Evaluate the definite integral: int_0^{2a} f( | vedclass

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(B) We are given the integral $I = \int_0^{2a} f(x) dx$. Using the property of definite integrals,we can split the interval: $I = \int_0^a f(x) dx + \

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$\int_0^a (f(x) + f(2a - x)) dx$

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(B) We are given the integral $I = \int_0^{2a} f(x) dx$. Using the property of definite integrals,we can split the interval: $I = \int_0^a f(x) dx + \int_a^{2a} f(x) dx$. In the second integral,let $x = 2a - t$. Then $dx = -dt$. When $x = a$,$t = a$,and when $x = 2a$,$t = 0$. Substituting these into the second integral: $\int_a^{2a} f(x) dx = \int_a^0 f(2a - t) (-dt) = \int_0^a f(2a - t) dt = \int_0^a f(2a - x) dx$. Thus,$I = \int_0^a f(x) dx + \int_0^a f(2a - x) dx = \int_0^a (f(x) + f(2a - x)) dx$. Comparing this with the given options,option $(b)$ is the correct representation.

List-$I$	List-$II$
$I. \int_{-1}^1 x\|x\| dx$	$(a) \frac{\pi}{2}$
$II. \int_0^{\pi/2} \left(1 + \log \left(\frac{4+3\sin x}{4+3\cos x}\right)\right) dx$	$(b) \int_0^a 2f(x) dx$
$III. \int_0^a f(x) dx$	$(c) \int_0^a [f(x) + f(-x)] dx$
$IV. \int_{-a}^a f(x) dx$	$(d) 0$
	$(e) \int_0^a f(a-x) dx$

Evaluate the definite integral: $\int_0^{2a} f(x) dx$

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