Question

If h(x) = [f(x) + g(x)][g(x) – f(x)] where f is an odd and g is an even function the ∫h(x)dx for x ∈ [–π/2, π/2] is equal to....... 

(a) 0 

(b) (π/2) 

(c) ^(π/2)∫₀h(x)dx 

(d) 2 ^(π/2)∫₀h(x)dx

Answer 1

The correct option (d) 2 ^(π/2)∫₀h(x)dx  

Explanation:

h(x) = [f(x) + g(x)][g(x) – f(x)] 

h(– x) = [f(– x) + g(– x)][g(– x) – f(– x)] 

∵ f(x) is odd and g(x) is even 

∴ h(– x) = [– f(x) + g(x)][g(x) + f(x)] 

∴ h(– x) = h(x) --------- even function 

∴ I = ^(π/2)∫_[(–π)/2]h(x)dx = 2^(π/2)∫₀h(x)dx.