Question

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?

Answer 1

`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]}