# 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?

Feb 14, 2017

$0.4291$ 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

$\int \left(f - g\right) \mathrm{dx}$, for x from 0 to 1.136

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

$= \left[\left(x - \frac{1}{2} \cos \left(2 x\right)\right) - \left(2 {e}^{\frac{x}{2}}\right)\right]$, between 0 and 1.136

$= \left(1.136 - \frac{1}{2} \cos \left(2.272 r a \mathrm{di} a n\right) - 2 {e}^{\frac{1.136}{2}}\right) - \left(- \frac{5}{2}\right)$

$= 0.4291$ 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]}