What is the integral from 0 to 1 of 2pix(33-33x) ?

The integral from 0 to 1 of 2pix(33-33x) is 11pi

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integral from 0 to 1 of 2pix(33-33x)

\int_{0}^{1} 2 π x (33 - 33 x) d x

11 π

Decimal Notation

34.55751 \dots

Solution steps

\int_{0}^{1} 2 π x (33 - 33 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 (33 - 33 x) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} (33 - 33 x) - \int - \frac{33 x ^{2}}{2} d x]_{0}^{1}

= 2 π [\frac{1}{2} x^{2} (33 - 33 x) - \int - \frac{33 x ^{2}}{2} d x]_{0}^{1}

\int - \frac{33 x ^{2}}{2} d x = - \frac{11 x ^{3}}{2}

= 2 π [\frac{1}{2} x^{2} (33 - 33 x) - (- \frac{11 x ^{3}}{2})]_{0}^{1}

Simplify

2 π [\frac{1}{2} x^{2} (33 - 33 x) - (- \frac{11 x ^{3}}{2})]_{0}^{1} : 2 π [\frac{1}{2} (33 x^{2} - 22 x^{3})]_{0}^{1}

= 2 π [\frac{1}{2} (33 x^{2} - 22 x^{3})]_{0}^{1}

Compute the boundaries:

\frac{11}{2}

= 2 π \frac{11}{2}

Simplify

= 11 π

x \to 0 lim (\frac{\frac{4}{x + 4} - 1}{x})

\int_{0}^{1} \frac{73}{x ^{5}} d x

x \to - 1 lim (\frac{in x}{x - 1})

\frac{d y}{d x} = 2 x y^{2} + 72 x

f (x) = 4 x^{2} + 3 x

What is the integral from 0 to 1 of 2pix(33-33x) ?
The integral from 0 to 1 of 2pix(33-33x) is 11pi