Adding two integrals with different limits of integration

[oddjobmj]

Homework Statement

Interpret the integrals (from 0 to 4)∫ (3x/4) dx + (from 4 to 5)∫ (sqrt(25-x^2)) dx as areas and use the result to express the sum above as one definite integral. Evaluate the new integral.

Homework Equations

The Attempt at a Solution

I see that I could integrate them separately and add the two values to find the total area. The issue I am having is writing them as one integral. I suspect that one way would be to re-write one with the equivilant function but using the limits of integration of the other. However, the only way I've ever changed limits of integration (and all I can find using searches) is when you're doing a u-substition and I'm not sure if the same methodology applies here.

Can I simply add '4' to the first function, 3x/4, to move it over 4 units to the right and then stick that whole thing, 3x/4 + 4, into the other integral between 4 and 5 (along with sqrt(25-x^2)? That would look something like:

the integral from 4 to 5 of (sqrt(25-x^2)+ 3x/4 + 4) dx

Thank you,
Odd

 

[Mark44]

Homework Statement

Interpret the integrals (from 0 to 4)∫ (3x/4) dx + (from 4 to 5)∫ (sqrt(25-x^2)) dx as areas and use the result to express the sum above as one definite integral. Evaluate the new integral.

Homework Equations

The Attempt at a Solution

I see that I could integrate them separately and add the two values to find the total area. The issue I am having is writing them as one integral. I suspect that one way would be to re-write one with the equivilant function but using the limits of integration of the other. However, the only way I've ever changed limits of integration (and all I can find using searches) is when you're doing a u-substition and I'm not sure if the same methodology applies here.

Can I simply add '4' to the first function, 3x/4, to move it over 4 units to the right

No. Adding 4 to the formula of a function shifts the graph up, not right.

and then stick that whole thing, 3x/4 + 4, into the other integral between 4 and 5 (along with sqrt(25-x^2)? That would look something like:

the integral from 4 to 5 of (sqrt(25-x^2)+ 3x/4 + 4) dx

Thank you,
Odd

I think that you are missing the point of this exercise. The first part of this problem asks you to interpret the two integrals as areas, which implies that you should sketch a graph of each function being integrated. It doesn't seem to me that you have done that.

To express the sum as a single integral is easy to do if you are working with polar integrals.

 

[oddjobmj]

I did graph them and it's easy to visualize the sum. I believe you when you suggest using polar integrals but I have -no idea- how to go about that. I just watched a video on double integrals using polar coordinates to see if it shed some light but I think I'm more confused than before now. I'll keep looking for more on that topic but would greatly appreciate any hints on how to get started.

I suspect the solution is simple but I feel that I'm simply missing a basic idea here.

 

[Dick]

I did graph them and it's easy to visualize the sum. I believe you when you suggest using polar integrals but I have -no idea- how to go about that. I just watched a video on double integrals using polar coordinates to see if it shed some light but I think I'm more confused than before now. I'll keep looking for more on that topic but would greatly appreciate any hints on how to get started.

I suspect the solution is simple but I feel that I'm simply missing a basic idea here.

What does the region you are finding the total area of look like? Can you describe it geometrically?

 

[oddjobmj]

Sure, a straight line starting at the origin up and to the right (described by 3x/4) and when x=4 the function curves downwards and ends up at (5,0) (described by sqrt(25-x^2). I did this by graphing them both separately and visualizing the combination. Maybe that's skipping a step you're looking for me to take?

EDIT: Sorry, I described the line. The region would be the area between that line and the x-axis.

 

[Dick]

Sure, a straight line starting at the origin up and to the right (described by 3x/4) and when x=4 the function curves downwards and ends up at (5,0) (described by sqrt(25-x^2). I did this by graphing them both separately and visualizing the combination. Maybe that's skipping a step you're looking for me to take?

EDIT: Sorry, I described the line. The region would be the area between that line and the x-axis.

