What is the integral from (-1)/8 to 1/8 of sin(2pix) ?

The integral from (-1)/8 to 1/8 of sin(2pix) is 0

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integral from (-1)/8 to 1/8 of sin(2pix)

\int_{\frac{- 1}{8}}^{\frac{1}{8}} sin (2 π x) d x

0

Solution steps

\int_{\frac{- 1}{8}}^{\frac{1}{8}} sin (2 π x) d x

Apply u-substitution

: \int_{- \frac{π}{4}}^{\frac{π}{4}} sin (u) \frac{1}{2 π} d u

= \int_{- \frac{π}{4}}^{\frac{π}{4}} sin (u) \frac{1}{2 π} d u

Take the constant out:

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

= \frac{1}{2 π} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} sin (u) d u

Use the common integral:

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

= \frac{1}{2 π} [- cos (u)]_{- \frac{π}{4}}^{\frac{π}{4}}

Compute the boundaries:

0

= \frac{1}{2 π} \cdot 0

Simplify

= 0

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What is the integral from (-1)/8 to 1/8 of sin(2pix) ?
The integral from (-1)/8 to 1/8 of sin(2pix) is 0