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Integral Calculus

The value of the integral from 0 to (pi/2) of [f(x)/f(x)+f((pi/2)-x)] is?

Profile image of s adhithya s
16 Years agoGrade
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1 Answer

Profile image of AskiitiansExpert Abhinav Batra
16 Years ago

Dear s adhitya

Suppose I=(0 to π/2)∫f(x)dx/f(x) + f(π/2-x)

then I = (0 to π/2)∫f(π/2-x)dx/f(π/2-x) + f(x)   (since (0 to a )∫f(x)=(0 to a) ∫ƒ(a-x))

 

Adding the two integrals

2I =(0 to π/2)∫f(x)dx/f(x) + f(π/2-x) + (0 to π/2)∫f(π/2-x)dx/f(π/2-x) + f(x)

2I=(0 to π/2)∫{f(x)+f(π/2-x)}dx/f(x) + f(π/2-x)

2I=(0 to π/2)∫dx

2I=π/2

I=π/4

 

ALL THE BEST

Abhinav Batra