At the end of this lesson you will be able to
The Mean-Value Theorem for Definite Integrals is stated as Theorem 2 at the top of page 352. Open your book to that page and that theorem.
What is the theorem saying? The theorem is stating that, if you have continuity on the closed interval [a, b], then there must a point x = c in the closed interval [a, b] such that f (c) * (b - a) is the same value as the definite integral over the same closed interval [a, b].
Now if f (c) ≥ 0, then f (c) is a vertical directed distance at x = c and (b - a) is a horizontal directed distance and f (c) * (b - a) is the area of a rectangle. Of course, if f (c) < 0, then we have the negative of the area for f (c) * (b - a). The functional value f (c) is called the "mean-value" or "average" of f (x). Therefore, upon dividing by (b - a), we have the expression for the f (c) given in the theorem at the top of page 352.
In other words, the Mean-Value Theorem describes "average-value" for a continuous, dependent variable f(x), giving us an application for the definite integral.
Now you can go to the Session 23 assignments to continue your assignment.
At the end of this lesson you will be able to
You will find the Fundamental Theorem of Integral Calculus, Part 1 in the middle of page 354 as Theorem 3. Open your book to that page and that theorem.
First the hypothesis is stating that you must have a continuous function f on closed interval [a, b]. Also, you have a F (x) equal to a definite integral defined as in the conclusion. The independent variable of F (x) is exactly the same as the upper limit for the definite integral and the independent variable in the definite integral is a "dummy" variable "t" in the integrand f(t)dt.
The derivative of F (x) is exactly the same as f (x)! The derivative of F (x) with respect to the upper limit of integration is exactly the same as f (t) evaluated at t = x. It does not matter what the lower limit of integration is just so the function f is continuous in the closed interval [a, b].
Wow! Think about this. It is amazing!
Now you can go to the Session 23 assignments to continue your assignment.
At the end of this lesson you will be able to
FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS, PART 2
Now the Fundamental Theorem of Integral Calculus, Part 2 is found at the top of page 358. Let us open our textbook to that page.
What does the hypothesis say? First it states that we have a function f that is continuous on a closed interval [a, b]. Therefore, you can ONLY use this theorem if f is continuous on a closed interval [a, b]. Second the theorem states there is a function F such that the derivative F ' (x) = f (x). In other words, you must be able to find an antiderivative F (x) of f (x) to use this theorem.
If both parts of the hypothesis is met, then we can apply the conclusion to find the value of a definite integral. Notice this theorem states the definite integral is calculated by evaluating the antiderivative F (x) at the upper end point and subtracting the evaluation of the antiderivative F (x) at the lower end point. Thus, if the hypothesis is satisfied, the definite integral depends only on F (b) - F (a) and not on any intervening evaluations. This is remarkable! The definite integral in this case only depends on the endpoint values of the antiderivative!
This is the MOST IMPORTANT theorem and concept of all integral calculus. Therefore, you must immediately learn the hypothesis and conclusion. Instead of using the definition of the definite integral, we can use this Fundamental Theorem of Integral Calculus to evaluate a definite integral, if the hypothesis is satisfied.
Fundamental Theorem of Integral Calculus is extremely powerful. It states that the definite integral can be calculated without taking limits, without calculating Riemann sums, and often rather easily as long as two conditions exist: (1) the antiderivative of f can be found and (2) the function f is continuous on a closed interval.
EVALUATING DEFINITE INTEGRALS
To evaluate a definite integral using the Fundamental Theorem of Integral Calculus, we are going to do the following steps:
- Find an antiderivative of the integrand, the simplest antiderivative we can find.
- Evaluate this antiderivative at the upper endpoint.
- Evaluate this antiderivative at the lower endpoiint.
- Subtract the two evaluations.
Instead of writing F(b) - F(a), we usually use the special symbolism that you see at the upper left margin of page 359 in your textbook.
NOTE ON THE CONSTANT OF INTEGRATION "C":
For an indefinite integral understanding, look at page 314. Indefinite integrals are antiderivatives. Notice in the middle of that page, antiderivatives are not unique and differ be a constant. Therefore, to indicate that for an indefinite integral we could have any constant, we write a "C" for the constant
For a definite integral understanding, look at page 343. Notice that the definite integral is a limit of a sum. There is no "C" involved here.
The symbol for the definite integral is on page 344. Because it looks similar to the symbol for an indefinite integral, some students become confused at this point. However as I mentioned above, they are two entirely different concepts.
To evaluate the definite integral we use the Fundamental Theorem of Integral Calculus on page 358. It requires us to find antiderivatives and indefinite integrals. Look at the proof in the middle of page 358. You will see that the "C" in the antiderivative G(b) + C and the antiderivative G(a) + C subtract out. Thus, there are C's in definite integrals, but C - C - 0.
Now you can go to the Session 23 assignments to continue your assignment.