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(C) By the property of definite integrals,we can split the interval $[0, 2a]$ into $[0, a]$ and $[a, 2a]$.Thus,$\int_{0}^{2a} f(x) \, dx = \int_{0}^{a

Vedclass · Accepted Answer

$\int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(2a - x) \, dx$

Vedclass · Answer

(C) By the property of definite integrals,we can split the interval $[0, 2a]$ into $[0, a]$ and $[a, 2a]$.Thus,$\int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{a}^{2a} f(x) \, dx$.In the second integral,let $x = 2a - t$,then $dx = -dt$.When $x = a$,$t = a$,and when $x = 2a$,$t = 0$.So,$\int_{a}^{2a} f(x) \, dx = \int_{a}^{0} f(2a - t) (-dt) = \int_{0}^{a} f(2a - t) \, dt = \int_{0}^{a} f(2a - x) \, dx$.Therefore,$\int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(2a - x) \, dx$.

$\int_{0}^{2a} f(x) \, dx = $

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