Let f: R (0, ) be a continuous function such that f(x)f( - x) = 4 for all x R . The value of the integral ₀^ x 2 + f(x) dx is equal to

Step-by-step solution

Let I = ₀^ x 2 + f(x) dx (1) Using the property ₀^a g(x) dx = ₀^a g(a-x) dx , we get: I = ₀^ ( - x) 2 + f( - x) dx I = ₀^ x 2 + f( - x) dx Given f(x)f( - x) = 4 , we substitute f( - x) = 4 f(x) : I = ₀^ x 2 + 4 f(x) dx I = ₀^ f(x) x 2f(x) + 4 dx I = 1 2 ₀^ f(x) x f(x) + 2 dx Multiplying by 2 on both sides: 2I = ₀^ f(x) x f(x) + 2 dx (2) From equation (1), multiplying by 2 yields: 2I = ₀^ 2 x f(x) + 2 dx (3) Adding equations (2) and (3): 4I = ₀^ f(x) x + 2 x f(x) + 2 dx 4I = ₀^ (f(x) + 2) x f(x) + 2 dx 4I = ₀^ x dx