What is the integral from 0 to l of lsin((npix)/l) ?

The integral from 0 to l of lsin((npix)/l) is (l^2(-(-1)^n+1))/(pin)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to l of lsin((npix)/l)

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

\frac{l ^{2} ( - ( - 1 ) ^{n} + 1 )}{πn}

Solution steps

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

Take the constant out:

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

= l \cdot \int_{0}^{l} sin (\frac{nπ x}{l}) d x

Apply u-substitution

= l \cdot \int_{0}^{nπ l} \frac{sin ( \frac{u}{l} )}{πn} d u

Take the constant out:

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

= l \frac{1}{πn} \cdot \int_{0}^{nπ l} sin (\frac{u}{l}) d u

Apply u-substitution

= l \frac{1}{πn} \cdot \int_{0}^{πn} sin (v) l d v

Take the constant out:

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

= l \frac{1}{πn} l \cdot \int_{0}^{πn} sin (v) d v

Use the common integral:

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

= l \frac{1}{πn} l [- cos (v)]_{0}^{πn}

Simplify

l \frac{1}{πn} l [- cos (v)]_{0}^{πn} : l^{2} \frac{1}{πn} [- cos (v)]_{0}^{πn}

= l^{2} \frac{1}{πn} [- cos (v)]_{0}^{πn}

Compute the boundaries:

- (- 1)^{n} + 1

= l^{2} \frac{1}{πn} (- (- 1)^{n} + 1)

Simplify

= \frac{l ^{2} ( - ( - 1 ) ^{n} + 1 )}{πn}

What is the integral from 0 to l of lsin((npix)/l) ?
The integral from 0 to l of lsin((npix)/l) is (l^2(-(-1)^n+1))/(pin)