Last time (on Dragonball Z...) we learned a neat little formula that we could substitute parts in to help evaluate more complicated integrals, and God bless whoever figured that out. There are still some integrals, mainly with more complex trig functions, that don't quite work. For example, if we have
⌠cos³x dx, then the only real substitution we could make would be to set u to cos x. If we did that, our du would be -sinx, which doesn't help us because we don't have one in there to get a du out of. If we want to integrate with cosine, then we have to have a sine function in there to make our du work, and vice versa with integrating a sine function.
But, being the clever mathematicians we are, we can take out of the cosine functions and call it cos²x(cos x). We know that cos²x = 1-sin²x, so this becomes (1-Sin²x)(cos x)dx. Now that we have sines and cosine, we can set u=sin x and get (1-u²)cos x dx. Now that our du is cos x, we get ⌠(1-u²)du, which is perfectly doable.
Since, as I said in the last entry, I'm pressed for time and need to get on with it, I'm going to jump straight to the strategy summaries.
For sine and cosine functions:
1. If the power of cosine is odd, save one of them out and change the remaining even powers of cosine to powers of (1-sin²x), set your u to sin x and du to cos x, and go from there.
2. If the power of sine is odd, save one of them and change the remaining to (1-cos²x) and do the same as above.
3. If both of them are even, use the half-angle identities instead in whatever path you feel works best.
For tangent and secant functions:
1. If the power of secant is even, save a factor of sec²x (make sure you're saving a squared factor, not just sec x, which isn't terribly useful) change the rest to a power of (1+tan²x), with u as tan²x and du as sec²x.
2. If the power of tangent is odd, pull out a (sec x)(tan x) then use (sec²x-1) for the rest of them, then set your u to sec x and tan x as your du.
For products of sine and cosine functions with constants A and B:
1. sinAcosB = .5[Sin(A-B) + sin(A+B)
2. sinAsinB = .5[cos(A-B) - cos(A+B)]
3. cosAcosB = .5[cos(A-B) + cos(A+B)]
The last three weren't proven in the textbook, but he promises that you get that if you solve with integration by parts, so I'll just take his word for it.
That's all for now. I might go back into more depth later, but I have a crapton to do.
Saturday, February 26, 2011
7.1 Integration by Parts
There's still a few back in chapter 6 I didn't get to, but with the psychotic weather we've been having and the subsequent schedule adjustment, we've been skipping around and passing on entire sections. I'll go back and do the other parts eventually, but for now I have to press on.
First off, we learn Integration by Parts.
Since the fundamental theorem linked together integration and differentiation, each rule we've learned for differentiation must also have a corresponding rule of integration. The Product Rule of differention (d/dx(f(x)g(x) = f(x)g'(x)+g(x)f'(x)) is mirrored in Integration by the rule of integration by parts.
The above formula for the Product rule, when substituted for integration, can be stated a few different ways:
⌠[f(x)g'(x)+f'(x)g(x)]dx = f(x)g(x)
or, since the integral of products is the product of the integrals,
⌠[(f(x)g'(x)]dx + ⌠[f'(x)g(x)]dx = f(x)g(x)
and we can re-arrange this a little further to say that:
⌠[f(x)g'(x)dx = f(x)g(x) - ⌠[f'(x)g(x)dx]
That rearranged version of it is the formula for integration by parts. Since that's quite a few letters in there, we can make a little prettier by substituting it for other terms. So we'll say f(x) = u, g(x) = v, and thus f'(x) = du and g'(x) = dv.
So we can restate our formula in the easier to read
⌠u(dv) = uv - ⌠v du
I think that people did it like this just so it would be annoying to write for people whose handwriting is not precise enough to readily tell the difference between u and v.
In general, when you get an ugly-looking integral, pick one of the parts of it to be u and another part to be dv. Find your du and v, then plug it into the above formula and you'll be able to integrate. You may have to loop this several times to end up with your answer, but it works. There are more details but I'm kinda pressed for time so I have to move on 7.2.
First off, we learn Integration by Parts.
