If _ - /2 ^ /2 x^3 x + 1 + ^2 x dx = ^2 2 , then the value of the constant is :

Step-by-step solution

Let I = _ - /2 ^ /2 x^3 x + 1 + ^2 x dx We can split the integral into two parts: I = _ - /2 ^ /2 x^3 x 1 + ^2 x dx + _ - /2 ^ /2 1 + ^2 x dx Let f(x) = x^3 x 1 + ^2 x . Since f(-x) = (-x)^3 (-x) 1 + ^2(-x) = -f(x) , the function is odd. Therefore, its integral over the symmetric interval [- /2, /2] is 0 . The second term is an even function, so we can write: I = 2 ₀^ /2 1 1 + ^2 x dx Dividing the numerator and denominator by ^2 x , we get: I = 2 ₀^ /2 ^2 x ^2 x + ^2 x dx I = 2 ₀^ /2 ^2 x 1 + 2 ^2 x dx Let t = x ,