I would add to Bob's comment that the lower limit of integration matters very much . . . please don't tell your students it doesn't matter! Consider the following three antiderivatives of f(t) = 1/l(t-2)(t-4)]:
F1(x) = integral (f(t), t, 10, x) F2(x) = integral (f(t), t, 4, x) F3(x) = integral (f(t), t, 0, x)
If you differentiate all three of these functions you get the same derivative f(x) = 1/[(x-2)(x-4)] so the lower limit doesn't matter. Or does it?
Consider the function F1(x). What is its domain? The domain of a function consists of the valid x-values which in this case is x > 4. Why is the domain restricted in this way? Clearly, f(t) is not defined at t = 2 or t = 4 so why isn't the domain all real numbers except for 2 and 4? The reason is that the integration is not defined across vertical asymptotes. If you look back at the definition of a Riemann integral I suspect you will find that your book insists on f being continuous on [a, b]. Therefore, in order for the integral to make sense we must stop at the vertical asymptote t = 4 and so the derivative is y = 1/[(x-2)(x-4)] on x > 4.
The domain of F2(x) is quite different. If we keep the integration valid we must stay in the t-interval 2 < t < 4 so F2(x) is only defined for 2 < x < 4 and, therefore, that is the only place where F2 can have a derivative.
Finally, the domain of F3 is x < 2 for similar reasons.
Here is one more reason why the lower limit matters even if f(t) is continuous for all real numbers. If f(t) = t then f is continuous for all real numbers and so F(x) = integral (f(t), t, k, x) is defined for all reals and its derivative is also defined for all reals. However, there are infinitely many antiderivatives of f(t) and this one is F(x) = t^2/2 - k^2/2. That is, the lower limit produces a vertical shift in the antiderivative. The lower limit is important because it distinguishes one antiderivative from another. This F is the only antiderivative for which F(k) = 0.
I hope that helps.
Alan Lipp 19 Payson Avenue Easthampton, MA 01026 413-529-3278
The Williston Northampton School inspires students to live with passion, purpose, and integrity.
-----Original Message----- From: Wilder Bob [mailto:Bob.Wilder@appo.k12.de.us] Sent: Wednesday, February 15, 2012 11:11 AM To: AP Calculus Subject: RE: [ap-calculus] FTC in plain language
Thanks, Melanee ... and I would be careful about saying 'does not matter'.!
Here's the way I present the idea ....
Consider some specific examples. For example, int( t^2, t=1, t=x ).
Evaluate that using the other part of the FTC: You get (t^3)/3 and need to evaluate that between 1 and x. You arrive at -- (x^3)/3 - 1/3. I call these things semi-definite integrals to convey the idea that they look like definite integrals, but are actually functions ... And then I tell them not to use that phrase themselves ....
So. What is the derivative with respect to x of this function? It is .... x^2 !! The derivative of the 1/3 is 0 ....
Do a couple more .... int( t^2, t = 2, t=x ). Or t=3 to x. Or t= (-1) to x ....
In each case you get a function related to x^2 ... only the y-intercept has changed ....
So these things are specific anti-derivatives. The lower limit determines which particular antiderivative of the entire family. The derivative of that 'filled in' constant is just 0....
So ... the lower limit DOES matter when we are just considering the semi-definite integral. But when I follow that up with a derivative, the lower limit has no effect on the answer (if the lower limit is a number ...).
Hope this helps !!
Bob Wilder Middletown High School
-----Original Message----- From: Melanee Dismuke [mailto:email@example.com] Sent: Tuesday, February 14, 2012 10:07 PM To: AP Calculus Subject: [ap-calculus] FTC in plain language
Can someone please explain to me, in layman's terms, why the lower limit of integration doesn't matter when using the Fundamental Theorem of Calc to evaluate the definite integral as a function? I get the "what" of the FTC, but I can't get a handle on the "why" of the irrelevance of the value of a.