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Chapter 5 22 - 454 U CHAPTEH5 INTEGRALS(c 9(17 fox...

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Unformatted text preview: 454 U CHAPTEH5 INTEGRALS (c) 9(17) : fox cos(t2)dt. Using an integration command. we ((1) We sketch the graph of 9' using the find that 9(0) : 0‘ 9(02) 2 0200‘ 9(04) m 0399. method of Example | in Section 2.9. The g(0.6) % 0.592. g(0.8) m 0.768. g(1.0) m 0.905. graphs of 9’(m) and f (2:) look alike. so g(1.2) m 0.974. g(1.4) a: 0.950. g(1.6) x: 0826. we guess that g’(m) = We)- g(1.8) x 0.635. and g(2.0) av. 0.461. 4. In Problems 1 and 2, we showed that ifg (x) : f: f(t) dt. then g'(:c) : flat). for the functions f(t) : 2t + 1 and f (t) = 1 + t2. In Problem 3 we guessed that the same is true for f (t) = cos(t2), based on visual evidence. So we conjecture that g’(a:) = f (as) for any continuous function f. This turns out to be true and is proved in Section 5.3 (the Fundamental Theorem of Calculus). 5.3 The Fundamental Theorem of Calculus 1. The precise version of this statement is given by the Fundamental Theorem of Calculus. See the statement of this theorem and the paragraph that follows it at the end of Section 5.3. 2. (a) g(m) : f; f(t) dt. so 9(0) 2 f0” f(t)dt : 0. 9(1) : fol f(t) dt 2 % -1~1 [area oftriangle] : 5 9(2) : f02 f(t)dt : f01 f(t)dt+ fff(t)dt [below the m—axis] _1 1 _ _§—§.1-1!0. 9(3)29(2)+f§f(t)dt:o_.;..1.1:1 9(4) :g(3)+f§f(t)dt:—§+§-1~1:0. NIH 9(5) : 51(4) —— ff f(t)dt : 0 + 1.5 : 1.5. 9(6) : 9(5) —— f: f(t) dt : 1.5 + 2.5 : 4. (b) 9(7) : 9(6) —— f; f(t) dt p: 4 + 2.2 [estimate from the graph] 2 6.2. (c) The answers from part (a) and part (b) indicate that g has a minimum at m = 3 and a maximum at m : 7. This makes sense from the graph of f since we are subtracting area on 1 < a: < 3 and adding area on 3 < w < 7. ...
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