What is the integral from 0 to 2 of sin(npix) ?

The integral from 0 to 2 of sin(npix) is 0

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integral from 0 to 2 of sin(npix)

\int_{0}^{2} sin (nπ x) d x

0

Solution steps

\int_{0}^{2} sin (nπ x) d x

Apply u-substitution

: \int_{0}^{nπ 2} \frac{sin ( u )}{πn} d u

= \int_{0}^{nπ 2} \frac{sin ( u )}{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}^{nπ 2} sin (u) d u

Use the common integral:

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

= \frac{1}{πn} [- cos (u)]_{0}^{nπ 2}

Compute the boundaries:

0

= \frac{1}{πn} \cdot 0

Simplify

= 0

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What is the integral from 0 to 2 of sin(npix) ?
The integral from 0 to 2 of sin(npix) is 0