What is the integral from 0 to 1 of 2pix(18-18x) ?

The integral from 0 to 1 of 2pix(18-18x) is 6pi

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integral from 0 to 1 of 2pix(18-18x)

\int_{0}^{1} 2 π x (18 - 18 x) d x

6 π

Decimal Notation

18.84955 \dots

Solution steps

\int_{0}^{1} 2 π x (18 - 18 x) 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}^{1} x (18 - 18 x) d x

Apply Integration By Parts

: [9 x^{2} (- x + 1) - \int - 9 x^{2} d x]_{0}^{1}

= 2 π [9 x^{2} (- x + 1) - \int - 9 x^{2} d x]_{0}^{1}

\int - 9 x^{2} d x = - 3 x^{3}

= 2 π [9 x^{2} (- x + 1) - (- 3 x^{3})]_{0}^{1}

Simplify

2 π [9 x^{2} (- x + 1) - (- 3 x^{3})]_{0}^{1} : 2 π [- 6 x^{3} + 9 x^{2}]_{0}^{1}

= 2 π [- 6 x^{3} + 9 x^{2}]_{0}^{1}

Compute the boundaries:

3

= 2 π 3

Simplify

= 6 π

\int sin^{2} (x) cos (x) ln (sin (x)) d x

\int sin^{5} (x) \cdot cos^{2} (x) d x

\frac{d}{d x} (x^{3} - 2 x + 1)

\frac{\partial}{\partial x} (2 tan (x y))

y^{^{'}} + a y = 0

What is the integral from 0 to 1 of 2pix(18-18x) ?
The integral from 0 to 1 of 2pix(18-18x) is 6pi