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

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

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

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

\frac{a ^{2}}{4}

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

\int_{1}^{9} 9 x^{2} d x

y = \frac{88 t}{99 t - 88}

\frac{d}{d x} (x - y^{2})

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

y^{^{'}} = \frac{3 x ^{2} + y ^{2}}{x y}

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