What is the integral of (3sqrt(y)-2y-3) ?

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

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integral of (3sqrt(y)-2y-3)

\int (3 y - 2 y - 3) d y

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

Solution steps

\int 3 y - 2 y - 3 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 y d y - \int 2 y d y - \int 3 d y

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

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

\int 3 d y = 3 y

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

Add a constant to the solution

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

\frac{d}{d x} ((8 x - 7)^{- 5})

x \to 0 lim (1 + x - 1)

y \to 0 lim (\frac{sin ( 5 y )}{7 y})

\int x lo g_{e} (x) d x

x \to 0 lim (sin^{x} (x))

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