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Connection between definite and indefinite integrals?

by tahayassen
Tags: connection, definite, indefinite, integrals
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tahayassen
#1
Jan7-13, 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]
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cytochrome
#2
Jan7-13, 10:26 PM
P: 162
Quote Quote by tahayassen View Post
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]
They are both the same thing, it's just that definite integrals involve plugging in those numbers (before you plug the limits of integration in, you have to solve the indefinite integral). You subtract because you want the numerical value between the two limits of integration for the function that you integrated.
tahayassen
#3
Jan8-13, 09:49 AM
P: 273
Quote Quote by cytochrome View Post
They are both the same thing, it's just that definite integrals involve plugging in those numbers (before you plug the limits of integration in, you have to solve the indefinite integral).
What about the constant? Isn't that why the indefinite integral is an infinite number of functions? So would the definite integral with a lower limit of 0 and an upper limit of a variable also be an infinite number of functions?

Quote Quote by cytochrome View Post
You subtract because you want the numerical value between the two limits of integration for the function that you integrated.
Why do you mean by limit of integration?

jgens
#4
Jan8-13, 11:03 AM
P: 1,622
Connection between definite and indefinite integrals?

Quote Quote by cytochrome View Post
They are both the same thing, it's just that definite integrals involve plugging in those numbers (before you plug the limits of integration in, you have to solve the indefinite integral).
This is wrong. The indefinite integral denotes that you want to find the family of anti-derivatives, while the definite integral denotes that you want to go through all of that Riemann/Darboux sum business. The Fundamental Theorem of Calculus establishes that, for continuous functions, we can compute the definite integral by finding an anti-derivative of the integrand and then do the whole evaluation thing at the end points.
tahayassen
#5
Jan8-13, 12:31 PM
P: 273
I just had an eureka moment while watching khan academy. I'll explain as soon as I get home.
mathman
#6
Jan8-13, 02:41 PM
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Quote Quote by tahayassen View Post
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]
The one on the left = 3t2, while the one on the right = 3t2 + C, where C is arbitrary (C may = 0, making the integrals =, but it doesn't have to be).
pwsnafu
#7
Jan8-13, 07:12 PM
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Quote Quote by mathman View Post
The one on the left = 3t2, while the one on the right = 3t2 + C, where C is arbitrary (C may = 0, making the integrals =, but it doesn't have to be).
Actually, the one on the left is 6t2.
HallsofIvy
#8
Jan9-13, 08:08 AM
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Quote Quote by mathman View Post
The one on the left = 3t2, while the one on the right = 3t2 + C, where C is arbitrary (C may = 0, making the integrals =, but it doesn't have to be).
Quote Quote by pwsnafu View Post
Actually, the one on the left is 6t2.
No, mathman is correct. Surely you know the indefinite integral [itex]\int tdt= (1/2)t^2+ C[/itex]
pwsnafu
#9
Jan9-13, 08:58 AM
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Quote Quote by HallsofIvy View Post
No, mathman is correct. Surely you know the indefinite integral [itex]\int tdt= (1/2)t^2+ C[/itex]
You can't have the variable of integration in the limits of a definite integral. The only valid way to parse ##\int_0^t 6t## is if t is constant (hence integrating the constant from 0 to t).
Whovian
#10
Jan9-13, 04:41 PM
P: 643
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.
tahayassen
#11
Jan9-13, 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?
HallsofIvy
#12
Jan10-13, 08:37 AM
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Quote Quote by pwsnafu View Post
You can't have the variable of integration in the limits of a definite integral. The only valid way to parse ##\int_0^t 6t## is if t is constant (hence integrating the constant from 0 to t).
While, for clarity, one shouldn't use the same variable in the limit as in the integral, there is no mathematical ban against it:
[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.
Whovian
#13
Jan10-13, 01:34 PM
P: 643
Quote Quote by tahayassen View Post
How did you know that if you choose your lower bound to be 0, then your C is also 0?
Since it's easy to show that [itex]\displaystyle\int_0^t\left(6u\right)\cdot\mathrm{d}u=3t^2+0[/itex], so choosing a lower bound of 0 gives a C of 0.

(If we had, instead, written the integral as [itex]3t^2-1+C[/itex], for a lower bound of 0, C would be 1.)


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