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

The integral from-2 to 2 of xsin((npix)/2) is -(8(-1)^n)/(pin)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-2 to 2 of xsin((npix)/2)

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

- \frac{8 ( - 1 ) ^{n}}{πn}

Solution steps

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

Apply u-substitution

: \int_{- πn}^{πn} \frac{4 u sin ( u )}{π ^{2} n ^{2}} d u

= \int_{- πn}^{πn} \frac{4 u sin ( u )}{π ^{2} n ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{4}{π ^{2} n ^{2}} \cdot \int_{- πn}^{πn} u sin (u) d u

Apply Integration By Parts

: [- u cos (u) - \int - cos (u) d u]_{- πn}^{πn}

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) - \int - cos (u) d u]_{- πn}^{πn}

\int - cos (u) d u = - sin (u)

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) - (- sin (u))]_{- πn}^{πn}

Simplify

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) + sin (u)]_{- πn}^{πn}

Compute the boundaries:

- 2 π (- 1)^{n} n

= \frac{4}{π ^{2} n ^{2}} (- 2 π (- 1)^{n} n)

Simplify

= - \frac{8 ( - 1 ) ^{n}}{πn}

\int_{4}^{10} \frac{1}{x - 2} d x

\int_{5}^{2} \frac{( 12 - x ^{3} )}{x ^{4}} d x

\int_{1}^{3} e^{3 x} d x

\int_{0}^{2} 3 x^{2} + 2 x + 1 d x

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

What is the integral from-2 to 2 of xsin((npix)/2) ?
The integral from-2 to 2 of xsin((npix)/2) is -(8(-1)^n)/(pin)