What is the integral from 0 to 9 of pi(2sqrt(y))^2 ?

The integral from 0 to 9 of pi(2sqrt(y))^2 is pi162

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integral from 0 to 9 of pi(2sqrt(y))^2

\int_{0}^{9} π (2 y)^{2} d y

π 162

Decimal Notation

508.93800 \dots

Solution steps

\int_{0}^{9} π (2 y)^{2} d y

Take the constant out:

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

= π \cdot \int_{0}^{9} (2 y)^{2} d y

Apply Integration By Parts

: [4 y^{2} - \int 4 y d y]_{0}^{9}

= π [4 y^{2} - \int 4 y d y]_{0}^{9}

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

= π [4 y^{2} - 2 y^{2}]_{0}^{9}

Simplify

= π [2 y^{2}]_{0}^{9}

Compute the boundaries:

162

= π 162

\frac{d y}{d x} + 4 y - e^{- x} = 0

\frac{e ^{- s}}{s ^{2} + 2 s + 2}

0

y^{^{'}} = ry (s - y)

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

What is the integral from 0 to 9 of pi(2sqrt(y))^2 ?
The integral from 0 to 9 of pi(2sqrt(y))^2 is pi162