{3}{2}}

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

Add a constant to the solution

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

What is the integral of 6rsqrt(4-r^2) ?
The integral of 6rsqrt(4-r^2) is -2(-r^2+4)^{3/2}+C