We now must find the antiderivative of f '(x). Even though C is unspecified, the anti-derivative of C will be Cx. Also, a second constant of integration, which we'll call D is required. Thus, we have that the antiderivative of f '(x) = 92x2 − cos x + C is

The antiderivative of a function, also known as the integral, is the reverse of taking the derivative.

Given the function f '(x) = 92x^2 - cos(x) + C, we can find the antiderivative by integrating each term separately.

1. The antiderivative of 92x^2 is (92/3)x^3. This comes from the power rule of integration, which states that the integral of x^n dx is (1/(n+1))x^(n+1).

2. The antiderivative of -cos(x) is -sin(x). This is because the derivative of sin(x) is cos(x), so the antiderivative of cos(x) is sin(x).

3. The antiderivative of a constant C is Cx. This is because the derivative of Cx is C.

4. Finally, when finding the antiderivative, we add a constant of integration, which we'll call D. This is because the derivative of a constant is zero, so when we reverse the process, we add a constant.

So, the antiderivative of f '(x) = 92x^2 - cos(x) + C is (92/3)x^3 - sin(x) + Cx + D.