## ggrree Group Title 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. 2 years ago 2 years ago

1. TuringTest Group Title

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

2. TuringTest Group Title

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

3. TuringTest Group Title

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

4. TuringTest Group Title

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

5. TuringTest Group Title

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

6. TuringTest Group Title

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

7. TuringTest Group Title

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

8. TuringTest Group Title

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

9. ggrree Group Title

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?

10. TuringTest Group Title

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

11. TuringTest Group Title

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

12. TuringTest Group Title

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...

13. TuringTest Group Title

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!)

14. TuringTest Group Title

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!

15. ggrree Group Title

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).

16. ggrree Group Title

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

17. ggrree Group Title

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

18. TuringTest Group Title

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...

19. TuringTest Group Title

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

20. ggrree Group Title

That's totally okay! Thanks!!!

21. TuringTest Group Title

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

22. TuringTest Group Title

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...

23. TuringTest Group Title

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

24. TuringTest Group Title

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...

25. TuringTest Group Title

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

26. ggrree Group Title

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

27. TuringTest Group Title

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

28. TuringTest Group Title

also I think your first integral is$\int\frac13du$

29. TuringTest Group Title

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

30. ggrree Group Title

YOu ARE BEAUTIFUL

31. TuringTest Group Title

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

32. ggrree Group Title

33. TuringTest Group Title

sure, they're good ones

34. ggrree Group Title

You are so awesome! Thanks a googleplex!

35. TuringTest Group Title

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

36. ggrree Group Title

Okay! Donne dealios!

37. TuringTest Group Title

lol

38. ggrree Group Title

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

39. TuringTest Group Title

welcome, anytime :D