What is the integral from 0 to 2 of pi(4-x^2)^2 ?

The integral from 0 to 2 of pi(4-x^2)^2 is pi 256/15

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integral from 0 to 2 of pi(4-x^2)^2

\int_{0}^{2} π (4 - x^{2})^{2} d x

π \frac{256}{15}

Decimal

53.61651 \dots

Solution steps

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

Expand

(4 - x^{2})^{2} : 16 - 8 x^{2} + x^{4}

= π \cdot \int_{0}^{2} 16 - 8 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} 16 d x - \int_{0}^{2} 8 x^{2} d x + \int_{0}^{2} x^{4} d x)

\int_{0}^{2} 16 d x = 32

\int_{0}^{2} 8 x^{2} d x = \frac{64}{3}

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

= π (32 - \frac{64}{3} + \frac{32}{5})

Join

32 - \frac{64}{3} + \frac{32}{5} : \frac{256}{15}

= π \frac{256}{15}

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

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

n = 1 \sum \infty 8 (- 1)^{n} n x^{n}

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

\int \frac{x + 3}{x ^{2} + 6 x ^{4}} d x

What is the integral from 0 to 2 of pi(4-x^2)^2 ?
The integral from 0 to 2 of pi(4-x^2)^2 is pi 256/15