# FTC part 2 and the global minimum

I have a problem with applying FTC part 2. Using the function graph, I need to find the minimum value of the function g(x). The graph of the function f(x).

The function g(x) is given in the form:

$$g(x) = \int_2^x f(t)\, dt$$

I found that the function g(x) has an extremum at points 0, 4/3, 4. At point 4, the global maximum of the function is observed. I'm trying to explore integrals at two other points.

$$g(0) = \int_2^0 f(t)\, dt = -\int_0^2 f(t)\, dt = \frac{\pi}{4} -\frac{1}{2}$$.

$$g(4/3) = \int_2^\frac{4}{3} f(t)\, dt = -\int_\frac{4}{3}^2 f(t)\, dt = -\frac{2}{3}$$.

So the global minimum is $$-\frac{2}{3}$$, but the answer is not correct. Where was the mistake?

Remember that the global extrema don't necessarily have to occur at critical points. They can also occur at the endpoints/boundaries. It's fairly obvious that $$g(4) > 0$$, so all that's left is checking $$g(-2)$$ and comparing it to $$g\left(\dfrac{4}{3}\right)$$, since you've already eliminated $$g(0)$$:
$$g(-2) = \int_2^{-2} f(t) \mathrm dt = -\int_{-2}^{2} f(t) \mathrm dt = -\left[\frac{1}{4}\pi(2)^2-\frac{1}{4}\pi(1)^2-\frac{1}{2}\left(\frac{1}{3}\right)(1)+\frac{1}{2}\left(\frac{2}{3}\right)(2)\right] = -\left[\frac{3}{4}\pi+\frac{1}{2}\right] = -\frac{1}{2}-\frac{3}{4}\pi$$
Clearly, $$g(-2)$$ is less than $$g\left(\dfrac{4}{3}\right)$$, so the global minimum is $$g(-2) = -\dfrac{1}{2}-\dfrac{3}{4}\pi$$.