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Show that $\int\limits_0^af(x)g(x)dx=2\int\limits_0^af(x)dx,$ if \(f\) and \(g\) are defined as \(f(x) = f(a - x)\) and \(g(x) + g(a - x)=4\)

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  • (i)$\int \limits_a^bf(x)dx=F(b)-F(a)$
  • (ii)$\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
Given $I=\int \limits_0^a f(x)g(x)dx-----(1)$
Now applyimg property $\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
$I=\int \limits_0^a f(a-x)g(a-x)dx$
$I=\int \limits_0^a f(x)g(a-x)dx------(2)$
Adding equ (1) and equ (2)
$2I=\int \limits_0^a f(x)g(x)+f(x)g(a-x)dx$
$I=\int \limits_0^a f(x)\{g(x)+g(a-x)\}dx$
But $ g(x)+g(a-x)=4$
Therefore$ 2I=\int f(x) \times 4dx$
$2I=4 \int f(x)dx$
Therefore $I=2\int f(x)dx$


answered Feb 14, 2013 by meena.p

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