What is the integral from 0 to 3 of pi(2sqrt(7y))^2 ?

The integral from 0 to 3 of pi(2sqrt(7y))^2 is pi126

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integral from 0 to 3 of pi(2sqrt(7y))^2

\int_{0}^{3} π (2 7 y)^{2} d y

π 126

Decimal Notation

395.84067 \dots

Solution steps

\int_{0}^{3} π (2 7 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}^{3} (2 7 y)^{2} d y

Apply Integration By Parts

: [28 y^{2} - \int 28 y d y]_{0}^{3}

= π [28 y^{2} - \int 28 y d y]_{0}^{3}

\int 28 y d y = 14 y^{2}

= π [28 y^{2} - 14 y^{2}]_{0}^{3}

Simplify

= π [14 y^{2}]_{0}^{3}

Compute the boundaries:

126

= π 126

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

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 13, y^{^{'}} (0) = b

\int 18 tan^{17} (x) sec^{2} (x) d x

x \to 9 lim (\frac{3 - x}{9 x - x ^{2}})

What is the integral from 0 to 3 of pi(2sqrt(7y))^2 ?
The integral from 0 to 3 of pi(2sqrt(7y))^2 is pi126