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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.
470
M 5 COS X ka
2K: (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[lnsin(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+13ln429 —13]+C The term —(—— 13) can be absorbed into the constant of integration, which gives the alternative form
of the answer 1212 [429 +13ln4z9 —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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 Spring '07
 Wyzlouzil

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