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

The integral from 0 to 2 of pi(x^2-L)^2 is pi(2L^2-(16L)/3+32/5)

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integral from 0 to 2 of pi(x^2-L)^2

\int_{0}^{2} π (x^{2} - L)^{2} d x

π (2 L^{2} - \frac{16 L}{3} + \frac{32}{5})

Solution steps

\int_{0}^{2} π (x^{2} - L)^{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}^{2} (- L + x^{2})^{2} d x

Expand

(- L + x^{2})^{2} : L^{2} - 2 L x^{2} + x^{4}

= π \cdot \int_{0}^{2} L^{2} - 2 L x^{2} + x^{4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= π (\int_{0}^{2} L^{2} d x - \int_{0}^{2} 2 L x^{2} d x + \int_{0}^{2} x^{4} d x)

\int_{0}^{2} L^{2} d x = 2 L^{2}

\int_{0}^{2} 2 L x^{2} d x = \frac{16 L}{3}

\int_{0}^{2} x^{4} d x = \frac{32}{5}

= π (2 L^{2} - \frac{16 L}{3} + \frac{32}{5})

\frac{2 x + 1}{x + 3}

y = \frac{1}{x}, y = 4 x, y = \frac{1}{4} x

f (t) = t cos (t) - sin (t)

\frac{d}{d x} (\frac{6}{x ^{4}} + \frac{1}{x ^{2}})

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

What is the integral from 0 to 2 of pi(x^2-L)^2 ?
The integral from 0 to 2 of pi(x^2-L)^2 is pi(2L^2-(16L)/3+32/5)