What is the integral from 0 to 4 of pi(16-x^2) ?

The integral from 0 to 4 of pi(16-x^2) is pi 128/3

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integral from 0 to 4 of pi(16-x^2)

\int_{0}^{4} π (16 - x^{2}) d x

π \frac{128}{3}

Decimal

134.04128 \dots

Solution steps

\int_{0}^{4} π (16 - 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}^{4} 16 - x^{2} 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}^{4} 16 d x - \int_{0}^{4} x^{2} d x)

\int_{0}^{4} 16 d x = 64

\int_{0}^{4} x^{2} d x = \frac{64}{3}

= π (64 - \frac{64}{3})

Join

64 - \frac{64}{3} : \frac{128}{3}

= π \frac{128}{3}

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

\frac{d}{d x} (arcsin (\frac{1}{e ^{x}}))

\frac{d}{d x} (\frac{2 + sin ( x )}{x})

f (x) = 2 x^{3} - 5 x, at x = - 1

\frac{1}{( t ^{5} + 4 ) ^{4}}

What is the integral from 0 to 4 of pi(16-x^2) ?
The integral from 0 to 4 of pi(16-x^2) is pi 128/3