What is the integral from 0 to 13/6 of pi(13/6-x)^2 ?

The integral from 0 to 13/6 of pi(13/6-x)^2 is (2197pi)/(648)

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integral from 0 to 13/6 of pi(13/6-x)^2

\int_{0}^{\frac{13}{6}} π (\frac{13}{6} - x)^{2} d x

\frac{2197 π}{648}

Decimal Notation

10.65135 \dots

Solution steps

\int_{0}^{\frac{13}{6}} π (\frac{13}{6} - 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}^{\frac{13}{6}} (\frac{13}{6} - x)^{2} d x

Apply u-substitution

: \int_{\frac{13}{6}}^{0} - u^{2} d u

= π \cdot \int_{\frac{13}{6}}^{0} - u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= π (- \int_{0}^{\frac{13}{6}} - u^{2} d u)

Take the constant out:

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

= π (- (- \int_{0}^{\frac{13}{6}} u^{2} d u))

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{\frac{13}{6}}

= π (- (- [\frac{u ^{3}}{3}]_{0}^{\frac{13}{6}}))

Simplify

= π [\frac{u ^{3}}{3}]_{0}^{\frac{13}{6}}

Compute the boundaries:

\frac{2197}{648}

= π \frac{2197}{648}

Simplify

= \frac{2197 π}{648}

\frac{\partial}{\partial x} (sin (π (x - 6 y)))

\int π (4 x)^{2} d x

\int \frac{x ^{2}}{9 + x ^{2}} d x

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

x = - 2, x = 1, y = 2 x^{2} + 4, y = 0

What is the integral from 0 to 13/6 of pi(13/6-x)^2 ?
The integral from 0 to 13/6 of pi(13/6-x)^2 is (2197pi)/(648)