How do you find the area between #f(x)=2-1/2x# and #g(x)=2-sqrtx#?

Question

Emily Ammerman · Accepted Answer

The area is #4/3# square units.

Start by finding the intersection points.

#{(y = 2- 1/2x), (y = 2 - sqrt(x)):}#

#2 - sqrt(x) = 2 - 1/2x#

#1/2x = sqrt(x)#

#(1/2x)^2 = (sqrt(x))^2#

#1/4x^2 = x#

#1/4x^2 - x = 0#

#x(1/4x - 1) = 0#

#x = 0 and 4#

#:.y = 2 - sqrt(x)#

#y = 2 - sqrt(0) and y = 2 - sqrt(4)#

#y = 2 and 0#

The intersection points are hence #{0, 2}# and #{4, 0}#.

Now, we do a rudimentary sketch of the graphs on one grid (only if you don't have access to a graphing calculator, of course).

Our process here is to subtract the area of the lower graph (#y = 2-sqrt(x)#), from the area of the upper graph (#y = -1/2x + 2#). This will all be in the interval #0 ≤ x ≤ 4#.

We start by finding the area of the upper graph.

#int_0^4(-1/2x + 2)dx#

#=>(-1/4x^2 + 2x)|_0^4#

#=>-1/4(4^2) + 2(4) - (-1/4(0)^2 + 2(0))#

#=>-4 + 8 - 0#

#=> 4#

Now for the lower graph.

#int_0^4(2 - sqrt(x))dx#

#=>(2x - 2/3x^(3/2))|_0^4#

#=> 2(4) - 2/3(4)^(3/2) - (2(0) - 2/3(0)^(3/2))#

#=>8 - 16/3#

#=>8/3#

We now subtract the first area from the second:

#4 - 8/3 = 4/3#

Hence, the area of the region between #f(x) = 2 - 1/2x# and #g(x) = 2 - sqrt(x)# is #4/3" units"^2#.

Emily Ammerman · Answer

undefined To find the area between the curves $ f(x) = 2 - \frac{1}{2}x $ and $ g(x) = 2 - \sqrt{x} $, you need to compute the definite integral of $ f(x) - g(x) $ from the points where the curves intersect. First, find the intersection points by setting $ f(x) = g(x) $ and solving for $ x $. Then integrate $ f(x) - g(x) $ over the interval between those intersection points.