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Let R be the region in the first quadrant enclosed between the xaxis 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.
 2 years ago
 2 years ago
Let R be the region in the first quadrant enclosed between the xaxis 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.
 2 years ago
 2 years ago

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TuringTestBest ResponseYou've already chosen the best response.1
always a good idea to have a picture in front of you when you do these problems
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
dw:1331329952729:dwhere is just the region R under y=xx^2 I like to just write out the integral for R first just as a sort of starting point:\[R=\int_{0}^{1}xx^2dx\]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
now we want to draw in our line that will divide R into a ratio of 1:7dw:1331330236307:dwthis 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
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
to find the intersection point we set the equations for the two graphs equal\[xx^2=mx\to x=0,1m\]so we confirm that they intersect at the origin, and we get the xcoordinate of our point P, which is x=m1 we are now ready to make our integrals
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
dw:1331330760501:dwnotice the upper bound of our integral will be x=1m since we found that top be the intersection point for the area of the top portion the integral will then be\[R_u=\int_{0}^{1m}(xx^2)mxdx\]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
*sorry, typo, that should be 1m in the drawing
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
for the area of the bottom portion we will need to split the integral up accordinglydw:1331331073448:dwthe integrals will be divided at x=1m, and will be\[R_l=\int_{0}^{1m}mxdx+\int_{1m}^{2}xx^2dx\]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
we know that the ratio of these areas is 1:7, so we have the following relationship\[R_l=7R_u\]\[\int_{0}^{1m}mxdx+\int_{1m}^{2}xx^2dx=7\int_{0}^{1m}(xx^2)mxdx\]integrate and solve for m
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.1
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?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
oh man, you're making me do the messy part? ok, hold on...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
ok that was painful but I think I got it...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
this is all about simplifying things in a clever way\[\int_{0}^{1m}mxdx+\int_{1m}^{2}xx^2dx=7\int_{0}^{1m}(xx^2)mxdx\]\[\int_{0}^{1m}mxdx+\int_{1m}^{2}xx^2dx=7\int_{0}^{1m}(1m)xx^2dx\]from here on I will have to write the left and right hand sides on separate lines because they are so long Left side:\[\frac12m(1m)^2+\frac12(2^2)\frac13(2^3)\frac12(1m)^2+\frac13(1m)^3\]Right side:\[=\frac72(1m)^3\frac72(1m)^3\]now multiply it all by 6 to get rid of the fractions...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
left:\[3m(1m)^2+3(2^2)2(2^3)2(2^3)3(1m)^2+2(1m)^3\]right:\[=21(1m)^314(1m)^3\]now simplify (do not expand anything!)
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
now I can write left and right on the same line again :)\[3m(1m)^23(1m)^24+2(1m)^3=7(1m)^2\]\[(3m3)(1m)^24=5(1m)^3\]notice the trick that we can do on the left here:\[(3m3)(1m)^2=3(1m)(1m)^2=3(1m)^3\]so the problem becomes\[3(1m)^34=5(1m)^3\]\[8(1m)^3=4\]\[1m=\frac1{\sqrt[3]2}\]\[m=1+\frac1{\sqrt[3]2}\]whew!
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.1
Thank you, but the answer comes up as 1/2. Earlier up, you integrated from 1m to 2, rather than from 1m to 1, also. 2 posts ago, when you retyped the left side and the right side again, you added an additional 2(2^3).
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.1
We still can't get the answer to come up as 1/2 though. :( But everything else you've done is beautiful.
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.1
Thanks for taking all this time. We really appreciate it. :)
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
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...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
'didn't don't' lol I can't type for some reason..
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.1
That's totally okay! Thanks!!!
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
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
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
taking it from where I multiplied everything by 6 and just looking at the left side\[3m(1m)^2+3(2^2)+3(1^3)2(1^3)3(1m)^2+2(1m)^3\]\[3m(1m)^2+13(1m)^2+2(1m)^3\]\[3(1m)^3+2(1m)^2+1\]so the 4 in my work above become a 1...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
that 3(2^3) in the first line above is another typo :/
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
also the last line has a 2 in the exponent where it should be 3\[3(1m)^3+2(1m)^2+1\]so yeah, follow it through the same way as before and you get to\[3(1m)^3+1=5(1m)^3\]\[8(1m)^3=1\]\[1m=\frac12\]\[m=\frac12\huge\checkmark\]tadah...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
wow I cannot seem to correct my typos the correction I meant to make above was\[3(1m)^3+2(1m)^3+1\]but the work is correct
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.1
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 ) \[ \int\limits_{3\pi/4}^{\pi/4}3+ \int\limits\limits_{1}^{1} x^3/(1+x^6) \]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
you are in luck, look at the second integrand: it is odd
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
also I think your first integral is\[\int\frac13du\]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
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
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
practice is all, lol that trick has cracked some tough nuts
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.1
May we ask you more questions please?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
sure, they're good ones
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.1
You are so awesome! Thanks a googleplex!
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
you're welcome, and thanks for asking good questions you should really post a separate thread though, long ones get slow
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.1
Thank you TuringTest! We need to go now, but we really appreciated all your help! :)
 2 years ago
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