First slide
Evaluation of definite integrals
Question

If f:RR, g:RR are continuous functions then the value of the integral

π/2π/2[(f(x)+f(x))(g(x)  g(x))] dx is 

Easy
Solution

Let F(x) =(f(x)+f(-x))(g(x)-g(-x)) so F(x)=(f(x)+f(x))(g(x)g(x))=F(x).

Hence  π/2π/2f(x)dx=0. (since F(x) is odd function)

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