What is the integral of 2piy(8-y^{3/2}) ?

The integral of 2piy(8-y^{3/2}) is 2pi(4y^2-2/7 y^{7/2})+C

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integral of 2piy(8-y^{3/2})

\int 2 π y (8 - y^{\frac{3}{2}}) d y

2 π (4 y^{2} - \frac{2}{7} y^{\frac{7}{2}}) + C

Solution steps

\int 2 π y (8 - y^{\frac{3}{2}}) d y

Take the constant out:

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

= 2 π \cdot \int y (8 - y^{\frac{3}{2}}) d y

Expand

y (8 - y^{\frac{3}{2}}) : 8 y - y^{\frac{5}{2}}

= 2 π \cdot \int 8 y - y^{\frac{5}{2}} 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

= 2 π (\int 8 y d y - \int y^{\frac{5}{2}} d y)

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

\int y^{\frac{5}{2}} d y = \frac{2}{7} y^{\frac{7}{2}}

= 2 π (4 y^{2} - \frac{2}{7} y^{\frac{7}{2}})

Add a constant to the solution

= 2 π (4 y^{2} - \frac{2}{7} y^{\frac{7}{2}}) + C

f (x) = (ln (x))^{3}, at x = 11

f (x) = x^{3} + 2 x, at x = 1

\frac{d}{d x} (\frac{x ^{3} + 3 x + 2}{x ^{2} - 1})

x^{2} y^{^{''}} - 5 x y^{^{'}} + 13 y = 0

\frac{d}{d x} (sin (cos (x^{2} + 3)))

What is the integral of 2piy(8-y^{3/2}) ?
The integral of 2piy(8-y^{3/2}) is 2pi(4y^2-2/7 y^{7/2})+C