﻿ Calculus confusion
23 May, 2013, 01:08:45
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ruler501
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 « on: 13 May, 2011, 01:35:51 » 0

I have started teaching myself calculus through many sources(I don't remember all  of them). I understand most of the stuff with derivatives, but what I don't really get is integrals. i understand how to do the basics just whenever it gets to more complicate problems I get completely lost. Could someone please help me by explaining this.

The hardest ones that for me are ones with fractions and lots of terms.

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 « Reply #1 on: 13 May, 2011, 02:08:42 » 0

I'll assume that you've already been through the fundamental thm of calc, ie integral=antiderivative (more or less)

So basically, integrals are bitchy things to evaluate, and you have to try and figure out which method might be best for each problem (not like derivatives at all)...Some basic guidelines/methods (u and v are letters I will use to notate functions in terms of x)

-Power rule: you should know this one already, integral(x^n)dx=x^(n-1)/(n-1)
-factoring out constants: integral(a*u)dx=a(integral(u)dx)
-natural logs: integral(1/x)dx=lnx
-partial fraction decomposition: this is back from algebra, but you can use it to break large rational functions into natural log pieces
-u substitution. Let u= something and then solve for du and substitute appropriately.
For example, integral(x/(x^2+1))dx Let u=x^2+1 and du (the deriv. of u)=2x. Then, we can rewrite as integral(1/2u)du and solve from there (it becomes ln(u)/2, now substitute back in to get ln(x^2+1)/2)
-Product rule: simply, the integral of the product of two functions u and dv can be rewritten as uv-integral(vdu) (where dv is the derivative of v, etc)
 « Last Edit: 13 May, 2011, 02:13:22 by squidgetx » Logged

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ruler501
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 « Reply #2 on: 13 May, 2011, 02:35:25 » 0

Could you explain a little more on substitution and the product rule?

Also, how do you evaluate it with limits on the upper bounds?
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 « Reply #3 on: 13 May, 2011, 03:57:55 » 0

I'll assume that you've already been through the fundamental thm of calc, ie integral=antiderivative (more or less)

So basically, integrals are bitchy things to evaluate, and you have to try and figure out which method might be best for each problem (not like derivatives at all).

No doubt Integrals are a pain to do, but they *can* be solved algorithmically if both the integrand and the integral are elementary. That's a bit beyond this discussion, though.

@Ruler, for an integral on the interval [A,B], you may find the anti-derivative of f(x) and the definite integral is equal to f(B)-f(A) according to the fundamental theorem of Calculus. However, be careful that the anti-derivative can be found. The vast majority of functions have what are called non-elementary anti-derivatives. These are basically anti-derivatives that exist, but cannot be written down nor found with any of the methods listed by squidgetx.
 « Last Edit: 13 May, 2011, 04:03:11 by Qwerty.55 » Logged

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ruler501
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 « Reply #4 on: 13 May, 2011, 04:01:32 » 0

I like derivatives more they have a set formula for how to find them. This seems too much like guess work to me.

Is there an algorithm for solving integrals that will work most of the time?
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 « Reply #5 on: 13 May, 2011, 04:04:47 » 0

There's only one known [and published] algorithm for computing indefinite integrals, called the Risch Algorithm. How it works is doctoral level stuff, though. It's such a complicated [pseudo-]algorithm that no one has ever actually managed to program the full thing, only the simpler Transcendental case of the algorithm.

However, you can approximate definite integrals quite easily using many different algorithms.
 « Last Edit: 13 May, 2011, 04:06:26 by Qwerty.55 » Logged

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ruler501
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 « Reply #6 on: 13 May, 2011, 04:41:30 » 0

Whats a good method for finding integrals then?
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 « Reply #7 on: 13 May, 2011, 05:06:10 » 0

If you want the honest answer, the best method is just to use a pre-made mathematical tool like Mathematica to find anti-derivatives. However, since you're probably doing this for a class, that's not really an option. On the other hand, integrals for schools tend to be solvable using the methods taught in class, so those are what you should study. The best advice I have on that front is to remember that what you're doing is finding an anti-derivative, so applying the rules of differentiation in reverse can help. Otherwise, learn the methods from squidgetx's list.
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 « Reply #8 on: 13 May, 2011, 05:08:06 » 0

I doing this on my own to teach myself it. My school put me in algebra 1 and refused to let me test out.

I don't understand the last two on squidgetx's list.
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 « Reply #9 on: 13 May, 2011, 05:24:57 » 0

Note: Log(x)==Ln(x) in older notation and that's what I use here.

U substitution: Let's take an example, ∫Log(x)/x dx. Notice that this is equal to ∫(1/x)Log(x) dx and that 1/x is the derivative of Log(x). With U substitution, let U=Log(x) and dU = 1/x dx. Thus, the integral is now ∫ U dU, which is easily solvable as (U²)/2+C. Since that's equivalent to the integral of the original integrand, substitute Log(x) for U and you get Log(x)²/2+C

UV substitution: It can be proven that for an integral ∫U dV, the anti-derivative is equal to UV-∫V dU where U and V are functions. Basically: ∫U dV = UV-∫V dU. What you have to do is figure out U and dV. Then you differentiate U and integrate the simpler equation V. The answer is U*V - the integral of V*dU.

Ex: ∫[Log(Sin(x))]/Cos(x)² dx.

U=Log(Sin(x))
dV=1/Cos(x)²

dU=1/Tan(x)
V=Log(Sin(x))

Thus, ∫[Log(Sin(x))]/Cos(x)² dx = Log(Sin(x))*Tan(x)-∫[1/Tan(x)]*Tan(x) dx

If we then solve the integral on the right (it's equal to 1, therefore it integrates to x+C), we get ∫[Log(Sin(x))]/Cos(x)² dx = Log(Sin(x))*Tan(x)-X+C
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