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

The integral from 0 to a/2 of xsin((npix)/a) is (a^2sin((pin)/2))/(pi^2n^2)-(a^2cos((pin)/2))/(2pin)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to a/2 of xsin((npix)/a)

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

\frac{a ^{2} sin ( \frac{πn}{2} )}{π ^{2} n ^{2}} - \frac{a ^{2} cos ( \frac{πn}{2} )}{2 πn}

Solution steps

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

Apply Integration By Parts

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

= [- a \frac{1}{πn} x cos (\frac{πn x}{a}) - \int - a \frac{1}{πn} cos (\frac{πn x}{a}) d x]_{0}^{\frac{a}{2}}

\int - a \frac{1}{πn} cos (\frac{πn x}{a}) d x = - a \frac{a}{π ^{2} n ^{2}} sin (\frac{πn x}{a})

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

Simplify

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

Compute the boundaries:

\frac{a ^{2}}{π ^{2} n ^{2}} sin (\frac{πn}{2}) - \frac{a ^{2}}{2 πn} cos (\frac{πn}{2})

= \frac{a ^{2}}{π ^{2} n ^{2}} sin (\frac{πn}{2}) - \frac{a ^{2}}{2 πn} cos (\frac{πn}{2})

Simplify

= \frac{a ^{2} sin ( \frac{πn}{2} )}{π ^{2} n ^{2}} - \frac{a ^{2} cos ( \frac{πn}{2} )}{2 πn}

\int_{- 2}^{2} 2 - x d x

\int_{- 2}^{3} 3 x^{2} d x

\int_{0}^{2} (6 x - 3) (4 x^{2} + 4) d x

\int_{1}^{2} \frac{5}{x ^{3}} d x

\int_{1}^{8} x e^{x^{2}} d x

What is the integral from 0 to a/2 of xsin((npix)/a) ?
The integral from 0 to a/2 of xsin((npix)/a) is (a^2sin((pin)/2))/(pi^2n^2)-(a^2cos((pin)/2))/(2pin)