The Fundamental Theorem of Calculus

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Presentation transcript:

The Fundamental Theorem of Calculus Lesson 7.4

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

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

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 !

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

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

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

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

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

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

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

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

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