What is the integral of pi(sqrt(y)+2)^2 ?

The integral of pi(sqrt(y)+2)^2 is pi((y^2)/2+8/3 y^{3/2}+4y)+C

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integral of pi(sqrt(y)+2)^2

\int π (y + 2)^{2} d y

π (\frac{y ^{2}}{2} + \frac{8}{3} y^{\frac{3}{2}} + 4 y) + C

Solution steps

\int π (y + 2)^{2} d y

Take the constant out:

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

= π \cdot \int (y + 2)^{2} d y

Expand

(y + 2)^{2} : y + 4 y + 4

= π \cdot \int y + 4 y + 4 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

= π (\int y d y + \int 4 y d y + \int 4 d y)

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

\int 4 y d y = \frac{8}{3} y^{\frac{3}{2}}

\int 4 d y = 4 y

= π (\frac{y ^{2}}{2} + \frac{8}{3} y^{\frac{3}{2}} + 4 y)

Add a constant to the solution

= π (\frac{y ^{2}}{2} + \frac{8}{3} y^{\frac{3}{2}} + 4 y) + C

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

\frac{d ^{2}}{d x ^{2}} (e^{6 x})

\int \frac{x ^{2}}{e ^{x^{3}}} d x

\frac{d}{d x} (ln (16))

x \to 6 lim (\frac{7 x - 4}{2 x - 8})

What is the integral of pi(sqrt(y)+2)^2 ?
The integral of pi(sqrt(y)+2)^2 is pi((y^2)/2+8/3 y^{3/2}+4y)+C