What is the integral from 0 to 1 of 2pix(21-21x) ?

The integral from 0 to 1 of 2pix(21-21x) is 7pi

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integral from 0 to 1 of 2pix(21-21x)

\int_{0}^{1} 2 π x (21 - 21 x) d x

7 π

Decimal Notation

21.99114 \dots

Solution steps

\int_{0}^{1} 2 π x (21 - 21 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 (21 - 21 x) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} (21 - 21 x) - \int - \frac{21 x ^{2}}{2} d x]_{0}^{1}

= 2 π [\frac{1}{2} x^{2} (21 - 21 x) - \int - \frac{21 x ^{2}}{2} d x]_{0}^{1}

\int - \frac{21 x ^{2}}{2} d x = - \frac{7 x ^{3}}{2}

= 2 π [\frac{1}{2} x^{2} (21 - 21 x) - (- \frac{7 x ^{3}}{2})]_{0}^{1}

Simplify

2 π [\frac{1}{2} x^{2} (21 - 21 x) - (- \frac{7 x ^{3}}{2})]_{0}^{1} : 2 π [\frac{1}{2} (21 x^{2} - 14 x^{3})]_{0}^{1}

= 2 π [\frac{1}{2} (21 x^{2} - 14 x^{3})]_{0}^{1}

Compute the boundaries:

\frac{7}{2}

= 2 π \frac{7}{2}

Simplify

= 7 π

(t - 1) cos (4 t - 4)

\frac{\partial}{\partial x} (e^{- t} \cdot cos (π x))

\frac{2}{x - 8} (10.1)

\int r 1 - r^{2} d r

7 x^{2} - 3 x, x^{3} - 7 x + 2

What is the integral from 0 to 1 of 2pix(21-21x) ?
The integral from 0 to 1 of 2pix(21-21x) is 7pi