Midterm1 Jan 2011

# Midterm1 Jan 2011 - 1(15 points The graph below represents...

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Unformatted text preview: 1) (15 points) The graph below represents the velocity v(t) at time t in m / sec of a particle moving in a straight line during the time interval [0,10], W) with the tick marks on the axes marked off in sec and m/sec respectively. [In case it is not clear: the graph passes through the points (1, —2), (4, 1) and (6,1)] t (a) Use this graph to complete the following table specifying the location x(t) = / v(s)ds of the 0 particle at time t. (b) Compute /5 v(t)dt. /ﬁmnm=xwy—um=—a5+a5=1 (0) Determine the displacement (i.e. net change in position) of the particle during the time interval [0,8]- —1.5 8 displacement = / v(t)dt = 56(8) 0 (d) Determine the total (i.e. absolute) distance the particle traveled during the time interval [0,7]. The total distance traveled is the area between the graph and the \$~aXlSZ 2+2+05+2+05=7 2) (18 points) (a) Express the following sum 677 27r 27r 67T 477 27: cos ———+—— ——+cos —+— —+ 3 n n 3 n 67r 67r 271r 671' 27m 27r cos —3-+-; --+...+cos —+—— in the form Zleai. n (6% 2m) 27r 2 cos —— + —~ —— i=1 3 n n (b) Find a deﬁnite integral f: f (:1?) d3: for which the above sum is a right Riemann sum. 4-70 M 5 COS X ka 2-K: (c) Use (b) to ﬁnd the limit of the above sum as n ,-—> oo. 3) (36 points) Find the following integrals. Show your work. (Numerical answers obtained by a calculator will receive no credit.) (a) fﬂgz’fdt 6t _ 2 12t _ 6t /(e 13) dt 2/6 266 +169dt e7t 65* 26 -1: 169 _7t “ n “ :38 ‘ Te + C (b) /cot(7x) In I sin(7x)[ d3: -We substitute u = In I sin(7:v)|. Differentiating we obtain day = ﬁlﬁ’ Thus the integral becomes /00t(7:r)ln | sin(7a:)| d2: : i/udu = iuz = —1-[ln|sin(7:c)l]2 + C 7 14 14 ' We substitute u = 429 — 13. Differentiating we get dac— —- 36 —8“. We obtain [4392—1361: 36/73d“ , ' Solvin for 29 in terms of u, we get z9 = l u + 13 . Substituting this back into the integral we obtain g 4 . 1 u+13 1 — 1 144 u du:1—44[u+ 31nluH 144 [4.2 ~13+13ln|429 —13|]+C The term —(—— 13) can be absorbed into the constant of integration, which gives the alternative form of the answer 1212 [429 +13ln|4z9 —13|] + C 31/? (d) f (3:) dx where —2 :1: ifo3 «184x? ifzr23 [Hintz interpret the integral as a difference of two areas] f0?) = The integral fom f(:c) d2: represents the area of an octant of a circle of radius 3\/§, while I112 f (ac) dot represents the negative of the area of a right triangle of side 2. Consequently 31/5 18 f(:c) dx = 7r — ~ g : 5.068518347057703 0 8 2 4) (15 points) Let x C(x) = x2 /6 . (5f(s) — 9) ds Suppose that f(7) = 1 and j: f(3) d3 = 6. Find G"(7). Applying the product rule and the Fundamental Theorem of Calculus, we obtain G’(:v) = 2:131 / (5 f(s) — 9) d3 + x2(5f(x) — 9) 6 Plugging in a: = 7, we get 0(7) = 14 (5 /67 f(s) ds — 9 [67013) + 49(5f(7) — 9) = 14 (5(6) — 9(7 — 6)) + 49(5(1) — 9) = 98 5) (16 points) Find the total area enclosed between the graphs of w = 83/2 '— y + 3/3 and an 2 3/2 + 7y3. Express your answer as a sum of two integrals that do not involve absolute values. DO NOT EVALUATE TH E Ls. “M First ﬁnd the intersection points between :1: = f (y) = 8y2 — y + y3.and x = g(y) = y2 + 7313. 0 = 9(y) - f(y) = 6113 — 7y2 + y = y(6y — 1)(y - 1) Thus the intersection points are in left to right order 0, g and 1. The enclosed area is 1 1 1/6 / lg(y)~f(y)ldy=/ l6y3—7y2+yldy+// l6y3—792+yldy 0 O 1 6 Since 63/3 + 7y2 + y = y(6y — 1)(y — 1) is positive on the interval (0, %) and negative on the interval (é, 1), this area is 1/6 1 / (61/3 - 73/2 + 3;) dz; — (63/3 — 73/2 + y) dy 0 1/6 This evaluates to 51 67 24 552 + 552'— 15 '589 9W9 ...
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