## anonymous one year ago Find the area of the region bounded by the functions f(x) = x^4 and g(x) = 2 − x^2.

1. anonymous

i've found the area bounded by the x-axis and f(x) its 3.771123

2. anonymous

would i have to integrate with respect to the y -axis on this one?

3. IrishBoy123

have you plotted this can you see the area enclosed by these curves?

4. anonymous

yep iv graphed it, i know what it looks like

5. IrishBoy123

you are looking for the area bounded by the curves so you need to know where they meet to establish the limits of integration

6. IrishBoy123

you could draw it using the Draw function here

7. anonymous

okay dont know how to draw function but the graphs seem to come together at x =1 and x=-1

8. IrishBoy123

so firstly integrate the functions within those limits and see what comes out

9. blackstreet23

It seems you need to use the method of slicing

10. anonymous

which function f(x)or g(x)

11. blackstreet23

this looks like a disk

12. anonymous

is their a formula to this method ?

13. IrishBoy123

@blackstreet23 are you deliberately trying to take this of course? this is a simple integration.

14. blackstreet23

no i just want to help

15. welshfella

|dw:1438814403525:dw|

16. blackstreet23

ohh yeah i am sorry it is not revolving haha

17. welshfella

intersections at (-1'1) and (1,1)

18. anonymous

yes okay i got that now what ....?

19. blackstreet23

$\int\limits_{?}^{?}( 2-x^2)-(x^4)$

20. blackstreet23

if i am right this is because it is suppose the areas under the curve

21. IrishBoy123

|dw:1438814688554:dw| "just" integrate

22. anonymous

okay so for the boundris it would just be -1 to 1 right ?

23. blackstreet23

if you see on the drawing that one of the users drew the upper function is 2-x^2an

24. blackstreet23

and the lower function is X^4

25. anonymous

yay okay that gave me 44/15

26. IrishBoy123

yes, that is right limits are $$-1 \le x \le +1$$

27. welshfella

integrate 2 - x^2 between the limits 0 and 1 integrate x^4 between limits 0 and 1 subtract the second result from the first then as the area is symmetrical about the y-axis multiply this result by 2

28. anonymous

why do i have to multiply by 2 ?

29. welshfella

because i integrate between 0 and 1 but its more direct to use limits -1 and 1

30. blackstreet23

but why should it be integrated separately if the bounded region is above the x-axis?

31. IrishBoy123

he changed the limits from -1, 1 to 0,1 as they are symmetrical, then x2 : but you can just do -1, 1 and fly with it

32. anonymous

okay the answer is 44/15 either way

33. welshfella

yes - i dont know why i did it the longer way Irish Boy is right both methods will gave same answer

34. anonymous

okay thank you very much :D

35. welshfella

yw

36. IrishBoy123

$$\huge A = \int_{-1}^{1} 2 - x^2 - x^4 \ dx$$ $$\huge = [2x - \frac{x^3}{3} - \frac{x^5}{5} ]_{-1}^{1}$$ $$\huge = 2 [2x - \frac{x^3}{3} - \frac{x^5}{5} ]_{0}^{1}$$ $$\huge = 2 [2 - \frac{1}{3} - \frac{1}{5} ] = 2(\frac{30 - 5 - 3}{15}) = 2( \frac{22}{15}) = \frac{44}{15}$$ $$\large \checkmark$$