If int_{-1}^{1} f(x) , dx = 0,then | vedclass

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(B) The property of definite integrals states that for an odd function $f(x)$,where $f(-x) = -f(x)$,the integral over a symmetric interval $[-a, a]$ i

Vedclass · Accepted Answer

$f(-x) = -f(x)$

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(B) The property of definite integrals states that for an odd function $f(x)$,where $f(-x) = -f(x)$,the integral over a symmetric interval $[-a, a]$ is zero.Specifically,$\int_{-a}^{a} f(x) \, dx = 0$ if $f(x)$ is an odd function.In this case,the interval is $[-1, 1]$,so $\int_{-1}^{1} f(x) \, dx = 0$ implies that $f(x)$ is an odd function,which satisfies the condition $f(-x) = -f(x)$.Therefore,the correct option is $(b)$.

If $\int_{-1}^{1} f(x) \, dx = 0$,then

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