Area bounded by curves-Integral

[Justabeginner]

Homework Statement

Find the area bounded by the curves, y= √x, y= (5-x)/4, and y= (3x-8)/2

Homework Equations

The Attempt at a Solution

I found the intersection between each of the three curves to each other. Not sure what exactly the area bounded is. Is it the small triangular area underneath all three with (5,0) as one vertex, and (3,1/2) as another or?

(3, 1/2)
(16/9, -4/3)
(1,1)

^These are what I think the three intersection points are.

∫((5-x)/4) - √x with b= 1 but I'm not sure what the lower limit is. I'm really stuck. Guidance please?

 

[Dick]

Homework Statement

Find the area bounded by the curves, y= √x, y= (5-x)/4, and y= (3x-8)/2

Homework Equations

The Attempt at a Solution

I found the intersection between each of the three curves to each other. Not sure what exactly the area bounded is. Is it the small triangular area underneath all three with (5,0) as one vertex, and (3,1/2) as another or?

(3, 1/2)
(16/9, -4/3)
(1,1)

^These are what I think the three intersection points are.

∫((5-x)/4) - √x with b= 1 but I'm not sure what the lower limit is. I'm really stuck. Guidance please?

(16/9, -4/3) can't be an intersection point with y=sqrt(x). -4/3 is negative. Try that one again.

 

[Justabeginner]

Oops! It's 4/3 (dumb mistake).

 

[Dick]

Oops! It's 4/3 (dumb mistake).

No, -4/3 is the correct y value if x=16/9. So that's not an intersection. But there's another x value that is.

 

[Justabeginner]

So for my intersection point I get:

2* sqrt(x) = (3x-8)^2
4x= 9x^2 - 48x + 64
9x^2 - 52x + 64 =0
x= 4 or x= 16/9
x= 4 therefore y= 2

Is (4,2) the right one?

 

[Dick]

So for my intersection point I get:

2* sqrt(x) = (3x-8)^2
4x= 9x^2 - 48x + 64
9x^2 - 52x + 64 =0
x= 4 or x= 16/9
x= 4 therefore y= 2

Is (4,2) the right one?

Yes, it is. So you want the area of the sort of triangular region between those three points.

 

[Justabeginner]

So the three points are (1,1) (3, 1/2) and (4,2).

From 1 to 3, (5-x)/4 is greater than sqrt(x). So I would find the integral of ((5-x)/4 - sqrt(x)) on the interval (1,3)? And the other interval would be (3,4) and the integral on this interval would be (3x-8/2) - (5-x/4)?

Am I doing this right? And I know there should be a third integral but I can't figure it out..

 

[Dick]

So the three points are (1,1) (3, 1/2) and (4,2).

From 1 to 3, (5-x)/4 is greater than sqrt(x). So I would find the integral of ((5-x)/4 - sqrt(x)) on the interval (1,3)? And the other interval would be (3,4) and the integral on this interval would be (3x-8/2) - (5-x/4)?

Am I doing this right? And I know there should be a third integral but I can't figure it out..

No, there's no third integral. But you should always sketch a graph when you are doing a problem like this. You ideas about what is greater than what are wrong. But you've got the right idea how to set the integrals up.

 

[Justabeginner]

I did sketch them, but I guess I need to pay more attention.

So 5-x/4 is NOT greater than sqrt(x) on the interval from 1 to 3?
And 3x-8/2 is similarly not greater than 5-x/4 on (3,4)?

I don't understand how that holds true.. Please explain? :/

 

[Dick]

I did sketch them, but I guess I need to pay more attention.

So 5-x/4 is NOT greater than sqrt(x) on the interval from 1 to 3?
And 3x-8/2 is similarly not greater than 5-x/4 on (3,4)?

I don't understand how that holds true.. Please explain? :/

I'd be interested in seeing what your sketch looks like. 2 is in (1,3). If you put x=2 into (5-x)/4 you get 3/4. If you put it into sqrt(x) you get sqrt(2). Which is greater?

 

[Justabeginner]

sqrt(2) is greater. I realized that I didn't plug in the correct values! But using the same logic, 3x-8/2 is still greater than 5-x/4 if I'm not mistaken?

Therefore it would be sqrt(x) - ((5-x)/(4)) with a= 1 and b= 3.
And 3x-8/2 - ((5-x)/(4)) with a= 3 and b= 4?

 

[Dick]

sqrt(2) is greater. I realized that I didn't plug in the correct values! But using the same logic, 3x-8/2 is still greater than 5-x/4 if I'm not mistaken?

Therefore it would be sqrt(x) - ((5-x)/(4)) with a= 1 and b= 3.
And 3x-8/2 - ((5-x)/(4)) with a= 3 and b= 4?

In my graph it looks like the area you want is bounded above by sqrt(x) and below by ((5-x)/4) on the interval (1,3) and above by sqrt(x) and below by (3x-8)/2 on (3,4). The only regions I see that bounded above by one line and below by the other aren't part of the region you want to find the area of.

 

[Justabeginner]

I graphed it on a calculator, and I see what you are referring to. My calculation for the total area is 23/12. Is this correct? Thanks.

 

[Dick]

I graphed it on a calculator, and I see what you are referring to. My calculation for the total area is 23/12. Is this correct? Thanks.

Yes, that's correct.

 

[Justabeginner]

Thank you so much. Now I leave with a much better understanding of this concept than I arrived with! :)