53 - Section 5.3 FTC 1 the fundamental theorem of calculus...

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Unformatted text preview: Section 5.3 FTC 1 the fundamental theorem of calculus If is continuous on , then the function defined by is continuous on , differentiable on and Proof: We need to show that lim So Animate Point Move B -> D 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 -2 -1 -0. 5 1 2 3 4 5 6 7 8 -1 -1. 5 -2 MTH 173 section 5.3 notes page 1 Now the idea is that the integral region is , so As 8. , the derivative of By the FTC 1, 12. the derivative of By the FTC 1, 14. the derivative of here let then tan ln tan ln gives the area of the red region. The width of the red gives the average function value in the interval. tan , and let MTH 173 section 5.3 notes page 2 FTC 2 If is continuous on , then where is any antiderivative of , then , so is an antiderivative of . If is Proof: We know that if another antiderivative of , then now then Evaluate with FTC2 20. 2 . 26. why not? not continuous in the interval from to 38. arctan arctan arctan MTH 173 section 5.3 notes page 3 ...
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This note was uploaded on 09/17/2009 for the course MTH Calc taught by Professor Bush during the Spring '09 term at Northern Virginia.

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