What is the integral of xsqrt(4-x^2) ?

The integral of xsqrt(4-x^2) is -1/3 (4-x^2)^{3/2}+C

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integral of xsqrt(4-x^2)

\int x 4 - x^{2} d x

- \frac{1}{3} (4 - x^{2})^{\frac{3}{2}} + C

Solution steps

\int x 4 - 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 = 4 - x^{2}

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

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (4 - x^{2})^{\frac{3}{2}} : - \frac{1}{3} (4 - x^{2})^{\frac{3}{2}}

= - \frac{1}{3} (4 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (4 - x^{2})^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (\frac{a}{a - x})

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

ln (x + 6)

y^{^{'}} = 9 y^{2} + x y^{2}

\frac{d}{d x} (\frac{2 x}{x - 1})

What is the integral of xsqrt(4-x^2) ?
The integral of xsqrt(4-x^2) is -1/3 (4-x^2)^{3/2}+C