(a) For some given x, A(x) gives the signed area under f(t) on the interval [a, x]. Since f(t) is simple to integrate, and with a = 2, you can compute it exactly:
A(x) = ∫₂ˣ f(t) dt
A(x) = ∫₂ˣ (3t + 2) dt
A(x) = (3/2 t ² + 2t ) |₂ˣ
A(x) = 3/2 x ² + 2x - 10
which is parabolic and easy to graph.
(b) Differentiating A(x) is also easy if you use the "closed" form above:
A'(x) = (3/2 x ² + 2x - 10)'
A'(x) = 3x + 2
and this function tells you how the area under f(t) changes as you adjust the size of the integration interval.
You can get the same result using the fundamental theorem of calculus:
A'(x) = d/dx ∫₂ˣ f(t) dt = f(x) = 3x + 2