What is the integral of sqrt(3-y) ?

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

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integral of sqrt(3-y)

\int 3 - y d y

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

Solution steps

\int 3 - y d y

Apply u-substitution

= \int - u d u

Take the constant out:

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

= - \int u d u

Apply the Power Rule

= - \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 - y

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

Add a constant to the solution

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

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