What is the integral from 0 to l of xsin^2((2pix)/l) ?

The integral from 0 to l of xsin^2((2pix)/l) is (l^2)/4

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integral from 0 to l of xsin^2((2pix)/l)

\int_{0}^{l} x sin^{2} (\frac{2 π x}{l}) d x

\frac{l ^{2}}{4}

Solution steps

\int_{0}^{l} x sin^{2} (\frac{2 π x}{l}) d x

Apply Integration By Parts

: [\frac{1}{2} x (x - \frac{1}{4 π} l sin (\frac{4 π x}{l})) - \int \frac{1}{2} (x - \frac{1}{4 π} l sin (\frac{4 π x}{l})) d x]_{0}^{l}

= [\frac{1}{2} x (x - \frac{1}{4 π} l sin (\frac{4 π x}{l})) - \int \frac{1}{2} (x - \frac{1}{4 π} l sin (\frac{4 π x}{l})) d x]_{0}^{l}

\int \frac{1}{2} (x - \frac{1}{4 π} l sin (\frac{4 π x}{l})) d x = \frac{1}{2} (\frac{x ^{2}}{2} + l \frac{l}{16 π ^{2}} cos (\frac{4 π x}{l}))

= [\frac{1}{2} x (x - \frac{1}{4 π} l sin (\frac{4 π x}{l})) - \frac{1}{2} (\frac{x ^{2}}{2} + l \frac{l}{16 π ^{2}} cos (\frac{4 π x}{l}))]_{0}^{l}

Simplify

[\frac{1}{2} x (x - \frac{1}{4 π} l sin (\frac{4 π x}{l})) - \frac{1}{2} (\frac{x ^{2}}{2} + l \frac{l}{16 π ^{2}} cos (\frac{4 π x}{l}))]_{0}^{l} : [\frac{1}{32 π ^{2}} (8 π^{2} x^{2} - 4 π l x sin (\frac{4 π x}{l}) - l^{2} cos (\frac{4 π x}{l}))]_{0}^{l}

= [\frac{1}{32 π ^{2}} (8 π^{2} x^{2} - 4 π l x sin (\frac{4 π x}{l}) - l^{2} cos (\frac{4 π x}{l}))]_{0}^{l}

Compute the boundaries:

\frac{8 π ^{2} - 1}{32 π ^{2}} l^{2} + \frac{l ^{2}}{32 π ^{2}}

= \frac{8 π ^{2} - 1}{32 π ^{2}} l^{2} + \frac{l ^{2}}{32 π ^{2}}

Simplify

= \frac{l ^{2}}{4}

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What is the integral from 0 to l of xsin^2((2pix)/l) ?
The integral from 0 to l of xsin^2((2pix)/l) is (l^2)/4