# Area of 2 curves -Integration

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• October 9th 2013, 04:15 AM
SlipEternal
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
I contrast my set up with yours and I don't understand why the second set up ranges from 0,2. The point of intersection is at 0,3.

That was actually my mistake. I had a smudge on my monitor and when I first read your post, I thought that the functions were:
$y=x^2, y=3x, x=-1, x=2$

I missed the negative sign before the two for the last line. It is sheer coincidence that the integrals I set up happen to yield the same answer as the one on your answer sheet.
• October 9th 2013, 05:19 AM
votan
Re: Area of 2 curves -Integration
Here's what I did:
Attachment 29426

this is not 31/6. Either the answer in your sheet is incorrect, or you posted the limits incorrectly.
• October 9th 2013, 06:06 AM
Darrylcwc
Re: Area of 2 curves -Integration
I got 41/6. But the answer states 31/6!
This is shouldn't to be, by virtue, a straight to the point simple problem because of the abstraction. I don't fare very well in applicational calculus.
However, it's ridiculous that I'm always working out 41/6 but which the sheet states 31/6.

I typed the question word for word.
• October 9th 2013, 06:15 AM
votan
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
I got 41/6. But the answer states 31/6!
This is shouldn't to be, by virtue, a straight to the point simple problem because of the abstraction. I don't fare very well in applicational calculus.
However, it's ridiculous that I'm always working out 41/6 but which the sheet states 31/6.

I typed the question word for word.

I am sorry to tell you the answer in your sheet is incorrect. It is not the answer to the question.
• October 9th 2013, 06:19 AM
Darrylcwc
Re: Area of 2 curves -Integration
Quote:

Originally Posted by votan
I am sorry to tell you the answer in your sheet is incorrect. It is not the answer to the question.

I'm already starting to suspect that I have wasted almost 2 days working out the other same problems over and over again because of wrong answers where in fact the sheet is flawed.
• October 9th 2013, 06:56 AM
Darrylcwc
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
I'm already starting to suspect that I have wasted almost 2 days working out the other same problems over and over again because of wrong answers where in fact the sheet is flawed.

wait, something isn't right.
The answer ought to be 41/6 - 9/2. Why does your set up consist only of a x^2? It should be x^2-3x for int -1 to -2; and, 3x-x^2 for int 0 to 3;
• October 9th 2013, 08:01 AM
SlipEternal
Re: Area of 2 curves -Integration
You always integrate from left to right. That is how the definite integral is defined. If you integrate from right to left, you get negative the correct area.

So, if you integrate from -1 to -2, you are integrating from a bigger number to a smaller number. That is going right to left. You mean it should be the integral of x^2-3x from -2 to -1? Yes. That is correct.

The region bounded by y=3x and y=x^2 from x=0 to x=3 is NOT part of the region bounded by
y=x^2, y=3x, x=-2, and x=-1

Note that x=0 and x=3 are both bigger than x=-1, which is the right bound of the two lines (x=-2 and x=-1). So, if you include that area, you are included area that is NOT bounded by the given functions and lines.
• October 9th 2013, 08:12 AM
Darrylcwc
Re: Area of 2 curves -Integration
We might be confused by a case of semantics or stipulative definition.
Shouldn't the bigger value be at the top of the sigma?

Edit: In essence, if I move along the x-axis from right to left, I would obtain a negative value-small value minus larger value. So therefore, adding a minus sign outside of the sigma resolve this problem since area can never be negative much like a scalar value.
In moving from left to right, we would proceed the algebraic operation as per usual since we will always obtain a positive value
• October 9th 2013, 08:30 AM
Darrylcwc
Re: Area of 2 curves -Integration
On a tangent:
How should I deal with a function where f'(x) = ln(x) to be integrate along the x-axis from positive to negative and negative to positive?
• October 9th 2013, 08:33 AM
SlipEternal
Re: Area of 2 curves -Integration
Use integration by parts
• October 9th 2013, 08:56 AM
votan
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
We might be confused by a case of semantics or stipulative definition.
Shouldn't the bigger value be at the top of the sigma?

That's correct, -1 is begger than -2
• October 9th 2013, 09:00 AM
votan
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
On a tangent:
How should I deal with a function where f'(x) = ln(x) to be integrate along the x-axis from positive to negative and negative to positive?

Please post this in a different thread and give it a new title. Thanks
• October 9th 2013, 09:08 AM
votan
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
wait, something isn't right.
The answer ought to be 41/6 - 9/2. Why does your set up consist only of a x^2? It should be x^2-3x for int -1 to -2; and, 3x-x^2 for int 0 to 3;

You could verity the validity of the answer this way.

Consider the trpezoid bouded but small base 1 +3 = 4 these are the coordinates from the parabola and the line. It is +3 because we are using the distance between points.

The long base is 4 + 6 = 10

area = [(10 + 4)/2]*1 = 7 = 42/6

This area is slightly larger than 41/6 because the curvature of the pararable
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