Let f and g be the functions given by f(x)=1+sin(2x) and g(x)=e^(x/2). Let R be the shaded region in the first quadrant enclosed by the graphs of f and g.
A. Find the the area of R.
B. Find the value of z so that x=z cuts the solid R into two parts with equal area.
C. Find the volume of slid generated when R is revolved around the x-axis.
I would start by sketching it
http://www.wolframalpha.com/input/?i=plot+y+%3D+1+%2B+sin%282x%29%2C+y+%3D+e%5E%28x%2F2%29
So now for the hard part of finding their intersection.
The left one is obviously (0,1)
How do they expect you to solve
1 + sin(2x) = e^(x/2) ??, Have you learned any methods such as Newton's Method ??
Wolfram says:
http://www.wolframalpha.com/input/?i=solve++1+%2B+sin%282x%29+%3D+e%5E%28x%2F2%29
x = 1.13569..
(I tested it , it works)
so now you have to integrate
1 + sin(2x) - e^(x/2) from x = 0 to x = 1.13569
The integration itself should pose no problem, just a lot of buttonpushing after that.
So that would be part A
Man, this is a major question!
This looks like a major assignment, and I just can't justify giving you the solution.
I gave you a good start, so hang in there.
hint for B.
you have to set the two integrals from 0 to z equal to (1.13569 - z) and solve for z
hint for C
use "washers" , that is the outer radius is the 1+sin2x
the inner radius is the e^(x/2) curve
I got .429 u^2 for A.
I don't understand your hint for B..
For C i for 1.348 u^3. Is that right?
To find the area of the shaded region R enclosed by the graphs of f and g in the first quadrant, we need to find the points of intersection between the two functions and then integrate the difference between the functions over the interval of intersection.
A. Finding the points of intersection:
To determine where the graphs of f(x) and g(x) intersect, we set them equal to each other:
1 + sin(2x) = e^(x/2)
To solve this equation, we can make a substitution. Let u = e^(x/2), then the equation becomes:
1 + sin(2x) = u
We can rewrite the equation as sin(2x) = u - 1
Now, we can square both sides of the equation (since sin^2(2x) = (u-1)^2) to get rid of the sine function:
(sin(2x))^2 = (u - 1)^2
Using the trigonometric identity: sin^2(2x) = (1 - cos(4x))/2, we can rewrite it as:
(1 - cos(4x))/2 = (u - 1)^2
Simplifying the equation gives us:
1 - cos(4x) = 2(u - 1)^2
cos(4x) = 1 - 2(u - 1)^2
Now, we need to solve the equation for x. Take the inverse cosine (arccos) of both sides:
4x = arccos(1 - 2(u - 1)^2)
Finally, divide both sides by 4 to isolate x:
x = (1/4) * arccos(1 - 2(u - 1)^2)
Using this equation, we can find the points of intersection by solving for x when f(x) = g(x), substituting the value of u we obtain for each intersection point into f(x) or g(x), and calculating the difference over the interval of intersection.
B. Finding the value of z:
To find the value of z, we need to find the value of x that cuts the solid R into two equal parts. This means that the integral of f(x) - g(x) over the interval [0, z] should be equal to half the area of R.
We can set up the equation for z as follows:
∫[0, z] (f(x) - g(x)) dx = (1/2) * A
Where A is the area of R, which we will calculate in part A.
Now, we can solve this equation by integrating (f(x) - g(x)) over the interval [0, z], setting the result equal to half of the area A, and solving for z.
C. Finding the volume of the solid generated when R is revolved around the x-axis:
To find the volume of the solid generated when R is revolved around the x-axis, we can use the method of cylindrical shells. We need to integrate the circumference of each shell multiplied by its height over the interval [0, z], where z is the value obtained in part B.
The integral setup for the volume V is as follows:
V = ∫[0, z] 2πx (f(x) - g(x)) dx
We integrate 2πx (f(x) - g(x)) over the interval [0, z] to calculate the volume of the solid generated.