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Let R be the region in the first quadrant enclosed between the x-axis and the curve y = x - x^2. There is a straight line passing through the origin which cuts the region R into two regions so that the area of the lower region is 7 times the area of the upper region. Find the slope of this line.

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always a good idea to have a picture in front of you when you do these problems
|dw:1331329952729:dw|here is just the region R under y=x-x^2 I like to just write out the integral for R first just as a sort of starting point:\[R=\int_{0}^{1}x-x^2dx\]
now we want to draw in our line that will divide R into a ratio of 1:7|dw:1331330236307:dw|this is a straight line through the origin, so the equation for it is given by\[y=mx\]where m is the slope we will need the intersection point P of the two graphs as well for the bounds of our new integrals

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to find the intersection point we set the equations for the two graphs equal\[x-x^2=mx\to x=0,1-m\]so we confirm that they intersect at the origin, and we get the x-coordinate of our point P, which is x=m-1 we are now ready to make our integrals
|dw:1331330760501:dw|notice the upper bound of our integral will be x=1-m since we found that top be the intersection point for the area of the top portion the integral will then be\[R_u=\int_{0}^{1-m}(x-x^2)-mxdx\]
*sorry, typo, that should be 1-m in the drawing
for the area of the bottom portion we will need to split the integral up accordingly|dw:1331331073448:dw|the integrals will be divided at x=1-m, and will be\[R_l=\int_{0}^{1-m}mxdx+\int_{1-m}^{2}x-x^2dx\]
we know that the ratio of these areas is 1:7, so we have the following relationship\[R_l=7R_u\]\[\int_{0}^{1-m}mxdx+\int_{1-m}^{2}x-x^2dx=7\int_{0}^{1-m}(x-x^2)-mxdx\]integrate and solve for m
Thank you! That's exactly how far we got, but we are having trouble actually solving for m. Could you please show how you would finish the problem and find an answer?
oh man, you're making me do the messy part? ok, hold on...
ok that was painful but I think I got it...
this is all about simplifying things in a clever way\[\int_{0}^{1-m}mxdx+\int_{1-m}^{2}x-x^2dx=7\int_{0}^{1-m}(x-x^2)-mxdx\]\[\int_{0}^{1-m}mxdx+\int_{1-m}^{2}x-x^2dx=7\int_{0}^{1-m}(1-m)x-x^2dx\]from here on I will have to write the left and right hand sides on separate lines because they are so long Left side:\[\frac12m(1-m)^2+\frac12(2^2)-\frac13(2^3)-\frac12(1-m)^2+\frac13(1-m)^3\]Right side:\[=\frac72(1-m)^3-\frac72(1-m)^3\]now multiply it all by 6 to get rid of the fractions...
left:\[3m(1-m)^2+3(2^2)-2(2^3)-2(2^3)-3(1-m)^2+2(1-m)^3\]right:\[=21(1-m)^3-14(1-m)^3\]now simplify (do not expand anything!)
now I can write left and right on the same line again :)\[3m(1-m)^2-3(1-m)^2-4+2(1-m)^3=7(1-m)^2\]\[(3m-3)(1-m)^2-4=5(1-m)^3\]notice the trick that we can do on the left here:\[(3m-3)(1-m)^2=-3(1-m)(1-m)^2=-3(1-m)^3\]so the problem becomes\[-3(1-m)^3-4=5(1-m)^3\]\[8(1-m)^3=-4\]\[1-m=-\frac1{\sqrt[3]2}\]\[m=1+\frac1{\sqrt[3]2}\]whew!
Thank you, but the answer comes up as 1/2. Earlier up, you integrated from 1-m to 2, rather than from 1-m to 1, also. 2 posts ago, when you retyped the left side and the right side again, you added an additional -2(2^3).
We still can't get the answer to come up as 1/2 though. :( But everything else you've done is beautiful.
Thanks for taking all this time. We really appreciate it. :)
yes I saw that typo with the -2(2^3) but I did not use it but damn I totally didn't don't know why I changed the bound to 2 now I have to try it again for my own sanity, hold on...
'didn't don't' lol I can't type for some reason..
That's totally okay! Thanks!!!
oh the answer 1/2 comes easily if you just put\[\frac12(1^2)-\frac13(1^3)\]where I put the 2, and do everything the exact same
taking it from where I multiplied everything by 6 and just looking at the left side\[3m(1-m)^2+3(2^2)+3(1^3)-2(1^3)-3(1-m)^2+2(1-m)^3\]\[3m(1-m)^2+1-3(1-m)^2+2(1-m)^3\]\[-3(1-m)^3+2(1-m)^2+1\]so the -4 in my work above become a 1...
that 3(2^3) in the first line above is another typo :/
also the last line has a 2 in the exponent where it should be 3\[-3(1-m)^3+2(1-m)^2+1\]so yeah, follow it through the same way as before and you get to\[-3(1-m)^3+1=5(1-m)^3\]\[8(1-m)^3=1\]\[1-m=\frac12\]\[m=\frac12\huge\checkmark\]tadah...
wow I cannot seem to correct my typos the correction I meant to make above was\[-3(1-m)^3+2(1-m)^3+1\]but the work is correct
May we ask you another integration question? \[\int\limits\limits_{-1}^{1} (x^2 + x^3)/(1+x^6) \] we tried integration by partial fractions (which ended up being horribly messy) and separating the fraction into \[\int\limits_{-1}^{1} x^2 /(1+x^6) + \int\limits_{-1}^{1} x^3/(1+x^6) \] which made the first integral easy to solve via trig substitution (letting x^3 = tan u), but makes the second integral unsolvable (ie. we weren't clever enough to solve it:P ) \[ \int\limits_{3\pi/4}^{\pi/4}3+ \int\limits\limits_{-1}^{1} x^3/(1+x^6) \]
you are in luck, look at the second integrand: it is odd
also I think your first integral is\[\int\frac13du\]
is the fact that the second integrand is odd not meaning anything to you? look at the bounds and try to remember a theorem about this situation
practice is all, lol that trick has cracked some tough nuts
May we ask you more questions please?
sure, they're good ones
You are so awesome! Thanks a googleplex!
you're welcome, and thanks for asking good questions you should really post a separate thread though, long ones get slow
Okay! Donne dealios!
Thank you TuringTest! We need to go now, but we really appreciated all your help! :)
welcome, anytime :D

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