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

The integral from 0 to 2 of 2pix(4-x^2) is 8pi

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

\int_{0}^{2} 2 π x (4 - x^{2}) d x

8 π

Decimal Notation

25.13274 \dots

Solution steps

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

Expand

x (4 - x^{2}) : 4 x - x^{3}

= 2 π \cdot \int_{0}^{2} 4 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}^{2} 4 x d x - \int_{0}^{2} x^{3} d x)

\int_{0}^{2} 4 x d x = 8

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

= 2 π (8 - 4)

Simplify

= 8 π

\frac{1}{2} (6 t^{2} + t)^{- 3}

\frac{d}{d x} (e^{3 s i n^{2} (2 x)})

\frac{\partial}{\partial x} (7 e^{x} ln (y))

\frac{d}{d x} (sec^{3} (5 x))

arctan (6 x)

What is the integral from 0 to 2 of 2pix(4-x^2) ?
The integral from 0 to 2 of 2pix(4-x^2) is 8pi