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

The integral from 0 to 9 of pi(sqrt(x))^2 is (81pi)/2

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

\int_{0}^{9} π (x)^{2} d x

\frac{81 π}{2}

Decimal Notation

127.23450 \dots

Solution steps

\int_{0}^{9} π (x)^{2} d x

Take the constant out:

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

= π \cdot \int_{0}^{9} (x)^{2} d x

Simplify

(x)^{2} : x

= π \cdot \int_{0}^{9} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{9}

= π [\frac{x ^{2}}{2}]_{0}^{9}

Compute the boundaries:

\frac{81}{2}

= π \frac{81}{2}

Simplify

= \frac{81 π}{2}

2^{(2 - x^{4})}

\frac{d}{d x} ((3 x^{3} + 4) (x^{4} - 3 x))

f (x) = \frac{7}{x}, at x = 2

\int_{1}^{18} \frac{3}{z} d z

- x + 7 x^{2} - 15 x

What is the integral from 0 to 9 of pi(sqrt(x))^2 ?
The integral from 0 to 9 of pi(sqrt(x))^2 is (81pi)/2