Let R be the region in the first and second quadrants bounded above by the graph of #y=20/(1+x^2)# and below by the horizontal line y=2, how do you find the area?

1 Answer
Jan 30, 2018

Answer:

#:. R~~37.96#

Explanation:

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#