If int_0^{2a} f(x) , dx = 2 int_0^a f(x) , dx | vedclass

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(B) We are given the property $\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx$. By substituting $x = 2a - t$ in the second inte

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$f(2a - x) = f(x)$

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(B) We are given the property $\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx$. By substituting $x = 2a - t$ in the second integral $\int_a^{2a} f(x) \, dx$,we get $dx = -dt$. When $x = a$,$t = a$,and when $x = 2a$,$t = 0$. Thus,$\int_a^{2a} f(x) \, dx = \int_a^0 f(2a - t) (-dt) = \int_0^a f(2a - t) \, dt = \int_0^a f(2a - x) \, dx$. Substituting this back,we have $\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx = \int_0^a [f(x) + f(2a - x)] \, dx$. Given that $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$,it follows that $\int_0^a [f(x) + f(2a - x)] \, dx = \int_0^a 2f(x) \, dx$. This equality holds if $f(2a - x) = f(x)$.

If $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$,then:

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