## anonymous 4 years ago Find the area of the region bounded by the curves f(x)=(3x+1)^2 g(x)=2^0.4x i checked my work a lot of times but I couldn't get 8.3522

1. anonymous

Ok, I'm going to try this one.

2. anonymous

oh thanks. May be I better show you what i have done so you can find my mistakes

3. anonymous

Sounds good. I'm looking at the graphs right now.

4. anonymous

|dw:1327278455526:dw|

5. anonymous

is that what you get?

6. anonymous

$\int\limits_{-1.259}^{0.62} 2^{0.4x}-[(3x+1)^{2}-7]dx$

7. anonymous

$\left[ (2^{0.4x}/0.4\ln2 - 3x ^{3}-3x ^{2}+6x \right]$$\int\limits_{-1.259}^{0.62} 2^{0.4x}-9x ^{2}-6x+6 dx$$\int\limits_{-1.259}^{0.62}2^{0.4x}-[9x ^{2}+6x+1-7]dx$

8. anonymous

$\left[ 2^{0.4x}/0.4\ln2 - 3x ^{3}-3x ^{2}+6x \right]$

9. anonymous

Sorry, it's taking me some time to check the algebra/arithmetic. :) But so far your approach looks good.

10. anonymous

it's okay :) I can't find what's wrong here :(

11. anonymous

oh btw did I get the interval right? [-1.259, 0.62]

12. anonymous

Did you get those values by setting the equations equal to themselves and solving for x?

13. anonymous

No. I got it from calculator. :)

14. anonymous

Well, I have to admit that I was having difficulty solving the equation, but what I did was put it into wolframalpha.com$9x ^{2}+6x+1=2^{0.4x}$ It gave me x=0 and 1.246 E -8. This is the link: http://www.wolframalpha.com/input/?i=9x%5E2%2B6x%2B1%3D2%5E%28.4x%29

15. anonymous

well i just look at the intersections of f(x) and g(x) on calculator. I'm confused now.

16. anonymous

Ok, never mind what I said if that confuses you. When I look at my calculator I get [-0.64, 0].

17. anonymous

oh my bad. I typed a wrong equation f(x)=(3x+1)^2 -7 sorry sorry sorry

18. anonymous

No problem! Glad you found your mistake. Let's try it with these limits.

19. anonymous

:) I'm so sorry.

20. anonymous

Don't be! I just started this too and have made errors that are much worse :) It has helped me to work through this with you.

21. anonymous

hehe this problem takes me like more than one hour to find the answer that my teacher stated.

22. anonymous

my answer is 9.913 do you get the same one?

23. anonymous

Almost have it . . .

24. anonymous

This is my equation:$\int\limits_{-0.64bj }^{0}2^{0.4x}-(9x+6x+1) dx$ My answer is way off, I got 3.018

25. anonymous

you didn't change the equation, did you? cuz it's supposed to be 9x+6x+1-7

26. anonymous

My bad it is supposed to be$9x ^{2}$, but where did the -7 come from? I didn't see it in the original equation.

27. anonymous

I mean question.

28. anonymous

i just said that I typed a wrong equation f(x)=(3x+1)^2-7 :)

29. anonymous

Ok now I get it so it is$\int\limits\limits_{-0.64}^{0}2^{0.4x}-( 9x-3x+1-7) dx$which equals$\int\limits_{-0.64}^{0}2^{0.4x}-9x+3x+6dx$

30. anonymous

Is this right?

31. anonymous

yup but the interval is different cuz you changed the equation. You should check the intersections with your calculator too.

32. anonymous

Oh yes

33. anonymous

hey wait it is $2^{0.4x}-9x ^{2}-6x+6$

34. anonymous

You are right, I meant 6x. So are you saying that the minus sign is already distributed?

35. anonymous

Yes

36. anonymous

So I should take it out when I plug it in to my calculator.

37. anonymous

Yup. Did you find the interval?

38. anonymous

Here we go: [-1.277,0.63] . However, the first number is not exact.

39. anonymous

I got [-1.259, 0.62]

40. anonymous

hmmm this is why I think an equation is better if only I could solve it correctly.

41. anonymous

But I think we are in the ballpark. This was tricky with the x in the exponent.

42. anonymous

you're trying to find the interval by solving the equations? I can't find it by hand, you know.

43. anonymous

Ok, well I was estimating at first; I get the same answer that you did using the calc function.

44. anonymous

yes then we just need to integrate that equation. I started over and get 9.913

45. anonymous

The integration is fine. I used all of the digits for the intersection and got 6.702.