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?

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

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]}

To find the area of the region ( R ) enclosed by the graphs of the functions ( f(x) = 1 + \sin(2x) ) and ( g(x) = e^{x/2} ) in the first quadrant, you need to determine the points of intersection of the two functions, which will define the bounds of integration. Then, integrate the absolute difference of the two functions over those bounds.

1. Set ( f(x) = g(x) ) and solve for ( x ) to find the points of intersection. [ 1 + \sin(2x) = e^{x/2} ]

2. Solve for ( x ) to find the points of intersection.

3. Once you have the points of intersection, denote them as ( x_1 ) and ( x_2 ), where ( x_1 ) is the smaller value and ( x_2 ) is the larger value.

4. Integrate the absolute difference between the functions from ( x_1 ) to ( x_2 ). [ \text{Area} = \int_{x_1}^{x_2} |f(x) - g(x)| , dx ]

5. Evaluate the integral to find the area of the region ( R ).