What does the curve y=sqrt(25-x^2) look like geometrically? Answer that one and if you are still confused, I'll tell you the answer I'm fishing for.

 

[oddjobmj]

Half of an oval. More specifically, the 'upper' half of an oval with the longest axis parallel with to y-axis. Similar to a parabola.

I suspect that you'd want me to express this in a polar system. I don't know how to go about this :(

 

[Dick]

Half of an oval. More specifically, the 'upper' half of an oval with the longest axis parallel with to y-axis. Similar to a parabola.

I'd call your 'oval' a 'circle'. The region whose area you are finding is a sector of a circle. Hence, Mark44's hint about polar coordinates. And it's not a double integral. Look up 'area in polar coordinates'.

 

[oddjobmj]

The end point of the first segment is (5,arcsin(3/5)). Working on the second now.

 

[Dick]

The end point of the first segment is (5,arcsin(3/5)). Working on the second now.

You can probably stop working on it, because I don't know what you are talking about. The region is a region sliced out a circle that looks like a slice of pie. It has fixed radius and a beginning angle and an ending angle. Work on those instead.

 

[oddjobmj]

Haha, this is getting pretty silly. I appreciate all the help so far. I think the confusion comes from the image I have in my mind versus the image I should have and the image you are visualizing.

That's what I was thinking. You add the two areas of the lines given the different limits of integration.

What I think you're looking for now is something like this: (extracted from a youtube video from patrickJMT titled 'Finding Areas in Polar Coordinates').

What I'm having a hard time doing is visualizing the -addition- of the two functions while they sit on top of one another finding two different values for theta. That idea makes me think subtraction, not addition. Thoughts?

You can probably stop working on it, because I don't know what you are talking about. The region is a region sliced out a circle that looks like a slice of pie. It has fixed radius and a beginning angle and an ending angle. Work on those instead.

The point where the two lines meet in the first picture I posted is described in polar form as (5, arcsin(3/5)). arcsin(3/5) is the angle theta that the first function makes with the x-axis and 5 is the radius.

 

[Dick]

Haha, this is getting pretty silly. I appreciate all the help so far. I think the confusion comes from the image I have in my mind versus the image I should have and the image you are visualizing.That's what I was thinking. You add the two areas of the lines given the different limits of integration.

What I think you're looking for now is something like this: (extracted from a youtube video from patrickJMT titled 'Finding Areas in Polar Coordinates').

What I'm having a hard time doing is visualizing the -addition- of the two functions while they sit on top of one another finding two different values for theta. That idea makes me think subtraction, not addition. Thoughts?

You've got exactly the right polar coordinate reference. But forget about adding anything. Now you have a single simple region. Find it's area with a single simple integral in polar coordinates.

 

[oddjobmj]

That's the only issue I'm having. I've had the area of this function down about one minute into the problem by simply adding the areas of each respective integral.

The issue I'm having is writing out -one integral- to represent them both. I'm working on understanding how to write these two as one integral using polar equations but am not yet there.

 

[Dick]

That's the only issue I'm having. I've had the area of this function down about one minute into the problem by simply adding the areas of each respective integral.

The issue I'm having is writing out -one integral- to represent them both. I'm working on understanding how to write these two as one integral using polar equations but am not yet there.

Stop saying "these two areas". In polar coordinates you can write the integral expressing the area of the region as a single integral. That's the whole point. Don't do them separately! What's r as a function of theta in the combined region? Then what are the limits for theta?

 

[oddjobmj]

Stop saying "these two areas". In polar coordinates you can write the integral expressing the area of the region as a single integral. That's the whole point. Don't do them separately!

Like I said, that's exactly what I am trying to do... I see that as theta changes r changes as it follows the curve down. Then I see that if I integrate that function with respect to theta (multiplied by a half because it's a triangle not a rectangle) it will give me the overall area from theta at point 4 to theta at point 5.

