## anonymous 4 years ago Find the exact area between y=ln(x) and y= ln(x^2) when 1=/<x </=2

1. anonymous

How would you set up this equation?

2. anonymous

Hi handy dandy helpers

3. anonymous

|dw:1326752403115:dw|

4. myininaya

$\int\limits_{1}^{2}(\ln(x^2)-\ln(x)) dx$

5. anonymous

Could you explain why mininaya?

6. myininaya

because of your graph lol

7. anonymous

lol

8. anonymous

lol you're right any questions rl?

9. anonymous

Think i got it

10. anonymous

Thanks myin and no data

11. myininaya

it always helps to draw a picture first or to graph using your calculator

12. anonymous

You're welcome rl

13. myininaya

then do $\int\limits_{lowerlimit}^{upperlimit} (upperfunction -lowerfunction) dx$

14. anonymous

that is great mininaya.

15. anonymous

and the upper function wld be the one with x^2 obviously?

16. anonymous

But what if he don't have a graphing calculator at hand?

17. myininaya

right

18. myininaya

he will have to draw it with out one then

19. anonymous

No data i am not sam now LOL

20. anonymous

who is sam?

21. anonymous

I am a she at the prsesnt moment

22. myininaya

also you may come of course some these when the functions with change from upper to lower and then you will have to break the integral up

23. anonymous

what do u mean?

24. myininaya

for example |dw:1326753034373:dw|

25. myininaya

if i wanted the area btw a and b bounded by g and f

26. myininaya

$\int\limits_{a}^{c}(f-g) dx+\int\limits_{c}^{b}(g-f) dx$

27. myininaya

then i would set up like that since f is the upper from a to c and g is the upper from c to b

28. anonymous

oh ok i c so basically u will have to draw the function and then decide which goes first

29. myininaya

yep drawing or seeing the graph is totally important

30. myininaya

the upper goes first

31. anonymous

Ok thanks for the tip myin u r awesome :DDDD

32. myininaya

:)