What is the integral from 0 to L of sin((npix)/L) ?

The integral from 0 to L of sin((npix)/L) is (L(-(-1)^n+1))/(pin)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to L of sin((npix)/L)

\int_{0}^{L} sin (\frac{nπ x}{L}) d x

\frac{L ( - ( - 1 ) ^{n} + 1 )}{πn}

Solution steps

\int_{0}^{L} sin (\frac{nπ x}{L}) d x

Apply u-substitution

: \int_{0}^{n L π} \frac{sin ( \frac{u}{L} )}{πn} d u

= \int_{0}^{n L π} \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

= \frac{1}{πn} \cdot \int_{0}^{n L π} sin (\frac{u}{L}) d u

Apply u-substitution

: \int_{0}^{πn} sin (v) L d v

= \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

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

Use the common integral:

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

= \frac{1}{πn} L [- cos (v)]_{0}^{πn}

Compute the boundaries:

- (- 1)^{n} + 1

= \frac{1}{πn} L (- (- 1)^{n} + 1)

Simplify

= \frac{L ( - ( - 1 ) ^{n} + 1 )}{πn}

\int_{0}^{1} \frac{1}{4 + x} d x

\int_{1}^{4} (x^{2} - 9 x) d x

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

\int_{0}^{1} x (a + b x^{2}) d x

\int_{0}^{2} π (2 x^{2})^{2} d x

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