Chapter 5 47 - 10 11 12 13 14 1 CHAPTER 5 REVIEW 479 False...

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Unformatted text preview: 10. 11. 12. 13. 14. 1. CHAPTER 5 REVIEW 479 False. You can‘t take a variable outside the integral sign. For example. using f(ac) = 1 on [0. 1}. fol xf(a:) d3; = fol atdm 2 E332]; : i (a constant) whileajfo1 1dac : a: [as]; : 27-1: at (a variable). . False. For example. letf(w) : 372. Then jol x/m2 diL‘ = folmda: = %.but ”fol :52 dm : fl = f. . True by the Net Change Theorem. . True by Comparison Property 7 of the Integral in Section 5.2. . False. For example. let a = 0,1) = 1, f(ac) = 3. g(a:) = w. f(a:) > g(:c) for each cu in (0,1). but f'(a:) =0< 1 :g'(w)fora: E (0.1). . True. The integrand is an odd function that is continuous on [—17 1]. so the result follows from Theorem 5.5.7(b). True. ff5(am2 + ()3: + 0) dm 2 ffsflmg + C) data + f; bar dm : 2 [05(agc2 + c) dm [by 5.5.7(a)] + 0 [by 5.5.7(b)] False. The function f (m) = 1/234 is not bounded on the interval [—2. 1]. It has an infinite discontinuity at as = 0, so it is not integrable on the interval. (If the integral were to exist, a positive value would be expected. by Comparison Property 6 of Integrals.) False. See the remarks and Figure 4 before Example 1 in Section 5.2, and notice that y = a: — 3:3 < 0 for 1 < a: S 2. False. For example. the function y : |m| is continuous on R, but has no derivative at a: : 0. True by FTC]. — EXERCISES 6 (a) L6 = Z f(a:¢_1)A:1: [Ax : % : 1] i=1 :f($0)‘1+f($1)'1+f($2)'1 +f(ms) - 1+ flu) - 1 + m5) - 1 m2:3.5l4l2+( 1)l( 2.5) 8 The Riemann sum represents the sum of the areas of the four rectangles above the m—axis minus the sum of the areas of the two rectangles below the art—axis. 6 (b) M6 : Z f(i,;)Aa: [A93 : 6—39 :1] z 1 f(E1)‘1+f(532)‘1+f(33)'1 +f(54)'1+f(§5)'1+f(§6)'1 = f(0.5) + f(1.5) + f(2.5) + f(3.5) + f(4.5) + f(5.5) m 3 + 3.9 + 3.4 + 0.3 + (,2) + (~29) : 5.7 The Riemann sum represents the sum of the areas of the four rectangles above the za—axis minus the sum of the areas of the two rectangles below the :c—axis. ...
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