What is the integral from (-pi)/2 to pi/2 of g(x) ?

The integral from (-pi)/2 to pi/2 of g(x) is 0

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integral from (-pi)/2 to pi/2 of g(x)

\int_{\frac{- π}{2}}^{\frac{π}{2}} g (x) d x

0

Solution steps

\int_{\frac{- π}{2}}^{\frac{π}{2}} gx d x

Take the constant out:

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

= g \cdot \int_{\frac{- π}{2}}^{\frac{π}{2}} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{\frac{- π}{2}}^{\frac{π}{2}}

= g [\frac{x ^{2}}{2}]_{\frac{- π}{2}}^{\frac{π}{2}}

Simplify

= g [\frac{x ^{2}}{2}]_{- \frac{π}{2}}^{\frac{π}{2}}

Compute the boundaries:

0

= g \cdot 0

Simplify

= 0

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What is the integral from (-pi)/2 to pi/2 of g(x) ?
The integral from (-pi)/2 to pi/2 of g(x) is 0