Area between curves (Integral): between y = lx^2 - 4l and y =2
- Thread starter brunosamp
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Where are you stuck in the process? You started with the algebra, drawing a graph of the two curves, shading the three areas in question, and finding the intersection points. You noted that each of the functions was even, so you could use symmetry to cut your integral in half (and then multiply the resulting area by 2). And... then what?
Please reply with a clear listing of all of your thoughts and efforts so far, starting from the algebra mentioned above. Thank you!
Where are you stuck in the process? You started with the algebra, drawing a graph of the two curves, shading the three areas in question, and finding the intersection points. You noted that each of the functions was even, so you could use symmetry to cut your integral in half (and then multiply the resulting area by 2). And... then what?
Please reply with a clear listing of all of your thoughts and efforts so far, starting from the algebra mentioned above. Thank you!
Yes, i did all that. So, when doing the integral, i thought about doing it like
integral of (x^2-4 -2) going from 0 to sqrt of 2 + integral of 2 -(x^2-4) going from sqrt of 2 to 2 + integral of 2-(x^2-4) going from 2 to sqrt of 6.
Everything multiplying with two, because of the symmetry.
But, is wrong =/. What is that i'm doing wrong?
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In future, kindly please provide that information at the start.
No. Take another look at that graph. Is the top line really y = x^2 - 4? Or something else?
No, for the same reason.
This portion is correct.
pka
Elite Member
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The above quoted is the original problem and here is the graph. The question is about the area is bounded by two graphs: \(\displaystyle y=2~\&~y=|x^2-4|\).
So we need three integrals added together and doubled.
\(\displaystyle \displaystyle{\int_0^{\sqrt 2 } {\left( {2 - {x^2}} \right)} dx + \int_{\sqrt 2 }^2 {\left( {{x^2} - 2} \right)} dx + \int_2^{\sqrt 6 } {\left( {6 - {x^2}} \right)} dx}\)
I have not explained how those integrals were determined. Please post answers to these:
a) why are there three integrals?
b) how are the three intergrands gotten?
c) how are the limits of integrations determined?
d) why do we double the sum of the three integrals?