Sketch the region enclosed by the curves given below. Decide whether to integrate with respect to x or y. Then find the area of the region.
y=4cos(x),y=4−8x/π.
I thought it was the integral of 4cos(x)- the integral of 4- 8x/π on the interval 0 to π/2.
This gave me 4- π which isn't right..
Well, let's take a closer look at the problem. We have two curves, y = 4cos(x) and y = 4 - (8x/π).
To find the region enclosed by these curves, we need to determine the values of x that define the boundaries of the region. We can do this by setting the two equations equal to each other and solving for x.
4cos(x) = 4 - (8x/π)
Now, using my great math skills (and a little bit of help from a calculator), I found that the intersection points occur at x = 0 and x = π/2.
Since we're looking for the area, we need to decide which variable to integrate with respect to – x or y. In this case, it's easier to integrate with respect to x, so we'll go with that.
Now, let's set up the integral to find the area:
A = ∫[0,π/2] (4 - (8x/π) - 4cos(x)) dx
Simplifying this, we get:
A = ∫[0,π/2] (4 - 8x/π - 4cos(x)) dx
Now, you can go ahead and integrate this expression between the limits of integration. After calculating the integral, you'll find the correct area of the region.
I hope this helps clear up the confusion and gets you on the right track!
To sketch the region enclosed by the given curves, we need to find the points of intersection between them.
Setting the two equations equal to each other gives us:
4cos(x) = 4 - (8x/π)
Simplifying, we have:
cos(x) = 1 - (2x/π)
To find the points of intersection, we can solve this equation graphically or algebraically.
Graphically: Plot the two curves on a graphing calculator or graphing software and find their intersection points.
Algebraically:
1. Set Up:
Start by finding one complete period of the cosine function, which is 2π.
Set up the equation:
cos(x) = 1 - (2x/π)
2. Solve for x:
Since the cosine function repeats every 2π, we can rewrite the equation as:
cos(x) = 1 - (2(x - π)/π)
Solving for x, we get:
x = π ± arccos(1 - (2(x - π)/π))
Since we are only interested in the points of intersection within the interval [0, π/2], we can write:
x = π/2 - arccos(1 - (2(x - π)/π))
3. Equate the y-values:
Substitute the value of x obtained into either one of the original equations to find the corresponding y-values.
Substituting x = π/2 - arccos(1 - (2(x - π)/π)) into y = 4cos(x):
y = 4cos(π/2 - arccos(1 - (2(x - π)/π))
4. Determine the integration variable and limits:
To find the area of the region, we need to integrate the difference of the two curves with respect to x or y.
Since the given equations are defined in terms of x, we integrate with respect to x. However, it is important to note that sometimes it is more convenient to integrate with respect to y, depending on the curves.
In this case, we integrate with respect to x.
To find the limits of integration, we need to look for the x-values where the curves intersect within the given interval [0, π/2]. These x-values are obtained from the solutions of the equation in Step 2.
5. Calculate the area:
Set up the integral of the difference between the two curves and integrate:
Area = ∫[limits] (4cos(x) - (4 - (8x/π))) dx
Evaluate the integral using the limits obtained in Step 4.
By following these steps, you should be able to find the correct area of the region enclosed by the given curves.
To sketch the region enclosed by the given curves, let's first plot the two equations on a graph.
The first equation, y = 4cos(x), represents a cosine function. When we graph this equation, we'll get a wave-like curve oscillating between -4 and 4.
The second equation, y = 4 - (8x/π), is a linear equation representing a downward-sloping line with a y-intercept of 4 and a slope of -8/π.
Now, let's find the points where these two curves intersect to determine the limits of integration. To find the intersection points, we need to set the two equations equal to each other and solve for x.
4cos(x) = 4 - (8x/π)
To solve this equation, let's isolate the cosine term:
cos(x) = (4 - (8x/π)) / 4
cos(x) = 1 - (2x/π)
Now, we solve for x by taking the inverse cosine (or arccos) of both sides:
x = arccos(1 - (2x/π))
Unfortunately, this equation cannot be solved analytically for x. Therefore, we need to find the intersection points numerically using a graphing calculator or computational method.
By finding the approximate intersection points, we can determine the limits of integration for calculating the area of the enclosed region.
Once we determine the intersection points, let's denote them as x1 and x2, where x1 < x2.
To find the area of the region, we need to integrate the difference between the two curves. However, since the curves intersect at multiple points, we need to split the region into smaller segments based on these intersection points and integrate each segment separately.
For our example, suppose we find that the intersection points are x1 = 0.6 and x2 = 1.8.
Thus, the area of the region is given by the sum of the individual areas of each segment:
Area = ∫[x1 to x2] (4cos(x) - (4 - (8x/π))) dx + ∫[0 to x1] (4 - (8x/π)) dx + ∫[x2 to π/2] (4cos(x) - (4 - (8x/π))) dx
Here, we integrate with respect to x because we are considering the curves as functions of x.
Now, you can set up and evaluate the definite integrals to find the area of the region.
From 0 to π/2, 4cos(x) is above 8-4x/π, so the area would be
∫[0,π/2] (8-4x/π)-4cos(x) dx = 7π/2 - 4
But there is a second symmetric region between π/2 and π, so the area is double that, or 7π-8
see
http://www.wolframalpha.com/input/?i=plot+y+%3D+4cos%28x%29,+y+%3D+4-8x%2Fpi
Evidently you integrated wrong, and didn't see fit to show your work...