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If \(\rm \int_0^a \left[f(x)+f(-x)\right]dx=\int_{-a}^{\ \ a} g(x)\ dx\), then what is g(x) equal to?

1. f(x)
2. f(-x) + f(x)
3. -f(x)
4. None of the above.

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Correct Answer - Option 2 : f(-x) + f(x)


Definite Integrals:

\(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).

If f(x) = f(2a - x), then

\(\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx\).

A function f(x) is:

  • Even, if f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
  • Odd, if f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
  • Periodic, if f(np ± x) = f(x), for some number p and n ∈ Z.


We know that for an even function,

f(-x) = f(x) and \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).

For the given condition,

\(\rm \int_0^a \left[f(x)+f(-x)\right]dx=\int_{-a}^{\ \ a} g(x)\ dx\) to be true,

f(x) and g(x) both must be even functions,

i.e. f(x) = f(-x) and g(x) = f(x) + f(-x).

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