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CHAPTER 4 SECTION 4.4 THE FUNDAMENTAL THEOREM OF CALCULUS

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Figure 4.27

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First Fundamental Theorem of Calculus Given f is –continuous on interval [a, b] –F is any function that satisfies F’(x) = f(x) Then

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Theorem 4.9 The Fundamental Theorem of Calculus

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First Fundamental Theorem of Calculus Given f is –continuous on interval [a, b] –F is any function that satisfies F’(x) = f(x) Then

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First Fundamental Theorem of Calculus The definite integral can be computed by –finding an antiderivative F on interval [a,b] –evaluating at limits a and b and subtracting Try

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Guidelines for Using the Fundamental Theorem of Calculus

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Evaluate using your calculator.

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2 methods: MATH menu: fnInt Y= Menu: make CALC menu: : 0 : 3

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3.Find the area of the region bounded by

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Find intersection points for bounds.

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Mean Value Theorem for Integrals

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Definition of the Average Value of a Function on an Interval

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1.Find the average value of on [1,4]. Then find where f(x) obtains this average value.

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This value is obtained when…

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Find the value of c guaranteed by the Mean Value Theorem for the integral over the specified interval.

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Definite Integral diagrams

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1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. Second Fundamental Theorem:

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1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. The long way: Second Fundamental Theorem:

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1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant.

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The upper limit of integration does not match the derivative, but we could use the chain rule.

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The lower limit of integration is not a constant, but the upper limit is. We can change the sign of the integral and reverse the limits.

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Using the 2 nd Fundamental Theorem of Calculus:

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We don’t know how to integrate this function!

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AP QUESTION

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IN OTHER WORDS, 55 – 55 = 0. THE INTEGRAL REPRESENTS V(12)- V(0) WHICH = 55 -55. SINCE THE AREA UNDER THE CURVE FROM 0 TO 12 = 55, AND V(0) WAS GIVEN TO US AS AN INITIAL VALUE OF 55.

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IN OTHER WORDS, IF THE LEAST VELOCITY PLUS THE VELOCITY OVER THE INTERVAL WHERE V IS DECREASING IS GREATER THAN ZERO, V NEVER REACHES ZERO.

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WHICH IS THE SAME THING!!!!!!!!!!!!!!!!!!

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