1. The Fundamental Theorem of Calculus
Lesson 7.4

2. Definite Integral Recall that the definite integral was defined as
But … finding the limit is not often convenient
We need a better way!

3. Why?
Why is the area of the yellow rectangle at the end = a b

4. Fundamental Theorem of Calculus
Given function f(x), continuous on [a, b]
Let F(x) be any antiderivative of f
Then we claim that
The definite integral is equal to the difference of the two antiderivatives
Shazzam !

5. What About + C ? The constant C was needed for the indefinite integral
It is not needed for the definite integral
The C's cancel out by subtraction

6. You Gotta Try It Consider What is F(x), the antiderivative?
Evaluate F(5) – F(0)

7. Properties Bringing out a constant factor
The integral of a sum is the sum of the integrals

8. Properties Splitting an integral
f must be continuous on interval containing a, b, and c

9. Example
Consider
Which property is being used to find F(x), the antiderivative?
Evaluate F(4) – F(0)

10. Try Another What is Hint … combine to get a single power of x
What is F(x)?
What is F(3) – F(1)?

11. When Substitution Is Used
Consider
u = 4m3 + 2
du = 12m2
It is best to change the new limits to be in terms of u
When m = 0, u = 2
When m = 3, u = 110

12. Area Why would this definite integral give a negative area?
The f(x) values are negative
You must take this into account if you want the area between the axis and the curve

13. Assignment Lesson 7.4A Page 424 Exercises 1 – 43 odd Lesson 7.4B