Since the fundamental theorem linked together integration and differentiation, each rule we've learned for differentiation must also have a corresponding rule of integration. The Product Rule of differention (d/dx(f(x)g(x) = f(x)g'(x)+g(x)f'(x)) is mirrored in Integration by the rule of integration by parts.
The above formula for the Product rule, when substituted for integration, can be stated a few different ways:
⌠[f(x)g'(x)+f'(x)g(x)]dx = f(x)g(x)
or, since the integral of products is the product of the integrals,
⌠[(f(x)g'(x)]dx + ⌠[f'(x)g(x)]dx = f(x)g(x)
and we can re-arrange this a little further to say that:
⌠[f(x)g'(x)dx = f(x)g(x) - ⌠[f'(x)g(x)dx]
That rearranged version of it is the formula for integration by parts. Since that's quite a few letters in there, we can make a little prettier by substituting it for other terms. So we'll say f(x) = u, g(x) = v, and thus f'(x) = du and g'(x) = dv.
So we can restate our formula in the easier to read
⌠u(dv) = uv - ⌠v du
I think that people did it like this just so it would be annoying to write for people whose handwriting is not precise enough to readily tell the difference between u and v.
In general, when you get an ugly-looking integral, pick one of the parts of it to be u and another part to be dv. Find your du and v, then plug it into the above formula and you'll be able to integrate. You may have to loop this several times to end up with your answer, but it works. There are more details but I'm kinda pressed for time so I have to move on 7.2.
Wednesday, February 9, 2011
6.2 - Volumes
Another fairly short section, since this is the last section in 3D.
The chapter starts off reminding us what a cylinder is, in case you slept through middle school. They also introduce the term "Rectangular parallelepiped" which is a really pretentious way to say "box" because too many people started the chapter knowing what they were talking about.
But if we don't have an easy shape like a cylinder where we can just figure out the information we need from the axes, and instead have the weird-looking slug thing they want us to find the area, we do it similarly to how we find areas, only with cross-sections instead of rectangles. When we slice out a cross-section, we can call it a cylinder with the area of the section and a height of ▲x. Whatever ▲x we choose, if we add up all the cross sections, we get an approximation of the volume the object.
The approximation gets better and better as the number of cross-sections get higher, so, it stands to reason that, just like with our rectangles, the limit of the approximation as the number of cross sections approaches infinity is the volume.
The rest of the chapter just shows a few example problems, as well as proving with Calculus that the volume of a sphere is (4/3)πr^3. The only other main idea the examples show is that if, when rotating around the given axis, you get a shape with a hole in it, you just subtract the volume of the empty spot (found by treating as its own object) from the volume of the outer object.
Then, if the shape cannot be cut into disks, cut it into shapes that work. They showed a square-bottomed pyramid and a wedge cut out of a cylinder. Basically, find how to figure out the volume of a cross section, and use it instead.
Fairly short. Not too bad. The equations they used for the non-cyllindrical shapes were a bit confusing, but I think I'll figure out while working out some of the problems.
The chapter starts off reminding us what a cylinder is, in case you slept through middle school. They also introduce the term "Rectangular parallelepiped" which is a really pretentious way to say "box" because too many people started the chapter knowing what they were talking about.
But if we don't have an easy shape like a cylinder where we can just figure out the information we need from the axes, and instead have the weird-looking slug thing they want us to find the area, we do it similarly to how we find areas, only with cross-sections instead of rectangles. When we slice out a cross-section, we can call it a cylinder with the area of the section and a height of ▲x. Whatever ▲x we choose, if we add up all the cross sections, we get an approximation of the volume the object.
The approximation gets better and better as the number of cross-sections get higher, so, it stands to reason that, just like with our rectangles, the limit of the approximation as the number of cross sections approaches infinity is the volume.
The rest of the chapter just shows a few example problems, as well as proving with Calculus that the volume of a sphere is (4/3)πr^3. The only other main idea the examples show is that if, when rotating around the given axis, you get a shape with a hole in it, you just subtract the volume of the empty spot (found by treating as its own object) from the volume of the outer object.
Then, if the shape cannot be cut into disks, cut it into shapes that work. They showed a square-bottomed pyramid and a wedge cut out of a cylinder. Basically, find how to figure out the volume of a cross section, and use it instead.
