Region R is bounded by the functions f(x) = 2(x-4) + pi, g(x) = cos^-1(x/2 - 3), and the x axis.
a. What is the area of the region R?
b. Find the volume of the solid generated when region R is rotated about the x axis.
c. Find all values c for f(x) and g(x) in the closed interval p <= c <= q for which each function equals the average value in the indicated interval.
the curves intersect at (4,π)
a = ∫[0,4] 2(x-4)+π dx
+ ∫[4,8] arccos(x/2-3) dx
= x^2+(π-8)x [0,4]
+ x arccos(x/2-3) + 6 arcsin(x/2-3) - √(4-(x-6)^2)
= 4π-16 + 2π
= 6π-16
Ok, that makes sense. How would I find the answers to b and c though? I'm not even sure where to start...
P and Q are the x values on the graph I was given for this problem where the two functions intersect the x axis. They don't have any numerical assignments.
the average value over [0,8] is k=(6π-16)/8
There are two x values where this occurs
f(x) = k and g(x) = k
to see where these are visit wolframalpha.com and enter
plot y=2(x-4)+pi , y=arccos(x/2-3), y=(6*pi-16)/8, x=0..8
f(c) = (6π-16)/8 = 2(c-4)+π
c = 3 - π/8
g(c) = (6π-16)/8 = arccos(c/2-3)
c = 6+2cos(3π/4 - 2)
Newsflash: 2(x-4)+ pi doesn't cross the x-axis before 0 so you can't use zero for the integral
a. To find the area of region R, we need to find the points where the functions intersect and the integral of the difference between the functions over that interval.
Step 1: Find the points of intersection:
- Set f(x) equal to g(x) and solve for x:
2(x-4) + pi = cos^(-1)(x/2 - 3)
Step 2: Solve the equation to find the values of x:
2(x-4) + pi = cos^(-1)(x/2 - 3)
2x - 8 + pi = cos^(-1)(x/2 - 3)
cos(2x - 8 + pi) = x/2 - 3
Step 3: Solve for x:
cos(2x - 8 + pi) = x/2 - 3
2cos(2x - 8 + pi) = x - 6
2cos(2x - 8 + pi) - x + 6 = 0
At this point, you can use numerical methods or graphing techniques to find the approximate values of x where the functions intersect.
Step 4: Once you have the points of intersection, integrate the absolute difference of the functions over that interval to find the area:
Area = ∫|f(x) - g(x)| dx from x = a to x = b
b. To find the volume of the solid generated when region R is rotated about the x-axis, we can use the disk method.
Step 1: Determine the limits of the integral by finding the points of intersection as described in part a.
Step 2: Set up the integral for the volume using the formula:
Volume = π∫[f(x)]^2 dx from x = a to x = b
c. To find the values c for which f(x) and g(x) equal their average value in the indicated interval:
Step 1: Determine the average values of the functions f(x) and g(x) over the given interval.
Step 2: Set up and solve the equations f(c) = average value and g(c) = average value for the variable c, where average value is the value obtained in Step 1.
By solving these equations, you can find the values c for which each function equals the average value in the indicated interval.
you must have some idea where to start.
see wolframalpha.com for doing integrals
v = ∫π y^2 dx for each region.
= π∫[0,4](2(x-4)+π)^2 dx + π∫[4,8](arccos(x/2-3))^2 dx
for the avg value, not sure what p and q are supposed to be.