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

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

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integral of sqrt(y/4)

\int \frac{y}{4} d y

\frac{1}{3} y^{\frac{3}{2}} + C

Solution steps

\int \frac{y}{4} d y

Simplify

\frac{y}{4} : \frac{y}{4}

= \int \frac{y}{4} d y

Take the constant out:

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

= \frac{1}{4} \cdot \int y d y

4 = 2

= \frac{1}{2} \cdot \int y d y

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{2}{3} y^{\frac{3}{2}}

Simplify

\frac{1}{2} \cdot \frac{2}{3} y^{\frac{3}{2}} : \frac{1}{3} y^{\frac{3}{2}}

= \frac{1}{3} y^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} y^{\frac{3}{2}} + C

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