Thanks

 

[Dick]

Like I said, that's exactly what I am trying to do... I see that as theta changes r changes as it follows the curve down. Then I see that if I integrate that function with respect to theta (multiplied by a half because it's a triangle not a rectangle) it will give me the overall area from theta at point 4 to theta at point 5.

Thanks

How does r change? Do you still think it's 'oval'? It's not. It's a circle. Even if it weren't you could still write the area as a single integral. It would just be more complicated.

 

[oddjobmj]

When I plot that function it looks like an oval.

http://www.wolframalpha.com/input/?i=plot+y=sqrt(25-x^2)+x=0..5

I'm not saying you're wrong. I just don't know how else to think about it.

At this point I don't know how r changes but I'm sure I can find that out using one of the polar formulas. Finals + missing concepts = frustration. I will have to come back to this problem because it's driving me nuts.

Thanks again and I will report back here after I get some sleep.

 

[Dick]

When I plot that function it looks like an oval.

http://www.wolframalpha.com/input/?i=plot+y=sqrt(25-x^2)+x=0..5

I'm not saying you're wrong. I just don't know how else to think about it.

The axes in that plot have different scales. Don't trust a Wolfram Alpha picture.

 

[Mark44]

When I plot that function it looks like an oval.

http://www.wolframalpha.com/input/?i=plot+y=sqrt(25-x^2)+x=0..5

I'm not saying you're wrong. I just don't know how else to think about it.

The only reason the graph looks like part of an ellipse is that the x and y axes are scaled differently. Seriously, if you are working with integrals, you should know that the graph of y = [itex]\sqrt{25 - x^2}[/itex] is the upper half of a circle.

 

[oddjobmj]

One thing that happens when the human brain is on very little sleep, is stressed, and is overworked is it begins to overlook things. I defaulted to the easiest method I know to understand the shape of something and that is to graph it. Yes, I know the equation of a circle. I'm sorry that I missed the swap from x^2+y^2=r^2 to y=sqrt(25-x^2) but you don't have to be rude about it. It really doesn't help solve anything...

Why I missed it probably has something to do with the fact that up until one post before the mention of shape it didn't matter if it was a circle or oval in the way I was visualizing the information.

 

[Mark44]

I'm not being rude - I'm being realistic. Lack of sleep and high stress level notwithstanding, it is reasonable to expect that a student who is working with integrals would recognize a function whose graph is a semicircle.

 

[oddjobmj]

Great point there, Mark44. Glad to have your vote of confidence.

Getting back to the problem though:

I obviously had some issues with the polar representation. But that thought process lead me to a few conclusions so went back through without the use of polar equations and got this:

As y changes from 0 to 3 we can build rectangles (rather than triangles using theta) and simplify the problem significantly.

The vertical component of each rectangle is dy and the horizontal component is (sqrt(25-y^) - 4y/3). Integrating this between 0 and 3 (the implied limits of y) we get 8.04376 (link below for visual). Which is the same answer I got for the sum of the two integrals separately and the same as the answer in the back of the book.

http://www.wolframalpha.com/input/?i=integrate+from+0+to+3+(sqrt(25-y^2)-4y/3)

Although this seems to be the easiest way to get the answer I am very glad to have been reintroduced to the polar representation because the coverage of that section in this class was pretty much negligible.

Please forgive my ignorance, Mark44. I'll note here though that whether the shape is a circle or oval does not matter.

 

[Mark44]

Sure, you can use horizontal strips to integrate, but the simplest way, by far, is a polar integral.

The polar function is r = 5, with θ ranging from 0 to tan^-1(3/4).

The integral is
[tex]\int_0^{tan^{-1}(3/4)} \frac{25}{2}dθ = 12.5 \cdot tan^{-1}(3/4) ≈ 8.0438[/tex]

 

[oddjobmj]

Awesome, two ways to do the problem :) I agree now that the polar representation is simpler if you're familiar with the methodology. I wasn't if that's not already apparent.

Thanks guys!