Indefinite Integration 1 Question 37

38. Let f:RR and g:RR be continuous functions. Then, the value of the integral π/2π/2[f(x)+f(x)][g(x)g(x)]dx is

(1990, 2M)

(a) π

(b) 1

(c) -1

(d) 0

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Solution:

  1. Let I=π/2π/2[f(x)+f(x)][g(x)g(x)]dx

Let φ(x)=[f(x)+f(x)][g(x)g(x)]

φ(x)=[f(x)+f(x)][g(x)g(x)]

φ(x)=φ(x)

φ(x) is an odd function.

π/2π/2φ(x)dx=0



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