Fairly short. Not too bad. The equations they used for the non-cyllindrical shapes were a bit confusing, but I think I'll figure out while working out some of the problems.
Tuesday, February 1, 2011
6.1 - Areas Between Curves
There was actually a section before this, but it was basically an entire chapter spent saying "If the integral is weird, substitute parts for u until it makes sense." We skipped over the reading and just did problems from it in class, so I'm not going to go through the whole chapter for now. Maybe some other time if I'm bored and in a calculus mood. So I'm moving on to 6.1.
We've already learned how to find the area under a curve on a graph, so now we're moving on to putting two lines on a graph at the same time and finding the area between them in a given interval. If we estimate it, we can use the same rectangle trick we did for the area under the curve. The only difference is that, instead of f(xi) being the height, the height will be (Assuming f(x)>g(x) between our chosen a and b) f(xi)-g(xi) since that will be the distance between the two lines on the given point. So our approximation would be Σ[f(xi)-g(xi)]▲x.
So, just like with the area under the curve, if we choose n as our number of rectangles, then the area between these two curves is the limit of our approximation as n approaches infinity. Since that limit is the definite integral of f - g, then we can define the area as A = ⌠(a to b)[f(x) - g(x)]dx.
The book also feels it necessary to point out that, if g(x)=0, then we use the old definition, which makes sense because using g(x) as the 0 line changes nothing.
If two lines are given that intersect at two different points, and your a and b are not given, then they can be found by solving their equations. If you have funky equations that would take ages to do by hand, then a graphic calculator or computer program (which I need to look into getting) will work. We can use Newton's Method, a rootfinder, or the super scientific solution of zooming in really close and seeing where it intersects.
For application-minded folk like me, the area between velocity curves is the distance between the cars at a given point, which is pretty cool.
If you get twisty curves where the lines switch from top to bottom and you a get a DNA-looking twist, then you split the area into multiple regions, find them each individually, then add them together. We can also write a slightly modified area equation that say the area between two curves is the integral of the absolute value of f(x)-g(x).
Also, if the two lines are vertical, you can treat as a function of y, which results in the same process, but flipping the whole thing sideways.
And that's pretty much it. Not a terribly complicated chapter, since once I put the work in to understand the whole integral concept, it's pretty much just application and slightly different ways of thinking about it.
We've already learned how to find the area under a curve on a graph, so now we're moving on to putting two lines on a graph at the same time and finding the area between them in a given interval. If we estimate it, we can use the same rectangle trick we did for the area under the curve. The only difference is that, instead of f(xi) being the height, the height will be (Assuming f(x)>g(x) between our chosen a and b) f(xi)-g(xi) since that will be the distance between the two lines on the given point. So our approximation would be Σ[f(xi)-g(xi)]▲x.
So, just like with the area under the curve, if we choose n as our number of rectangles, then the area between these two curves is the limit of our approximation as n approaches infinity. Since that limit is the definite integral of f - g, then we can define the area as A = ⌠(a to b)[f(x) - g(x)]dx.
The book also feels it necessary to point out that, if g(x)=0, then we use the old definition, which makes sense because using g(x) as the 0 line changes nothing.
If two lines are given that intersect at two different points, and your a and b are not given, then they can be found by solving their equations. If you have funky equations that would take ages to do by hand, then a graphic calculator or computer program (which I need to look into getting) will work. We can use Newton's Method, a rootfinder, or the super scientific solution of zooming in really close and seeing where it intersects.
For application-minded folk like me, the area between velocity curves is the distance between the cars at a given point, which is pretty cool.
If you get twisty curves where the lines switch from top to bottom and you a get a DNA-looking twist, then you split the area into multiple regions, find them each individually, then add them together. We can also write a slightly modified area equation that say the area between two curves is the integral of the absolute value of f(x)-g(x).
Also, if the two lines are vertical, you can treat as a function of y, which results in the same process, but flipping the whole thing sideways.
And that's pretty much it. Not a terribly complicated chapter, since once I put the work in to understand the whole integral concept, it's pretty much just application and slightly different ways of thinking about it.
Subscribe to:
Posts (Atom)