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.

- ggrree

- jamiebookeater

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- TuringTest

always a good idea to have a picture in front of you when you do these problems

- TuringTest

|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:\[R=\int_{0}^{1}x-x^2dx\]

- TuringTest

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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## More answers

- TuringTest

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

- TuringTest

|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\[R_u=\int_{0}^{1-m}(x-x^2)-mxdx\]

- TuringTest

*sorry, typo, that should be 1-m in the drawing

- TuringTest

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\[R_l=\int_{0}^{1-m}mxdx+\int_{1-m}^{2}x-x^2dx\]

- TuringTest

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

- ggrree

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?

- TuringTest

oh man, you're making me do the messy part?
ok, hold on...

- TuringTest

ok that was painful but I think I got it...

- TuringTest

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...

- TuringTest

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!)

- TuringTest

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!

- ggrree

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).

- ggrree

We still can't get the answer to come up as 1/2 though. :(
But everything else you've done is beautiful.

- ggrree

Thanks for taking all this time. We really appreciate it. :)

- TuringTest

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...

- TuringTest

'didn't don't' lol
I can't type for some reason..

- ggrree

That's totally okay! Thanks!!!

- TuringTest

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

- TuringTest

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...

- TuringTest

that 3(2^3) in the first line above is another typo :/

- TuringTest

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...

- TuringTest

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

- ggrree

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 )
\[ \int\limits_{3\pi/4}^{\pi/4}3+ \int\limits\limits_{-1}^{1} x^3/(1+x^6) \]

- TuringTest

you are in luck, look at the second integrand:
it is odd

- TuringTest

also I think your first integral is\[\int\frac13du\]

- TuringTest

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

- ggrree

YOu ARE BEAUTIFUL

- TuringTest

practice is all, lol
that trick has cracked some tough nuts

- ggrree

May we ask you more questions please?

- TuringTest

sure, they're good ones

- ggrree

You are so awesome! Thanks a googleplex!

- TuringTest

you're welcome, and thanks for asking good questions
you should really post a separate thread though, long ones get slow

- ggrree

Okay! Donne dealios!

- TuringTest

lol

- ggrree

Thank you TuringTest! We need to go now, but we really appreciated all your help! :)

- TuringTest

welcome, anytime :D

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