your help would be appreciated
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Originally Posted by BlackRaven your help would be appreciated Because $f$ is an even function we have $\int_{ - 2}^2 {f(x)dx} = 2\int_0^2 {f(x)dx}$
I know. I just can't figure out how to find f(x) for the first integal... Is it just 16?
$\displaystyle \int_{-2}^2 f(x) - 8 \, dx = \int_{-2}^2 f(x) \, dx - \int_{-2}^2 8 \, dx = 2\int_0^2 f(x) \, dx - 2\int_0^2 8 \, dx = 2\int_0^2 f(x) \, dx - 32 = 8$ finish it ...