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

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

integral of sqrt(16-x^2)x

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

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

Solution steps

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

Apply u-substitution

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

Substitute back

u = 16 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{3} (3 - x)^{2} d x

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 t e^{2 t}

x \to + 0 + lim (3 x)

f (x) = x^{2} 4 x - 1

sin (5 t - \frac{π}{3}) \cdot 1 (5 t - \frac{π}{3})

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