## virtus 3 years ago find the area bounded by the graph y = 4x^2 -1, the x axis and the ordinates x = -1/2 and x=1

1. dumbcow

$A = \int\limits_{1/2}^{1}(4x^{2} -1) dx$

2. virtus

smartcow, i thought the ordinates were from -1/2 where did you get the 1/2 from?

3. dumbcow

well when i graphed it, the parabola from -1/2 to 1/2 is below the x-axis ?

4. virtus

so ....?

5. dumbcow

is it bounded by the x-axis above or below ?

6. virtus

|dw:1328176928988:dw|

7. virtus

|dw:1328176997411:dw|

8. virtus

so is the shaded the area? |dw:1328177068433:dw|

9. dumbcow

hmm i guess that makes more sense...why do they say it is bounded by the x-axis then

10. virtus

but when i find the area i get 0 or maybe i have done some stupid error lemme check

11. dumbcow

no i get 0 as well, if the limits are -1/2 to 1

12. dumbcow

thats because the area under the x-axis is negative

13. virtus

the answer in my solution book says the answer is 4/3 square units

14. dumbcow

hmm well the area from 1/2 to 1 is 2/3 are you sure its not x=-1 an x=1

15. virtus

i am sure it is x=1

16. virtus

perhaps the answer in the solution book is wrong then or the question is wrong

17. dumbcow

|dw:1328177655800:dw| $\int\limits_{-1}^{-1/2}(4x^{2} -1)dx + \int\limits_{1/2}^{1}(4x^{2} -1)dx = 2/3 + 2/3 = 4/3$ thats my best guess :)

18. virtus

WOAH!!!!!!!!!!!!!!!! that is brilliant thinking goodjob smartcow, thanks for the help greatly appreciated

19. dumbcow

20. FoolForMath

But the only solution for $$4x^2-1$$ is $$\pm\frac 12$$

21. virtus

genius what are you trying to say?

22. FoolForMath

The parabola intersects the x axis at $$\pm \frac 12$$

23. FoolForMath

This question is not correct in my opinion.

24. virtus

oh i see, i think i support your opinion

25. FoolForMath

See here, the area is not well defined : http://www.wolframalpha.com/input/?i=+y+%3D+4x%5E2+-1%3Bx%3D-1%2F2%3Bx%3D1%3Bx%3D0%3By%3D0

26. virtus

ya

27. FoolForMath

ya ya :D

28. FoolForMath

So I think dumcow's interpretation is not correct either.

29. virtus

ya ya :D

30. dumbcow

haha sorry i should have clarified , my drawing assumes the question is wrong at to get area of 4/3 the bounds must be x=-1 and x=1 and x-axis

31. virtus

no need to apologise! you have a case indeed hehehe

32. FoolForMath

But the bound x=-1 and x=1 makes no sense right?

33. virtus

ya

34. FoolForMath

ya ya :D

35. FoolForMath

I am leaving now, take care :)

36. virtus

farewell genius, and thanks for the help once again !