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 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(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.

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

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