Let f and g be the functions given by #f(x)=1+sin(2x)# and #g(x)=e^(x/2)#. Let R be the region in the first quadrant enclosed by the graphs of f and g. How do you find the area?

1 Answer
Feb 14, 2017

Answer:

#0.4291# areal units..

Explanation:

The points of intersection are given by

#y = 1+sin2x=e^(x/2). The y-intercept ( x = 0 ) for both are the same 1.

So, one common point is (0, 1). Glory to Socratic utility, the other is

approximated graphically to 4-sd as 1.136.

Now, the area is

#int (f-g) dx#, for x from 0 to 1.136

#int ((1+sin (2x) )-(e^(x/2)) dx#, for x from 0 to 1.136

#=[(x-1/2cos(2x))-(2e^(x/2))]#, between 0 and 1.136

#=(1.136-1/2cos(2.272radian)-2e^(1.136/2))-(-5/2)#

#=0.4291# areal units.

graph{(1+sin(2x)-y)(e^(x/2)-y)=0x^2 [-1, 9, -2.5, 2.5]}

graph{1+sin(2x)-e^(x/2) [1.13, 1.14, -.01, .01]}