Determining the area enclosed by inverse

  • Thread starter Saitama
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  • #26
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Good point.
You can calculate the area to the right of the y-axis, which is what you did.
The graph is not symmetrical, but it's close enough for our purposes.
You are only asked to give an integer approximation, which is what you did.

Good! :wink:
Thanks a lot ILS! :smile:

Now that the problem is solved, I was wondering if you could help me understand the solution given in the solution booklet? The solution booklet solves the problem in a single line and I am clueless about what the paper setters did. Here's what is written in booklet:

$$A=\left(4-\int_0^1(f(y)-1)dy \right)+\left|4-\int_{-1}^0 (1-f(y))dy\right|=\frac{9}{2}$$

Any idea what is going on above?
 
  • #27
LCKurtz
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Thanks a lot ILS! :smile:

Now that the problem is solved, I was wondering if you could help me understand the solution given in the solution booklet? The solution booklet solves the problem in a single line and I am clueless about what the paper setters did. Here's what is written in booklet:

$$A=\left(4-\int_0^1(f(y)-1)dy \right)+\left|4-\int_{-1}^0 (1-f(y))dy\right|=\frac{9}{2}$$

Any idea what is going on above?
That is an unnecessarily obscure way to set it up. The natural way would be just use the usual calculus formula for area between two curves:$$
A = \int_a^b y_{upper} - y_{lower}~dx$$on both pieces:$$
\int_{-1}^0 f(x) -(-3)~dx + \int_0^15-f(x)~dx$$
 
  • #28
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That is an unnecessarily obscure way to set it up. The natural way would be just use the usual calculus formula for area between two curves:$$
A = \int_a^b y_{upper} - y_{lower}~dx$$on both pieces:$$
\int_{-1}^0 f(x) -(-3)~dx + \int_0^15-f(x)~dx$$
Yep, agreed! :)

But it looks to me that the solution was trying to define a odd function g(x)=f(x)-1. Anyways, its quite difficult to catch that during the exam so simply setting up the integrals is way better.

Do you think it would be possible to solve the problem without looking at the plots? I couldn't get a feel for the problem until I looked at the graphs. Do you have any suggestions?

Thank you very much LCKurtz and ILS! I got to learn something new. :smile:
 
  • #29
LCKurtz
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Yep, agreed! :)

Do you think it would be possible to solve the problem without looking at the plots? I couldn't get a feel for the problem until I looked at the graphs. Do you have any suggestions?
My advice would be to NEVER set up an area integral without making a plot. The only exception would be where it is so simple that you already have a mental image, which amounts to the same thing.
 
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  • #30
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My advice would be to NEVER set up an area integral without making a plot. The only exception would be where it is so simple that you already have a mental image, which amounts to the same thing.
Thank you once again LCKurtz! :)
 

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