Day-8 - Math 1314 Day 8 The FUNDAMENTAL THEOREM OF CALCULUS...

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Math 1314 Day 8 The FUNDAMENTAL THEOREM OF CALCULUS (Think this might be important??) Suppose ( ) f x is a continuous function on the interval [a, b]. Then ( ) ( ) ( ) b a f x dx F b F a = - where F is any antiderivative of f . There’s a lot to this theorem. An informal proof can be found on page 986 in Section 14.4 of the Text. But I won’t be provin’ it here. We’ll use this notation: Here’s a very basic example: Example 1 : ( 29 3 1 2 3 x dx - +
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What happens to the C??? ( ) b a f x dx = Example 2 : Evaluate: - + 3 1 2 ) 7 4 3 ( dx x x
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Example 3 : Evaluate: 4 2 1 1 6 dx x x -
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Example 4 : Evaluate: + - 5 2 2 6 4 2 dx x x x
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Example 5 : Evaluate: dx x e x + - 3 0 ) 1 (
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Example 6 : Evaluate: ( 29 ( 29 dx x x 1 3 2 0 2 - -
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Example 7 : Find the area of the region under the graph of 2 3 ) ( x x x f - = over the interval [0, 3].
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We will need to use substitution to evaluate some problems: Example 8 : Evaluate ( 29 - 3 0 5 2 3 4 dx x x
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Example 9 : Evaluate 1 0 2 3 dx e x x
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Example 10 : Evaluate + 2 1 3 2 6 3 dx x x
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Applications Example 11 : A company purchases a new machine for which the rate of depreciation is
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Day-8 - Math 1314 Day 8 The FUNDAMENTAL THEOREM OF CALCULUS...

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