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

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

Popular >

integral of xsqrt(16-x^2)

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

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

Solution steps

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

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 - 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

\frac{d}{d s} (\frac{s ^{2} - 4 s + 20}{s ^{2} + 6 s + 8})

\frac{d}{d x} ((x - 3)^{3} + 3 (x - 3)^{2} + 6)

7 arcsin (x^{3})

y = (4 t - 1) (6 t - 3)^{- 1}

\int \frac{4 x ^{4} + x + 1}{x ^{5} + x ^{4}} d x

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