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?

Question

Samantha Adkison · Accepted Answer

#:. R~~37.96# 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#

Evelyn Agard · Answer

undefined To find the area of the region $R$, bounded above by the graph of $y = \frac{20}{1+x^2}$ and below by the horizontal line $y = 2$ in the first and second quadrants, you need to integrate the difference between the upper and lower functions with respect to $x$. The bounds of integration will be the x-values where the two functions intersect. The area can be calculated using the definite integral:

$$ A = \int_{a}^{b} \left( \frac{20}{1+x^2} - 2 \right) dx $$

Where $a$ and $b$ are the x-coordinates of the points where $y = \frac{20}{1+x^2}$ intersects $y = 2$. These points can be found by solving the equation:

$$ \frac{20}{1+x^2} = 2 $$

Once you find the x-values of the intersection points, you can integrate the expression over that interval to find the area.

Emma Aldridge · Answer

undefined To find the area of the region R bounded by the given curves, you can use the definite integral. The integral of the upper curve minus the lower curve over the given interval will give the area of the region. In this case, the upper curve is y = 20/(1 + x^2) and the lower curve is y = 2. So, the integral to find the area A is:

A = ∫[a, b] (20/(1 + x^2) - 2) dx

where [a, b] is the interval where the curves intersect. To find the intersection points, set the equations of the curves equal to each other and solve for x. Then, evaluate the integral over this interval to find the area of region R.