What is the integral of sqrt(16-r^2)r ?

The integral of sqrt(16-r^2)r is -1/3 (16-r^2)^{3/2}+C

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integral of sqrt(16-r^2)r

\int 16 - r^{2} r d r

- \frac{1}{3} (16 - r^{2})^{\frac{3}{2}} + C

Solution steps

\int 16 - r^{2} r d r

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 16 - r^{2}

= - \frac{1}{2} \cdot \frac{2}{3} (16 - r^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (16 - r^{2})^{\frac{3}{2}} : - \frac{1}{3} (16 - r^{2})^{\frac{3}{2}}

= - \frac{1}{3} (16 - r^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (16 - r^{2})^{\frac{3}{2}} + C

y^{^{''}} + 7 y^{^{'}} + 12 = 0

cos (x), - 0.25

x \to 2 + lim (\frac{1}{x})

\int 6 sin^{5} (x) d x

What is the integral of sqrt(16-r^2)r ?
The integral of sqrt(16-r^2)r is -1/3 (16-r^2)^{3/2}+C