How Do You Calculate Areas Enclosed by Curves and Lines?

[lionely]

Homework Statement

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~!

 

[Dick]

Homework Statement

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~!

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).

 

[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?

 

[Dick]

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

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.

 

[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....

 

[Dick]

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....

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.

 

[lionely]

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

 

[Dick]

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

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

 

[lionely]

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

 

[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] ?

 

[Dick]

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] ?

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.

 

[lionely]

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

then subtract them?

 

[Dick]

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

then subtract them?

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?

 

[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.

 

[Dick]

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.

The graph will tell you what to do.

 

[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 !

 

[Dick]

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 !

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!

 

[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.

 

[Dick]

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.

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.

 

[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.

 

[Dick]

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.

The area between x=0 and x=1 is 26, no doubt, yes? Between x=1 and x=3 the value of 26 is above and x^3-1 is on the bottom. So it's the integral of 26-(x^3-1) which is 34. You just add areas. The graph is supposed to tell you which function is on top and which is on the bottom.