Supposing f(x) = 20/(1+x^2) and g(x) = 2, then to find R is equivalent to finding int_a^bf(x) - g(x) dx where x=a and x=b are the two x-values at which the functions f and g intersect.
This gives the area between the two curves (one curve and one line, rather). Function f never goes below the x-axis, so the area is by definition restricted to the first two quadrants.
Start by integrating f(x) using the trigonometric substitution x=tan theta (recognising the Pythagorean Identity 1 + tan^2theta = sec^2theta
int 20/(1+x^2) dx -= 20*int 1/sec^2theta dx
Using the change of variable rule int f(x) dx = f(x) dx/(d theta) d theta we get dx/(d theta) = sec^2theta.
Hence 20*int 1/sec^2theta dx -= 20*int d theta
:. int f(x) dx = 20arctanx + c
int g(x) dx = 2x + c, of course.
Now, we must find where f and g intersect by letting f(x) = g(x).
20/(1+x^2) = 2
:. 20 = 2(1+x^2)
:. x^2 - 9 = 0
:. x= 3 or -3.
Hence: R = int_-3^3 f(x) dx - int_-3^3 g(x) dx
:. R = [20arctanx]_-3^3 - [2x]_3^-3
=20arctan3 - 20arctan(-3) - (6 + 6)
= 40arctan3 - 12 (exact answer).
:. R~~37.96
