- #1

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(Self study-stuck on this example)

Edit: the second step is where it loses me (1/2*cos(u)*2

- Thread starter Shinaolord
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- #1

- 92

- 4

(Self study-stuck on this example)

Edit: the second step is where it loses me (1/2*cos(u)*2

- #2

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First, you have to look at your function and determine u. In this case, your function is cos(2x) dx, so it is obvious that you should let u=2x.

Then, you must find du in terms of dx. From u=2x, you get du = 2* dx. Once this is done, you need to isolate dx in the equation. So we get dx = du/2.

Finally, we substitute du/2 for dx and u for 2x in the original integrand, getting integrate(cos(u) du/2), which is 1/2* integrate (cos(u) du), which is 1/2*sin(u) +C.

As the last step, you substitute 2x back for u. So you get 1/2*sin (2x) +C.

- #3

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Thank you! I could not find any way to even come up with that methodology. Now I can progress and prepare myself for Michigan technological university August 24th :)

My eternal gratitude, good sir.

MAy I ask the importance of Simpsons rule?

My eternal gratitude, good sir.

MAy I ask the importance of Simpsons rule?

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- #4

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And I learned to estimate integrals using the trapezoidal rule, but the wikipedia's article on simpson's rule is very easy to understand. I'm sure you can understand it, although its derivation is not so easy.

And I don't exactly understand what is meant by "importance". Everything in math is important-is the question asking why Simpson's rule is accurate?

- #5

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You might not need to know that the integral is 9*pi*sqrt(8). Maybe you just need to know that it's about 55 point something. This is important in engineering and applied physics.

Some integrals it is IMPOSSIBLE to find an antiderivative for, such as e^(x^2).

- #6

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I set u=tan2x

Thus du=sec^2(2x) dx

And dx=du/sec^2(2x)

And I can't figure out where to go

;( preparing for calculus is hard IMO.

- #7

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- #8

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But even then, your sec should cancel out and you should be left with a "u" only.

- #9

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Could it be that du/dx=sec^2(2x)? That's the only thing I can think may be wrong

- #10

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Now, you don't ALWAYS solve for dx. You are trying to solve for whatever the second part of the function is-in this case, sec^2(2x). The reason behind this is that when you are substituting, you are trying to eliminate x by using u. When you have u=tan(2x), the tan(2x) part has been taken care of, but not the sec^2(2x) dx. Thus, you are trying to find du in terms of sec^2(2x) dx, so that you can substitute du for it.

That said, du/2=sec^2(2x) dx. So by substitution, integrate tan(2x)*sec^2(2x)=1/2*integrate u du, which is equal to 1/4*u^2+c, which is equal to 1/4*tan^2(2x)+C.

- #11

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That's the thing I don't understand. That 2.

I feel as if it comes out of thin air. Please try to keep in mind I am still a "high school student",and I have a very limited understanding of the chain rule. Just the basics of it.

- #12

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"Actual physical application". Well, Simpson's rule is a way of numerically approximating integrals. So what you're asking for is applications of integrals in physics. If you go into physics like you say you want to. In a year or two, you will look back at this question and laugh. :rofl:

I'll give you a very basic example of an application of integration. Newton's Second law states that

ƩF=ma. The sum of the forces on an object is equal to the mass of the object times its acceleration. Now what is acceleration? It's v'. What's v? It's x'. Where v is velocity and x is position.

So his second law states that

ƩF=mx''. The sum of the forces on an object is equal to the mass of the object times the second derivative of its position. Now if I were to give you a "force function", and you wanted to find its position (x) function, you would need to integrate TWICE.

Simpson's Rule is cool because sometimes the functions can become quite complicated. You know the trouble you're having with integrating basic functions? Imagine if they gave you a function that was HUGE or one that WAS IMPOSSIBLE TO FIND AN ANTIDERIVATIVE FOR. You would still need to find the position, but you would have to use it using Simpson's rule.

On another note, not trying to put you down or anything, but one thing at a time. You are just out of high school. A PhD should be the last thing on your mind. Take things one step at a time, just worry about undergrad for the time being. I strongly encourage you to pursue physics and study all you can in it, but don't worry about a PhD. You have no idea what physics is like at that level (or even at an undergrad level, TBH) and it's a little presumptuous to say that you could handle or even WANT one. It's not saying that you're dumb, or anything negative about you personally, but keep your doors open.

- #13

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EDIT: Thank you so much for the basic, basic application of Simpson's rule. now I think/feel I have a vague but better idea of what it's used for. truly I love this forum :-)

- #14

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Similarly, the derivative of (2x)^2 is 4x times the derivative of 2x, which is 4x*2, which is 8x.

Note that (2x)^2 =4x^2, and d/dx 4x^2=8x, which confirms the result.

- #15

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Thank you the magic 2 has been found :-) rofl

Similarly, the derivative of (2x)^2 is 4x times the derivative of 2x, which is 4x*2, which is 8x.

Note that (2x)^2 =4x^2, and d/dx 4x^2=8x, which confirms the result.

- #16

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Maybe you should stop working on antiderivatives and work on strengthening your differentiation. Antiderivatives are to derivatives as division is to multiplication.

That's the thing I don't understand. That 2.

I feel as if it comes out of thin air. Please try to keep in mind I am still a "high school student",and I have a very limited understanding of the chain rule. Just the basics of it.

Division is guess and check based on what you know about multiplication. Same thing with computing antiderivatives. Unless your differentiation is strong, you won't be able to antidifferentiate well.

- #17

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- #18

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I strongly encourage you to work towards it. If it's what you want, go for it. Just realize things change a lot.

EDIT: Thank you so much for the basic, basic application of Simpson's rule. now I think/feel I have a vague but better idea of what it's used for. truly I love this forum :-)

In my area (mathematics), everything changes a lot from one step to the next. I am a 4th year undergrad, and am starting some graduate course work in the fall. Upper division is completely different from lower division. Some people love math until they hit their first upper division class, where everything gets flipped around, then hate it. Also, in studying for my graduate class, it's a lot different as well. Some people can do well in undergrad and hate grad. Or some people can hate undergrad and love grad. Hate lower div, love upper div. Different levels are just so different and require different approaches, levels of thinking.

In lower division, it's a lot about performing computations without error. And some people are really good at that. Some people aren't. Upper division is about proof writing, and some people hate that. Others love it. I really liked both levels, hard to say which I preferred. And my grad class is going to be a lot different as well, I don't know how I'll like that.

- #19

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I see. Well, thank you for the encouragement. I much appreciate it

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