Evaluate: int_{-a}^a f(x) dx - int_0^a f(-x) | vedclass

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(C) We know that the property of definite integrals states: $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$. By substituting $x = -t$ i

Vedclass · Accepted Answer

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

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(C) We know that the property of definite integrals states: $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$. By substituting $x = -t$ in the first integral $\int_{-a}^0 f(x) dx$,we get $\int_a^0 f(-t) (-dt) = \int_0^a f(-t) dt = \int_0^a f(-x) dx$. Therefore,$\int_{-a}^a f(x) dx = \int_0^a f(-x) dx + \int_0^a f(x) dx$. Substituting this into the given expression: $(\int_0^a f(-x) dx + \int_0^a f(x) dx) - \int_0^a f(-x) dx = \int_0^a f(x) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,the result is $\int_0^a f(a-x) dx$.

Evaluate: $\int_{-a}^a f(x) dx - \int_0^a f(-x) dx$

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