int_0^{2a} f(x) dx - int_a^{2a} f(x) dx = | vedclass

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(A) We are given the expression: $I = \int_0^{2a} f(x) dx - \int_a^{2a} f(x) dx$. Using the property of definite integrals,$\int_0^{2a} f(x) dx = \int

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$\int_0^a f(x) dx$

Vedclass · Answer

(A) We are given the expression: $I = \int_0^{2a} f(x) dx - \int_a^{2a} f(x) dx$. Using the property of definite integrals,$\int_0^{2a} f(x) dx = \int_0^a f(x) dx + \int_a^{2a} f(x) dx$. Substituting this into the expression,we get: $I = (\int_0^a f(x) dx + \int_a^{2a} f(x) dx) - \int_a^{2a} f(x) dx$. Simplifying the expression,the term $\int_a^{2a} f(x) dx$ cancels out. Therefore,$I = \int_0^a f(x) dx$.

$\int_0^{2a} f(x) dx - \int_a^{2a} f(x) dx =$

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