What is the integral of-2ysqrt(-y^2+2) ?

The integral of-2ysqrt(-y^2+2) is 2/3 (-y^2+2)^{3/2}+C

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integral of-2ysqrt(-y^2+2)

\int - 2 y - y^{2} + 2 d y

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

Solution steps

\int - 2 y - y^{2} + 2 d y

Take the constant out:

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

= - 2 \cdot \int y - y^{2} + 2 d y

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 = - y^{2} + 2

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

Simplify

- 2 (- \frac{1}{2} \cdot \frac{2}{3} (- y^{2} + 2)^{\frac{3}{2}}) : \frac{2}{3} (- y^{2} + 2)^{\frac{3}{2}}

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

Add a constant to the solution

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

f (x) = 1 - 7 x^{2}, (- 2, - 27)

\frac{\partial}{\partial x} (e^{x y} + \frac{x}{y})

y^{^{''}} + 13 y^{^{'}} + 42 y = 0

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\int \frac{x ^{6}}{6} d x

What is the integral of-2ysqrt(-y^2+2) ?
The integral of-2ysqrt(-y^2+2) is 2/3 (-y^2+2)^{3/2}+C