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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply u-substitution

= - 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

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

Substitute back

u = 2 - y^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{p ^{2} + 2 p} d p

a re a x = y^{2}, x = y + 6

\frac{d}{d x} (csc (x) cot (x))

d er i v a t i v e x^{2}

a re a f (x) = - x^{2} + 2, g (x) = - x - 4

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