What is the integral of sqrt(9-x^2)(-2x) ?

The integral of sqrt(9-x^2)(-2x) is 2/3 (9-x^2)^{3/2}+C

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integral of sqrt(9-x^2)(-2x)

\int 9 - x^{2} (- 2 x) d x

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

Solution steps

\int 9 - x^{2} (- 2 x) d x

Simplify

= \int - 2 x 9 - x^{2} d x

Take the constant out:

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

= - 2 \cdot \int x 9 - x^{2} d x

Apply u-substitution

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

= - 2 \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

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

Apply the Power Rule

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

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

Substitute back

u = 9 - x^{2}

= - 2 (- \frac{1}{2} \cdot \frac{2}{3} (9 - x^{2})^{\frac{3}{2}})

Simplify

- 2 (- \frac{1}{2} \cdot \frac{2}{3} (9 - x^{2})^{\frac{3}{2}}) : \frac{2}{3} (9 - x^{2})^{\frac{3}{2}}

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

Add a constant to the solution

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

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

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

f (x) = 6 x^{2} + x - 3

\frac{t}{2}

y = x, x = 9, y = 0

What is the integral of sqrt(9-x^2)(-2x) ?
The integral of sqrt(9-x^2)(-2x) is 2/3 (9-x^2)^{3/2}+C