What is the integral from 0 to 9 of pi(y^{3/2})^2 ?

The integral from 0 to 9 of pi(y^{3/2})^2 is (6561pi)/4

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integral from 0 to 9 of pi(y^{3/2})^2

\int_{0}^{9} π (y^{\frac{3}{2}})^{2} d y

\frac{6561 π}{4}

Decimal Notation

5152.99735 \dots

Solution steps

\int_{0}^{9} π (y^{\frac{3}{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}^{9} (y^{\frac{3}{2}})^{2} d y

Simplify

(y^{\frac{3}{2}})^{2} : y^{3}

= π \cdot \int_{0}^{9} y^{3} d y

Apply the Power Rule

: [\frac{y ^{4}}{4}]_{0}^{9}

= π [\frac{y ^{4}}{4}]_{0}^{9}

Compute the boundaries:

\frac{6561}{4}

= π \frac{6561}{4}

Simplify

= \frac{6561 π}{4}

n = 0 \sum \infty \frac{5 ^{n}}{4 ^{n}}

\frac{d}{d x} (\frac{x ^{\frac{7}{3}} + 3 x ^{3} - 8}{6 x ^{2}})

x \to 0 lim (\frac{x + tan ( x )}{x})

\frac{d}{d x} (x^{v})

\int \frac{1 + x + x ^{2}}{x} d x

What is the integral from 0 to 9 of pi(y^{3/2})^2 ?
The integral from 0 to 9 of pi(y^{3/2})^2 is (6561pi)/4