What is the integral of sqrt(-4y) ?

The integral of sqrt(-4y) is i 1/3 (-2y^{3/2}+6ysqrt(y))+C

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integral of sqrt(-4y)

\int - 4 y d y

i \frac{1}{3} (- 2 y^{\frac{3}{2}} + 6 y y) + C

Solution steps

\int - 4 y d y

- 4 y = 4 - y

= \int 4 - y d y

Take the constant out:

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

= 4 \cdot \int - y d y

4 = 2

= 2 \cdot \int - y d y

Apply Integration By Parts

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

= 2 (y - y - \int - \frac{y}{2 - y} d y)

\int - \frac{y}{2 - y} d y = \frac{i y ^{\frac{3}{2}}}{3}

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

Simplify

2 (y - y - \frac{i y ^{\frac{3}{2}}}{3}) : i \frac{1}{3} (- 2 y^{\frac{3}{2}} + 6 y y)

= i \frac{1}{3} (- 2 y^{\frac{3}{2}} + 6 y y)

Add a constant to the solution

= i \frac{1}{3} (- 2 y^{\frac{3}{2}} + 6 y y) + C

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What is the integral of sqrt(-4y) ?
The integral of sqrt(-4y) is i 1/3 (-2y^{3/2}+6ysqrt(y))+C