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Area under curve

  • Thread starter kenny87
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  • #1
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The question:

Let R and S be regions in the first quadrant. The region R is bounded by the x-axis and the graphs of y=2-x^3 and y=tanx. The region S is bounded by the y-axis and the graphs of y=2-x^3 and y=tanx.

a) Find the area of R
b) Find the area of S

I really need help starting this. I know that I will have to do some integration, but I'm just confused... please help!
 

Answers and Replies

  • #2
68
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first graph tanx and remember that tanx is bounded in the region -pi/2 to pi/2. But since we are only looking at the first Quadrant, we only care about the region around x=0 to x = pi/2. (we aren't necessairly going to integrate to pi/2) but that is where we are looking at the graph at. now we graph the function y=2-x^3 and find where they intersect.
Now for the area R i believe where we are considering the x-axis we will take stripes of a thickness dy, so we will be going horizontally and we will have our integral, tanx-2+x^3... and for our other region S which is with respect to the y, we get stripes of thickness dx, so we integrates the upper function minus the lower funciton, in this case it is 2-x^3-tanx..
all you need to do is figure out where they intersect, either by calculator or by hand. I think it is way easier by calculator, i am not sure anymore how to do it by hand. I hope it gives you some idea on how to do this problem.
 
  • #3
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You'll want to start by doing a quick graph of 2-x^3 and tan x. For (a), find where the two functions and the x-axis form a completely enclosed region. Use the points where these functions intersect each other and the x-axis as the limits for your integration, and simply integrate (you'll have to do two separate integrals and add them) to find the area of the region.
 
  • #4
23
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I guess I'm confused as to what function I am suppossed to integrate...
I don't understand why I need to be subtracting the two functions as georgeh said...
 
  • #5
68
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you subtract both functions to get the area "trapped" between the two curves. That area is the one you are interested in.
So you do upper minus lower or if its an integral that has a thickness dy, then you get left minus right to find the area trapped there.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
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You first learned to find "area under a curve" using the x-axis (f(x)= 0) as the lower boundary. When you have a second curve and want to find the area between them (this thread would have been better titled "area between curves) there are two ways to think.

1) Imagine those skinny rectangles, of width "dx", extending only from the upper curve, y= f(x), to the lower curve, y= g(x). The height of the rectangle is f(x)- g(x) so its area is (f(x)- g(x))dx. That's what you want to integrate:
[tex]A= \int (f(x)- g(x))dx[/tex]

2) Treat each curve as a separate problem. The area under the curve y= f(x) is [itex]\int f(x)dx[/itex] and the area under the curve y= g(x) is [itex]\int g(x)dx[/itex]. The area between the two curves is the area under y= f(x) minus the area under g(x):
[tex]A= \int f(x)dx- \int g(x)dx= \int (f(x)- g(x))dx[/tex]
 

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