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

The integral from 0 to 4 of 2pi(4-x)(16-x^2) is (640pi)/3

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

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

\frac{640 π}{3}

Decimal Notation

670.20643 \dots

Solution steps

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

Take the constant out:

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

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

Expand

(4 - x) (16 - x^{2}) : 64 - 4 x^{2} - 16 x + x^{3}

= 2 π \cdot \int_{0}^{4} 64 - 4 x^{2} - 16 x + x^{3} 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

= 2 π (\int_{0}^{4} 64 d x - \int_{0}^{4} 4 x^{2} d x - \int_{0}^{4} 16 x d x + \int_{0}^{4} x^{3} d x)

\int_{0}^{4} 64 d x = 256

\int_{0}^{4} 4 x^{2} d x = \frac{256}{3}

\int_{0}^{4} 16 x d x = 128

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

= 2 π (256 - \frac{256}{3} - 128 + 64)

Simplify

2 π (256 - \frac{256}{3} - 128 + 64) : \frac{640 π}{3}

= \frac{640 π}{3}

\frac{d}{d x} (\frac{sin ( 3 x ) cos ( 2 x )}{3})

\int e^{x} e^{e^{x}} d x

n = 2 \sum \infty \frac{1}{n ^{2} + 6 n + 8}

- (3 x - 4)^{3}, 3

\int_{0}^{\frac{3 π}{2}} 5 ∣ sin (x) ∣ d x

What is the integral from 0 to 4 of 2pi(4-x)(16-x^2) ?
The integral from 0 to 4 of 2pi(4-x)(16-x^2) is (640pi)/3