What is the integral from 0 to 19/6 of pi(19/6-x)^2 ?

The integral from 0 to 19/6 of pi(19/6-x)^2 is (6859pi)/(648)

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integral from 0 to 19/6 of pi(19/6-x)^2

\int_{0}^{\frac{19}{6}} π (\frac{19}{6} - x)^{2} d x

\frac{6859 π}{648}

Decimal

33.25337 \dots

Solution steps

\int_{0}^{\frac{19}{6}} π (\frac{19}{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{19}{6}} (\frac{19}{6} - x)^{2} d x

Apply u-substitution

: \int_{\frac{19}{6}}^{0} - u^{2} d u

= π \cdot \int_{\frac{19}{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{19}{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{19}{6}} u^{2} d u))

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{\frac{19}{6}}

= π (- (- [\frac{u ^{3}}{3}]_{0}^{\frac{19}{6}}))

Simplify

= π [\frac{u ^{3}}{3}]_{0}^{\frac{19}{6}}

Compute the boundaries:

\frac{6859}{648}

= π \frac{6859}{648}

Simplify

= \frac{6859 π}{648}

f (x) = e^{8 x} cos (π x), at x = 0

\frac{\partial}{\partial z} (r cos (t))

1 - 2 e^{- 2 t}

\int e^{4 x} 1 + e^{2 x} d x

x \to 3 lim (2 x^{2} + 4 x + 1)

What is the integral from 0 to 19/6 of pi(19/6-x)^2 ?
The integral from 0 to 19/6 of pi(19/6-x)^2 is (6859pi)/(648)