What is the integral from 0 to 6 of pi((y^2)/3)^2 ?

The integral from 0 to 6 of pi((y^2)/3)^2 is (864pi)/5

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integral from 0 to 6 of pi((y^2)/3)^2

\int_{0}^{6} π (\frac{y ^{2}}{3})^{2} d y

\frac{864 π}{5}

Decimal Notation

542.86721 \dots

Solution steps

\int_{0}^{6} π (\frac{y ^{2}}{3})^{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}^{6} (\frac{y ^{2}}{3})^{2} d y

Apply Integration By Parts

: [\frac{y ^{5}}{9} - \int \frac{4 y ^{4}}{9} d y]_{0}^{6}

= π [\frac{y ^{5}}{9} - \int \frac{4 y ^{4}}{9} d y]_{0}^{6}

\int \frac{4 y ^{4}}{9} d y = \frac{4 y ^{5}}{45}

= π [\frac{y ^{5}}{9} - \frac{4 y ^{5}}{45}]_{0}^{6}

Join

\frac{y ^{5}}{9} - \frac{4 y ^{5}}{45} : \frac{y ^{5}}{45}

= π [\frac{y ^{5}}{45}]_{0}^{6}

Compute the boundaries:

\frac{864}{5}

= π \frac{864}{5}

Simplify

= \frac{864 π}{5}

\frac{d}{d x} ([arctan (x)] (1 + x^{2}))

x \to π lim (\frac{1}{sin ( x )})

\int \frac{x ^{2}}{( 1 - 16 x ^{2} ) ^{\frac{3}{2}}} d x

x \to 5 - lim (5 - x)

\int \frac{x ^{5} + 7}{x ^{2}} d x

What is the integral from 0 to 6 of pi((y^2)/3)^2 ?
The integral from 0 to 6 of pi((y^2)/3)^2 is (864pi)/5