What is the integral from 0 to l of sin((pix)/l) ?

The integral from 0 to l of sin((pix)/l) is (2l)/pi

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integral from 0 to l of sin((pix)/l)

\int_{0}^{l} sin (\frac{π x}{l}) d x

\frac{2 l}{π}

Solution steps

\int_{0}^{l} sin (\frac{π x}{l}) d x

Apply u-substitution

: \int_{0}^{π} \frac{l sin ( u )}{π} d u

= \int_{0}^{π} \frac{l sin ( u )}{π} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{l}{π} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{l}{π} \cdot 2

Simplify

= \frac{2 l}{π}

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What is the integral from 0 to l of sin((pix)/l) ?
The integral from 0 to l of sin((pix)/l) is (2l)/pi