Homework Help: Area under curve questions

1. Feb 5, 2013

lionely

1. The problem statement, all variables and given/known data
Find the areas enclosed by the following curves and straight lines:
c) y= (1/x2) -1 , y= -1 , x=1/2, and x=2

b) y = x3-1, the axes and y = 26

2. The attempt at a solution
Okay I sketched the curve and to me it looks like the curve occupies no area at y=-1 it's undefined there.

so isn't the area occupied from x = 1/2 to x =2

so ∫1/x2) -1 . dx
y= [(1/x) - x ] + c

( (1/2) - 2) - ( 2 - (1/2)) = -3
but the answer in my book says 3/2

and for the 2nd question
The points at which these curves intersect are (3,26)
∫x3 -1 . dx
y = (x4/4) - x = 17 1/4 ( x = 0 to 3)

∫26.dx = 26x = 78 sq units( from x 0 to 3)

78- 17/4 = 60 3/4 but in the back of my book it says 60 exactly, and I don't see what I did wrong..

Help is greatly appreciated~!!

2. Feb 5, 2013

Dick

The area between two curves f(x)>g(x) is the integral of f(x)-g(x). In the first case, f(x) is 1/x^2-1 and g(x) is (-1).

3. Feb 5, 2013

lionely

So you mean just to (1/x)-x -x? and that's the area?
But what do I put in for the values of x.. 1/2 and 2?

4. Feb 5, 2013

Dick

The function to integrate is ((1/x^2)-1)-(-1). Yes, put the x limits to 1/2 and 2. Your previous attempt simply ignored the y=(-1) for reasons that are somewhat obscure to me. It's true that the two curves don't cross. But you can't ignore the lower curve.

5. Feb 5, 2013

lionely

I thought that it had no area at y=-1 because when y=-1 the function is undefined .

So do the same thing for the question before?

>.> sorry it's not undefined I don't know what I was thinking about.................

6. Feb 5, 2013

Dick

There is no point where 1/x^2-1 equals (-1). Correct. That just means the two curves don't intersect each other. Doesn't mean you can ignore one.

7. Feb 5, 2013

lionely

When I use 1/x I keep getting -3/2

8. Feb 5, 2013

Dick

Show how you got that. Hint: the integral of 1/x^2 is not 1/x.

9. Feb 5, 2013

lionely

=.= my god .... didn't see it's -1/x forgot to divide by -1

10. Feb 5, 2013

lionely

but for the 2nd question I can't get the exact 60 I get 60 3/4
Isn't the integral [ (x^4/4) -27x] ?

11. Feb 5, 2013

Dick

You really need to draw a sketch of the region you are integrating over. You are working blind here. You don't integrate over that function for the whole interval [0,3]. At some point x^3-1 will cross the x-axis. That's supposed to be one of your boundaries. You need to split the integral up.

12. Feb 5, 2013

lionely

Okay so I integrate the region x=0 to x=1 and then x= 1 to x=3

then subtract them?

13. Feb 5, 2013

Dick

I would say add them. And what you are integrating in each region is different. What two different forms are they? Why do you say subtract them?

14. Feb 5, 2013

lionely

Oh because I was thinking about what you said area between curves is
f(x) - g(x) but now when I look at my sketch that wouldn't make sense.

15. Feb 5, 2013

Dick

The graph will tell you what to do.

16. Feb 5, 2013

lionely

I just don't understand! The integral is ((x^4/4) -27x) right? when I integrate from x=0 to x=1 I get -26 3/4

and when I do it for x=1 to x=3 I get -34.. I can't get the +60 !

17. Feb 5, 2013

Dick

What are the boundaries between x=0 and x=1? Hint: Neither of them is x^3-1. Look at your graph! The upper boundary is 26 and the lower boundary is the x-axis!!

18. Feb 5, 2013

lionely

I think I got it now... the integral for x=0 and x=1 is 26x so it's 26..

then for the other side of the graph it's (x^4/4 -x) for x=3 and x=1 and that's 18
then I find the one for 26x for x = 3 and x= 1 and I get 52

Then I subtract the areas and get 34 then I add it to the area bounded by the 26
and I got 60.

19. Feb 5, 2013

Dick

Yes, x=0 to x=1. It's 26. From x=1 to x=3 it's the integral of 26-(x^3-1) which gives 34. Then you just add them. But your way works too.

20. Feb 5, 2013

lionely

Umm 1 question I know this might be common sense.. but
on these graphs, the one that looks like it covers more area is the one that I will subtract whatever the function? Like y=26 looks like it covers more than x^3-1.. but maybe it looks that way cause of my sketch sometimes I can't trust my sketch that much cause my work is very untidy.

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