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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{x}{2} - \frac{l}{4 π} sin (\frac{2 π x}{l}) d x = \frac{x ^{2}}{4} + l \frac{l}{8 π ^{2}} cos (\frac{2 π x}{l})

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

(5.1) (3.6)

\int \frac{7 x + 2}{x ^{2} - 4} d x

\frac{d}{d x} (\frac{( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} )})

\int e^{x} sin^{2} (x) d x

\frac{d y}{d x} = \frac{x}{y}

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