Connection between definite and indefinite integrals?by tahayassen Tags: connection, definite, indefinite, integrals 

#1
Jan713, 09:51 PM

P: 273

I understand that the indefinite integral is like infinite definite integrals, but how come when we calculate the definite integral we simply substitute the two values into the indefinite integral and subtract? Why do we subtract? Why not add?
Also, there aren't the same thing, right? What's the difference between the two? [tex]\int _{ 0 }^{ t }{ 6t } \neq \int { 6t }[/tex] 



#2
Jan713, 10:26 PM

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#3
Jan813, 09:49 AM

P: 273





#4
Jan813, 11:03 AM

P: 1,623

Connection between definite and indefinite integrals? 



#5
Jan813, 12:31 PM

P: 273

I just had an eureka moment while watching khan academy. I'll explain as soon as I get home.




#6
Jan813, 02:41 PM

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#7
Jan813, 07:12 PM

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#8
Jan913, 08:08 AM

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#9
Jan913, 08:58 AM

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#10
Jan913, 04:41 PM

P: 642

Okay, I felt like clarifying this.
The first one we could say, correctly, is [itex]3t^2+C[/itex], where C is an arbitrary constant. We could also say, correctly, that it's [itex]\left(3t^2+1\right)+C[/itex], where C is an arbitrary constant. In fact, the point of the arbitrary constant is that [itex]3t^2[/itex], [itex]3t^2+1[/itex], [itex]3t^2+2[/itex], and [itex]3t^2\cdot\lim\limits_{a\to0^+}^{}\left( \displaystyle\int_a^1 \left(x^x\right)\cdot\mathrm{d}x\right)[/itex] are all equally valid antiderivatives (or indefinite integrals.) [itex]3t^2[/itex] isn't "special" in being the "fundamental" one which all others differ by a constant from. When we say "indefinite integral," we don't mean "integral with lower bound 0." We mean "integral from some arbitrary lower bound." Our C is going to depend on this lower bound (if we choose it to be 0, our C is 0, but this relation is false for integrating most functions.) It's probably best to say that [itex]\displaystyle\int\left(6t\right)\cdot\mathrm{d}t[/itex] is the set of all functions f such that, for all x, [itex]f\left(x\right)=3x^2+C[/itex], where C is an arbitrary constant. (Again, this is also the set of all functions f such that, for all x, [itex]f\left(x\right)=\left(3x^2+1\right)+C[/itex], where C is an arbitrary constant.) Once you've gotten all that out of the way and understand the previous paragraph, it should be pretty clear how the two are different. We can define [itex]\displaystyle\int_a^b\left(f\left(x\right)\right) \cdot\mathrm{d}x=F\left(b\right)F\left(a\right)[/itex], where F is any antiderivative of f. It's easy to show that we get the same value for any two antiderivatives F1 and F2, since if we define k to be the constant by which F1 and F2 differ, the k's cancel out and we're left with the same value regardless of what F1 and F2 we choose. Another point I want to drive home is that the only reason we use [itex]3x+C[/itex] instead of, say, [itex]3x+1+C[/itex] is because the former is the most convenient to write. I sort of explained this earlier, but it's pretty important if you want to know about the nature behind the C. 



#11
Jan913, 08:13 PM

P: 273

How did you know that if you choose your lower bound to be 0, then your C is also 0?




#12
Jan1013, 08:37 AM

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[itex]\int_0^t 6t dx= (6t)_0^t= 6t^2[/itex] [itex]\int_0^t 6t dt= (3t^2)_0^t= 3t^2[/itex] The real problem was that tahayassen didn't have a "dt" or "dx" in his integral so it really has no meaning. 



#13
Jan1013, 01:34 PM

P: 642

(If we had, instead, written the integral as [itex]3t^21+C[/itex], for a lower bound of 0, C would be 1.) 


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