f(x) defines a semicircle of radius 2 and centre (2,0), assuming the positive square root. It derives from the equation y^2=4x-x^2 or y^2-4=-(4-4x+x^2)=-(x-2)^2 or (x-2)^2+y^2=4, which is a circle of radius, r=2 centre at (2,0). f(x) is the semicircle above the x-axis. The area of the circle=πr^2=4π.
(a) A(x) is the area inside the semicircle at various points defined by x. The area of the semicircle is πr^2/2=2π and A(0)=0 because it's the starting point for the semicircle. A(2) is a quadrant so its area is half the area of the semicircle=π. A(4) is the area of the semicircle=2π. A(x)=∫√(4x-x^2) dx=∫√(4-(x-2)^2) dx.
Let x-2=2sin(u) for 0≤x≤4, or -π/2≤u≤π/2; u=arcsin((x-2)/2); √(4-(x-2)^2)=√(4-4sin^2(u))=√(4(1-sin^2(u))=√4cos^2(u)=2cos(u); dx=2cos(u)du; A(x)=∫2cos(u).2cos(u)du=∫4cos^2(u) du. cos(2u)=2cos^2(u)-1, so 4cos^2(u)=2(cos(2u)+1).
A(x)=∫2(cos(2u)+1) du=sin(2u)+2u=2sin(u)cos(u)+2u=(x-2)√(1-((x-2)/2)^2)+2arcsin((x-2)/2)=(x-2)√(4-x^2+4x-4)/2+2arcsin((x-2)/2)=((x-2)/2)√(4x-x^2)+2arcsin((x-2)/2). Now apply the limits {0,x}:
[((x-2)/2)√(4x-x^2)+2arcsin((x-2)/2)]{0,x}=((x-2)/2)√(4x-x^2)+2arcsin((x-2)/2)+π.
A(0)=0, so A(2)=π, A(4)=2π, as we obtained geometrically earlier.
(b) First, the graph of f(x):
Now A(x) (red) and f(x) (blue):
A(3)-A(1) represents the area under the blue curve between x=1 and x=3. In the plot of A(x) it can be seen by subtracting the values on an almost straight part of the curve for A(x). A(3) is about 5.05 and A(1) about 1.25. So A(3)-A(1)=3.8 approx.