If ₀^ k x^2+1 2x^2 - 2kx + k^2 + 2 dx = 4 , where k > 0 , then the value of k is

Step-by-step solution

Let I = ₀^ k x^2+1 2x^2 - 2kx + k^2 + 2 dx . Notice that the denominator can be rewritten as (x^2+1) + ((k-x)^2+1) . Thus, I = ₀^ k x^2+1 (x^2+1) + ((k-x)^2+1) dx (1) Using the property ₀^ a f(x) dx = ₀^ a f(a-x) dx , we replace x with k-x : I = ₀^ k (k-x)^2+1 ((k-x)^2+1) + (x^2+1) dx (2) Adding equations (1) and (2), we get: 2I = ₀^ k (x^2+1) + ((k-x)^2+1) (x^2+1) + ((k-x)^2+1) dx 2I = ₀^ k 1 dx = k I = k 2 Given that I = 4 , we have: k 2 = 4 k = 8 Answer: 8