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MATH124 CALCULUS II for EngineersLab Section:Quiz 1(12 pts)January 14, 2009Pleaseprintnames and IDs inink:NSID:Family Name:First Name:Student ID:INSTRUCTIONS:1. Time Limit: 30 minutes3. Closed book. Closed notes. No calculators.2. No cheating.4. Write clearly & legibly.1. (2 pt) Write12−24+38−416+· · ·+(−1)n−1n2nusing sigma notation.=ni=1(−1)i−1·i2i2. (4 pts) Consider the regionAabovey=x2−1, belowy= 0.(a) (2 pts) Find an equation forSnand clearly explain your answer. Do not simplify the equation.The regionAis symmetric about they-axis, so we can double the area betweenx= 0 andx= 1.Divide this interval intonequal subintervals of width1nand use the distance0−(x2−1) = 1−x2, betweeny= 0 andy=x2−1 for the heights of rectangles, to findSn= 2ni=11n1−i2n2= 2ni=11n−i2n3.(b) (2 pts) Given that the equation from (a) can be simplified toSn=n−n(n+ 1)(2n+ 1)6n22n,find the areaA.A= limn→∞Sn= limn→∞2nn−2n(n+ 1)(2n+ 1)6n3= limn→∞2−(n+ 1)(2n+ 1)3n2=2−limn→∞2n2+ 3n+ 13n2= 2−23=43.
3. (2 pts) Find the closed form value forni=1(3i+i).ni=1(3i+i) =ni=13i+ni=1i=ni=13i−1·3 +n(n+ 1)2= 3·3n−13−1+n(n+ 1)2=3n+1−32+n(n+ 1)2=3n+1−3 +n2+n2.4. (2 pts) Evaluate the lower Riemann sum off(x) =x2on [0,8] with 4 equal subintervals.x=8−04= 2x1= 0,x2= 2,x3= 4,x4= 6Therefore,L(f, P4) = 2(02+ 22+ 42+ 62) = 2(4 + 16 + 36) = 1125. (2 pts) Evaluate the upper Riemann sum off(x) = cosxon [0,π] with 4 equal subintervals.x=π−04=π4x1= 0,x2=π4,x3=π2,x4=3π4Therefore,U(f, P4) =π4(cos 0 + cosπ4+ cosπ2+ cos3π4) =π41 +√22+ 0−√22=π4Quiz Score:
MATH124 CALCULUS II for EngineersLab Section:Quiz 2(12 pts)January 21, 2009Pleaseprintnames and IDs inink:NSID:Family Name:First Name:Student ID:INSTRUCTIONS:1. Time Limit: 30 minutes3. Closed book. Closed notes. No calculators.2. No cheating.4. Write clearly & legibly.1. (1 pt) If53f(x)dx=−1 and54f(x)dx=−3, then43f(x)dx=43f(x)dx=53f(x)dx−54f(x)dx=−1−(−3) = 22. (3 pts) Evaluate the following integrals using the properties of the definite integral and interpretingintegrals as areas. Do not use anti-differentiation.(a) (1 pt)1−1x5cosxdx= 0Observe that cosx= 0 for anyxin [−1,1] and thatf(x) is an odd function over an intervalsymmetric about zero. Therefore, the integral is zero.(b) (2 pts)1−2(3x+ 4)dxThe intersection ofy= 3x+ 4 with thex-axis is the point−43,0.LetA1be the area of the right triangle below thex-axis, with sides 2 and−43−(−2) =23.ThenA1=122·23=23.LetA2be the area of the right triangle above thex-axis, with sides 7 and 1−−43=73.ThenA2=127·73=496.Therefore,1−2(3x+ 4)dx=A2−A1=496−23=456=152.3. (2 pts) Find the average values of the functionf(x) =√25−x2over the interval [0,5].Average value =15−05025−x2=1514π(5)2=5π4.
4. (1 pt) Findf(x) for the following function:f(x) = cosx+x0sec(t−1)dt.