# Thread: I want nothing more than to take calculus

1. Originally posted by Silvercord
ygf I think I understand everything you've said in at least the broadest terms.
That's what I intended. I'm not getting into specifics here... for sanity's sake. That stuff you can learn on your own.
Are integrals used to find the area/volume of any weird forumla
Yep.
(even graphed in 3d?).
It seems that integrals are only an approximation.

f(t) = t + 3
The slope of that line, it's derivative, is 1.
f'(t) = 1
f'(t) just means the derivative of f(t)

Say I wanted to find out the area under f'(t) from t=0 to t=3. Since the integral is the area under the curve, and the integral is opposite the derivative:
f(t) is the integral of f'(t)
f(3) - f(0) = 6 - 3 = 3

That's exact. No approximation there. For polynomials you often don't need approximations because some formula will give you the exact answer. Sometimes you need to approximate because there is no other way. (Logs are approximations.)

2. That's exact. No approximation there. For polynomials you often don't need approximations because some formula will give you the exact answer. Sometimes you need to approximate because there is no other way. (Logs are approximations.)
Well, sort of. An answer in the form of (log x) or (sin x) or something of that nature is considered to be exact, though a numerical approximation may be more useful from a practical standpoint. There are, however, functions that have no closed form integral - that is, their integral cannot be expressed in terms of elementary functions, and therefore, the integrals can only be approximated.

One that comes to mind in one-variable calculus is the integral of e^(x^2) dx.

3. Is the product rule limited to two multiplying terms?
Ok, suppose something like x^3 * y^3 can be expressed as xyxyxy. By using the product rule (without simplifying) directly onto it, can it be derived?

4. Is the product rule limited to two multiplying terms?
Ok, suppose something like x^3 * y^3 can be expressed as xyxyxy. By using the product rule (without simplifying) directly onto it, can it be derived?
Yes, the product rule is a binary operation. So you have to split xyxyxy up into x(yxyxy) and then split the part in the parentheses and so on, or something in the same nature. So your best bet is to just express it as x^3 * y^3.

5. Oh i see, thanks.
Another qu, what is the integral of 1/x dx ?
I know its something like y = loge x or something (not sure what it is exactly) but i dunno why.. First of all, what is e. I know its used heaps in logs, but its just some number originating from an infinite series of 1/1 + 1/2! + 1/3! ... etc but what advantage does it have in terms of logs? and umm, yeah, the integral question. How does 1/x become loge x?

6. Yeah, I agree that the hardest part about calculus is knowing the trig identities...
Well, to *do* differentiation/integration you certainly need to know trig identities. But just to *understand* the concept of a derivative or integral, you don't need that. Identities can always be looked up in a book, and if the frequency of your usage warrants it, you'll remember them without trying. IMHO it's wrong to just emphasize facts which can be looked up in a book.

7. Originally posted by Panopticon
Oh i see, thanks.
Another qu, what is the integral of 1/x dx ?
I know its something like y = loge x or something (not sure what it is exactly) but i dunno why.. First of all, what is e. I know its used heaps in logs, but its just some number originating from an infinite series of 1/1 + 1/2! + 1/3! ... etc but what advantage does it have in terms of logs? and umm, yeah, the integral question. How does 1/x become loge x?
Well, what function has a derivative of 1/x? You already have 1/x therefore, you need to find the antiderivative of it. This is ln(x).

8. Originally posted by Silvercord
I'm starting from the beginning, and I'm reading about limits and how to find the slope at a point using lines that are secant to the curve.
It's probably a typo, but you meant tangent lines.

Originally posted by Major_Small
if you can figure out:

(56x^6+48x^5-14x^4+468x^3+45x^2+189x-189)/(2x^6+18x^5-48x^4+165x^3-59x^2-486x+48)

assuming x==489488436498899878648897898 then you're good for calculus...

btw... the answer is around 28... (limits)
There's another way to do this without using L'Hospital's rule (yeah, I believe you have to stick the s in his name when you don't have the funny accent over the o...darn french) X in his example is insanely large. When X gets insanely large in a polynomial divided by another polynomial like that, only the terms raised to the highest power really affect it at all. This *technically* only works with X=infinity, but it'll do well enough here if you're not looking for an exact answer. "Simplify" the given problem to (56x^6)/(2x^6). (x^6) divides out, and you are left with 56/2, which equals 28. Tada.

