What is the integral from 0 to 4 of pi((y^2)/2)^2 ?

The integral from 0 to 4 of pi((y^2)/2)^2 is (256pi)/5

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integral from 0 to 4 of pi((y^2)/2)^2

\int_{0}^{4} π (\frac{y ^{2}}{2})^{2} d y

\frac{256 π}{5}

Decimal Notation

160.84954 \dots

Solution steps

\int_{0}^{4} π (\frac{y ^{2}}{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_{0}^{4} (\frac{y ^{2}}{2})^{2} d y

Apply Integration By Parts

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

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

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

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

Join

\frac{y ^{5}}{4} - \frac{y ^{5}}{5} : \frac{y ^{5}}{20}

= π [\frac{y ^{5}}{20}]_{0}^{4}

Compute the boundaries:

\frac{256}{5}

= π \frac{256}{5}

Simplify

= \frac{256 π}{5}

\int_{3}^{5} \frac{1}{4 + ( x - 3 ) ^{2}} d x

y^{^{'}} + 3 y = e^{x} y^{2}

x \to 4 + lim (\frac{x + 5}{x - 4})

9 y^{^{''}} + y = 0

y^{^{''}} + 4 y = t^{2} + 3 e^{t}

What is the integral from 0 to 4 of pi((y^2)/2)^2 ?
The integral from 0 to 4 of pi((y^2)/2)^2 is (256pi)/5