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Posted: 12/16/2005 12:41:39 PM EDT
[#1]
equal to F(b)-F(a), where F(x) is the anti-derivative of f(x)
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Posted: 12/16/2005 12:45:52 PM EDT
[#2]
So, say, for f(x)=(3x^2), the integral of f(x) over interval (0,2), since F(x) = x^3 (right?), will be 8? |
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Posted: 12/16/2005 12:46:02 PM EDT
[#3]
The flagon with the dragon, contains the pellet with the poison, bu the pessell from the castle contains the brew that is true.
Get it. Got it. Good. |
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Posted: 12/16/2005 12:47:43 PM EDT
[#4]
Yep |
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Posted: 12/16/2005 12:48:07 PM EDT
[#5]
Yes, your instinct is correct. Integration is nothing more than addition over an interval. [That's over simplified, but that is how the problem boils down to its essence.] |
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Posted: 12/16/2005 12:49:33 PM EDT
[#6]
Okay, thanks. I assume it gets more complicated when you start dealing with equations like [e^(2x-1)]/[sin(x)cos(x)] +2x^5, etc.
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Posted: 12/16/2005 12:53:45 PM EDT
[#7]
in that case, it's just finding the anti-derivative that's harder. The basic concept of integration remains the same. In the future, you'll learn new techniques of finding integrals, such as using trig properties, integration by parts, table method, etc... |
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Posted: 12/16/2005 12:54:17 PM EDT
[#8]
if you want to check your answer you can always put the equation on your graphing clac under the Y= section for graphing functions and then choose the clalc menu (on the TI-83 at least) and there is an integral choice (number 7 on TI-83+) and put in the lower limit (in this case 0) hit enter and then the upper limit (2) and hit enter, it will shade in that area and give you the answer at the bottom
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Posted: 12/16/2005 12:57:34 PM EDT
[#9]
Talking of calculus, you know what I just realized today?
The circumference of a circle is the first derivative of the area, and the surface area of a sphere is the first derivative of the volume. I feel stupid for not seeing it before, but I had just worked it out when doing a problem, and it hit me, like "duuuuuhhh..." |
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Posted: 12/16/2005 1:31:15 PM EDT
[#10]
yeah velocity is the time deriviative of position, and acceleration is the time derivative of velocity. wait till you start doing integration by parts, that always bugged me, i didn't figure it out complety till i took Partial differential equations. |
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Posted: 12/16/2005 1:38:55 PM EDT
[#11]
TI-89 will integrate for you (gives formula). I recommend not getting one, you'll never do homework and get a B in the class and ruin your 4.0 because exams allow not calculator of any kind... (I had the misfortune of not talking calc in high school) Additionally, there is numerical (approx) integration software for the 83. |
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Posted: 12/16/2005 7:50:43 PM EDT
[#12]
Yeah, we did that a little while ago, while learning the applications of derivation. And jerk is the d(A)/dt. As to the TI-89, it's required for my course (AP calc AB), so I bought one. It's handy, and very good. But I still end up doing most derivation by hand. |
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Posted: 12/16/2005 8:10:44 PM EDT
[#13]
My cat's breath smells like cat food
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Posted: 12/16/2005 8:15:06 PM EDT
[#14]
tag for fizzacist
 answer is yes, if I think I read the question correctly. It is the infinite sum though, not an integral sum, the difference between sums and integrals. |
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Posted: 12/16/2005 8:17:33 PM EDT
[#15]
AND QUIT RELYING ON YOUR DAMN TI CALCULATORS!!!
THINK, DAMMIT! |
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Posted: 12/16/2005 8:21:29 PM EDT
[#16]
That there is the reason I changed my major to Business.
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Posted: 12/16/2005 8:22:26 PM EDT
[#17]
My physics teacher graduated with his degree (in physics, not education, fortunately) in the 60s. His final exam in one of his second-year classes was 3 hours long. This was back when all they had was slide rules and their heads. The test had one problem, a 2-dimensional momentum problem. Not one student finished within the time limit. The professor expected this. I think I'll stick with my useful tools, thank you. EVOLVE, PLEASE. |
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Posted: 12/16/2005 8:37:29 PM EDT
[#18]
They also put a man on the moon using slide rules. You need to know the concepts before taking the easy way out. 
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Posted: 12/16/2005 8:47:15 PM EDT
[#19]
Personally, I miss those days as a consumer. Everything was over-engineered to hell-and-gone, and would therefore survive 'life' a little easier. |
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Posted: 12/16/2005 8:51:27 PM EDT
[#20]
Does an HP with RPN count?
 TC |
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Posted: 12/16/2005 8:55:42 PM EDT
[#21]
Exemption, of course! RPN Calculators are The Rule. It is only when drawing out the problem and asking for an answer is where the problem is. Real Life doesn't give you a neatly drawn out equation and ask you to integrate it. You need to "see" that yourself, then approximate with RPN or a crapload of (((((()))(()(((()))) on a Ti. |
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Posted: 12/16/2005 9:03:40 PM EDT
[#22]
Thanks, Prof. -brass-. I think you're onto something here.
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