intlimits_{ - a}^a {f(x),dx} = | vedclass

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Question

(A) Let $I = \int\limits_{-a}^a f(x)\,dx$. Using the property $\int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx$,we have $I = \int\limits_{-a}^

Vedclass · Accepted Answer

$\int\limits_0^a {[f(x) + f(-x)]\,dx}$

Vedclass · Answer

(A) Let $I = \int\limits_{-a}^a f(x)\,dx$. Using the property $\int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx$,we have $I = \int\limits_{-a}^a f(-x)\,dx$. Adding the two expressions for $I$: $2I = \int\limits_{-a}^a [f(x) + f(-x)]\,dx$. Since the integrand $g(x) = f(x) + f(-x)$ is an even function (because $g(-x) = f(-x) + f(x) = g(x)$),we can write: $2I = 2\int\limits_0^a [f(x) + f(-x)]\,dx$. Dividing by $2$,we get $I = \int\limits_0^a [f(x) + f(-x)]\,dx$. Thus,the correct option is $A$.

$\int\limits_{ - a}^a {f(x)\,dx} = $

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