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## Calculus 2

### Unit 1: Lesson 9

Fundamental theorem of calculus and definite integrals

# The fundamental theorem of calculus and definite integrals

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.B (LO)
,
FUN‑6.B.1 (EK)
,
FUN‑6.B.2 (EK)
,
FUN‑6.B.3 (EK)
There are really two versions of the fundamental theorem of calculus, and we go through the connection here. Created by Sal Khan.

## Want to join the conversation?

• This stuff is all about `a` and `b`. Why does it need `c` and `d` from the start? Are they some kind of the limit of `f(x)`?
• you need C and D to calculate the area of some form
• I'm confused about the use of the word "anti-derivative", isn't it the same as saying "integral"?
• An antiderivative is an indefinite integral.

There are other kinds of integrals. The main one you'll be dealing with is the definite integral. In more advanced mathematics there are some subtleties of definitions of various kinds of integrals. I will leave that to a professional mathematician because I am not fully-versed in the nuances.
• I am a bit confused with the two parameters x and t. Can someone clarify their relationship for me?
• lower case f(t) is a derivative....so it is a rate.....some unit per second or y per second or y per minute or whatever your t values are would then be a rate...so it is f(t) , the derivative...then integrate to get the antiderivative.......which then means that F(t) is the anti derivative, or as I like to think of it as the original function that you are being given the derivative of. Y and x coordinates would give you the function and then you would get the derivative but we are integrating which is the reverse order....hope that helps and did not confuse you more.
• Why was
``Capital F (x) = ∫ f( t )dt``
`` Capital F (x) = ∫ f( x ) dx ``
?
• Because the bounds of ∫ f(t) are [0,x]. That makes F(x) a function of x, not t.
• If F(x) = integral of f(t) with boundaries a to x,
then F(b) = integral of f(t) with boundaries a to b, right?
So, by the 2nd theorem, F(b) = F(b) - F(a) ? and F(a) = 0 ?!
• Given the way F(x) is defined, it should not be surprising that F(a) = 0, as it would be the integral of f(t) with boundaries a to a.
• I really don't understand why the second fund thr can still be used if the upper limit is not x by finding the upperbound derivative. Can someone please explain?

-Test tomorrow!
• Consider this: instead of thinking of the second fundamental theorem in terms of x, let's think in terms of u. A function for the definite integral of a function f could be written as
``       ⌠uF(u) = | f(t) dt       ⌡a``

By the second fundamental theorem, we know that taking the derivative of this function with respect to u gives us `f(u)`.

Now, what if `u = g(x)` where g(x) is any function of x? This means that
``⌠u          ⌠g(x)| f(t) dt = | f(t) dt = F(g(x))⌡a          ⌡a``

Then if take the derivative with respect to x of F(g(x)), which is the derivative of an integral with an upper bound other than x, we can just use the chain rule, which gives us `d/dx F(g(x)) = f(g(x)) * g'(x)` (given that F'(x) = f(x) ).

So we end up with `the derivative of the upper bound` multiplied by `the inner function (integrand) evaluated at the upper bound`.

(Credit to MIT on edX for the explanation.)
• what does the sign that looks like an elongated S signify?
• The "elongated S" is the integral symbol which signifies that we are integrating the function that comes after the symbol.
• At , it is stated that if f(t) is continuous at a given interval, then it differentiable at every point in it's domain.......
But in the differential calculus course, we learned that a continuous function is not necessarily differentiable...eg:- f(x) = |x| is not differentiable at x=0, even though it is continuous.....

• It is only claimed that the continuity of `ƒ` implies the differentiability of `F`.