What is the area enclosed by the curves y=√x, y=1/2x, and x=25?

[Gundown64]

I hope you guys can help. I feel like I know how to do this, but I keep getting the same answer no matter what I do, which is wrong.

Sketch the regions enclosed by the given curves.
y=√x, y=1/2x, x=25
Find its area.


The attempt at a solution
∫4,25 ((1/2x)-(√x))dx
[(1/4x²)-(2/3x^3/2)] integrated at 4 and 25

Basically I solve for 25, solve for 4, and subtract them. I end up with 297/4 as my answer everytime. What am I doing wrong?

 

[tylerc1991]

I hope you guys can help. I feel like I know how to do this, but I keep getting the same answer no matter what I do, which is wrong.

Sketch the regions enclosed by the given curves.
y=√x, y=1/2x, x=25


The attempt at a solution
∫4,25 ((1/2x)-(√x))dx
[(1/4x²)-(2/3x^3/2)] integrated at 4 and 25

Basically I solve for 25, solve for 4, and subtract them. I end up with 297/4 as my answer everytime. What am I doing wrong?

Why are you integrating from 4 to 25? I get the 25 part, but the curves do not intersect at 4.

 

[Mark44]

I hope you guys can help. I feel like I know how to do this, but I keep getting the same answer no matter what I do, which is wrong.

Sketch the regions enclosed by the given curves.
y=√x, y=1/2x, x=25


The attempt at a solution
∫4,25 ((1/2x)-(√x))dx
[(1/4x²)-(2/3x^3/2)] integrated at 4 and 25

Basically I solve for 25, solve for 4, and subtract them. I end up with 297/4 as my answer everytime. What am I doing wrong?

The problem is asking for a sketch of the region, not its area. In your problem description, you didn't say anything about finding the area of this region. Did you forget to add that bit of information?

In any case, your first step should be to graph the region so that you can get the correct limits of integration and the correct integrand.

 

[Gundown64]

Well, it looks like part of this may be my own fault for believing this graph? The homework claims this (attached document) is the correct graphing of the two curves, which intersect at (4,2). Which would mean integrating from 4 to 25, correct? Sorry, and it goes on to say (at the bottom) to find the area.

 

[SammyS]

Well, it looks like part of this may be my own fault for believing this graph? The homework claims this (attached document) is the correct graphing of the two curves, which intersect at (4,2). Which would mean integrating from 4 to 25, correct? Sorry, and it goes on to say (at the bottom) to find the area.

So, the task at hand is to sketch the region above y=\sqrt{x}\,, below \displaystyle y=\frac{1}{2}x\,, and to the left of x=25\,.

The sketch is done in your thumbnail (shown below).

Now, it seems you are trying to find the area of the shaded region.

 

[Gundown64]

So, the task at hand is to sketch the region above y=\sqrt{x}\,, below \displaystyle y=\frac{1}{2}x\,, and to the left of x=25\,.

The sketch is done in your thumbnail (shown below).

image

Now, it seems you are trying to find the area of the shaded region.

Correct. So, basically I found the area below 1/2x by finding the anti derivative and integrating from 4 to 25. Then I did the same for sqrt(x) and subtracted. Is that not correct?

 

[SammyS]

Correct. So, basically I found the area below 1/2x by finding the anti derivative and integrating from 4 to 25. Then I did the same for sqrt(x) and subtracted. Is that not correct?

That's OK.

You can do it all at once, but for these functions it doesn't really save any work.

\displaystyle \int_4^{25}\left( \frac{1}{2}x-\sqrt{x}\right)dx​

 

[Gundown64]

That's OK.

You can do it all at once, but for these functions it doesn't really save any work.

\displaystyle \int_4^{25}\left( \frac{1}{2}x-\sqrt{x}\right)dx​

That's how I did it, I just explained it the longer way. However, my answer, 297/4, is not right according to the online homework program.

 

[Mark44]

Well, it looks like part of this may be my own fault for believing this graph? The homework claims this (attached document) is the correct graphing of the two curves, which intersect at (4,2). Which would mean integrating from 4 to 25, correct? Sorry, and it goes on to say (at the bottom) to find the area.

Why isn't the piece between 0 and 4 included?

Sketch the regions enclosed by the given curves.
y=√x, y=1/2x, x=25
Find its area.

Sketch the regions[/color] ...

The last part is grammatically incorrect. The first part asks for regions, so the last part should say, "find their area."

BTW, when you write y = 1/2x, that is technically correct, but to be clearer, it could be written as y = (1/2)x. Many members here would interpret what you wrote as 1/(2x), which is not what you meant, and also incorrect.