## anonymous one year ago Find the between graphs of y=x^2+2 and y=1-x between x=0 and x=1. Can you guys please check if my equation is right, I will type it below.

1. Zale101

Find what?

2. Zale101

Area ?

3. anonymous

Find the area oops

4. Zale101

Makes sense

5. anonymous

[(1-x)-(x^2+2)]

6. anonymous

limits 0 to 1

7. anonymous

backwards

8. anonymous

parabola is above the line, not below it

9. anonymous

ooopsss I think my drawing is wrong

10. Zale101

which line?

11. anonymous

|dw:1438743560886:dw| i thought this is like this

12. anonymous

not if it is $$y=x^2+2$$

13. anonymous

upper strip - lower strip right?

14. anonymous
15. Zale101

You had y=x^2+2, the parabola is shifted up to 2 units Yes, i graphed mines a little off |dw:1438743695117:dw|

16. Zale101

|dw:1438743765513:dw|

17. Zale101

When you major the rectangle, the top graph is the parabola y=x^2+2 and the bottom of the rectangle is the line graph y=1-x

18. anonymous

Soooo it is [(x^2+2)-(1-x)]

19. Zale101

$\int\limits_{a}^{b}(Y(x)-y(x))dx$

20. Zale101

The limits of the integral are x=0 and x=1 So a=0 and b=1 Y=x^2+2 and y=1-x So: $\int\limits_{0}^{1} [(x^2+2)-(1-x)]dx$

21. anonymous

$[x^2+x-1]$ then this?

22. anonymous

then integrate?

23. Zale101

|dw:1438743978587:dw|

24. Zale101

|dw:1438744033419:dw|

25. anonymous

I got 1/6

26. anonymous

-1/6

27. Zale101

You integrate$\int\limits_{0}^{1}[x^2+x+1]dx$

28. anonymous

Did I got it right?

29. anonymous

11/6!!

30. Zale101

No because your integral is not correct due to not distributing appropriately.

31. Zale101

Yes ! 11/6 :)

32. anonymous

It is 11/6 right???

33. Zale101

Yeah

34. anonymous

Thank you for your help!! Hoping you can help me sometimes again. Thankyou!

35. Zale101

It was good to work with you!

36. Zale101

No problem