9. Originally posted by Panopticon
Oh i see, thanks.
Another qu, what is the integral of 1/x dx ?
I know its something like y = loge x or something (not sure what it is exactly) but i dunno why.. First of all, what is e. I know its used heaps in logs, but its just some number originating from an infinite series of 1/1 + 1/2! + 1/3! ... etc but what advantage does it have in terms of logs? and umm, yeah, the integral question. How does 1/x become loge x?
OK, here's a simple proof that the differential of ln(x) is 1/x, which hence proves that the integral of 1/x is ln(x) (the integral is the antiderivative).

First, we must define e^x. e^x is defined as the function which is its own derivative, so d(e^x)/dx = e^x. Now we notice that ln(x)is defined as the inverse function of e^x, so:

ln(e^x) = x.

Now we ask ourselves what the derivative of ln(x) is. Well to start, we will let x = e^y for some y.

We also know that ln(e^y) = y, differentiating this on both sides with respect to y and we get:

d(ln(e^y))/dy = d(y)/dy = 1

Then (x = e^y),

d(ln(x))/dy = 1
d(ln(x))/dx * dx/dy = 1

But dx/dy = d(e^y)/dy = e^y = x, (by our original definition of y (x = e^y)).

So d(ln(x))/dx * x = 1 and d(ln(x))/dx = 1/x.

Regards,

10. It's probably a typo, but you meant tangent lines.
Well, you take a secant line, and let the points of intersection get closer and closer, and let them approach 0 (i.e. the limit). That limit is the tangent line, but you can't have the tangent line without the secants.

There's another way to do this without using L'Hospital's rule (yeah, I believe you have to stick the s in his name when you don't have the funny accent over the o...darn french) X in his example is insanely large. When X gets insanely large in a polynomial divided by another polynomial like that, only the terms raised to the highest power really affect it at all. This *technically* only works with X=infinity, but it'll do well enough here if you're not looking for an exact answer. "Simplify" the given problem to (56x^6)/(2x^6). (x^6) divides out, and you are left with 56/2, which equals 28. Tada.
Still L'Hospital's rule (the 's' is just terrible)... It just uses a bit of common sense to solve more quickly.

11. Oh.... I don't get it cos I know jack about logs. But thanks anyway.
I guess ill have to wait till my class is up to there or ask my tutor.

12. It's probably a typo, but you meant tangent lines.
Nope, the book specifically says you use secant lines until you get to the limit, and the slope at the limit is the slope at that point (the tangent)

EDIT:
oops, Zach already said this above :
Well, you take a secant line, and let the points of intersection get closer and closer, and let them approach 0 (i.e. the limit). That limit is the tangent line, but you can't have the tangent line without the secants.

13. The great thing about the limit is being able to simulate exactness without any of the nasty side-effects (like division by zero).

When X gets insanely large in a polynomial divided by another polynomial like that, only the terms raised to the highest power really affect it at all. This *technically* only works with X=infinity, but it'll do well enough here if you're not looking for an exact answer.
It is the exact answer. (AFAIK)

One thing to keep in mind about limits is that they are not approximations. Even though they are not exact numbers, they are infinitely close. By "infinitely close" I mean that it can be proven that no matter how close to the number 'x' is, f(x) is also close to it.

14. Well, it still is an approximation in that case since we're evaluating x at a ridiculously large number, not takeing the limit as it goes to infinity.

Also remember that the limit of the function as the independent variable approaches a point is not necessarily equal to the function at that point (jump or point discontinuities for example).

15. Well, it still is an approximation in that case since we're evaluating x at a ridiculously large number, not takeing the limit as it goes to infinity.
The situation for that problem is for x as it approaches infinity. x != infinity because infinity is not a real number (for any real value infinity, there's a higher one that must be infinity instead).

There's no point to mess around with rediculously high values when you have limits. The whole point of using limits is to prove a problem for any value sufficiently large. Approximations can only guess at the answer.

Another way of looking at that problem is to divide every term by x^6, the highest degree factor. Since anything divided by infinity is 0 (as far as limits go), you're left with 56/2 == 28.

Also remember that the limit of the function as the independent variable approaches a point is not necessarily equal to the function at that point (jump or point discontinuities for example).
True.