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WUSTL MATH 132 - m132_E3sSP08

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Math 132 - Exam III - Spring 2008 1This exam contains 15 multiple choice questions and 2 hand graded ques-tions. The multiple choice questions are worth 5 points each and the handgraded questions are worth a total of 25 points. The latter questions will beevaluated not only for having the correct solutions but also for clarity. Pointsmay be taken for confusing and disorganized writing, even when the answer iscorrect.1. Evaluate the integralZ542x − 1x2− 5x + 6dxA) 3 ln(3/2)B) 5 ln 2C) ln(2/3) + 5 ln 3D) 3 ln(3) + 4 ln 2E) 5 ln(2) + 3 ln 5F) ln(2) + 5*G) 3 ln(2/3) + 5 ln 2H) 3 + 5 ln 2I) 3 ln(5)J) 4 ln(2/3) + 7 ln 2First note that x2−5x + 6 = (x −2)(x −3). We can expand the integrand intopartial fractions by solving for A and B:2x − 1x2− 5x + 6=Ax − 2+Bx − 3.Writing both sides over a common denominator and equating the numeratorsgives:2x − 1 = A(x − 3) + B(x − 2).This is easily solved and gives: A = −3, B = 5. Therefore,Z542x − 1x2− 5x + 6dx =Z54„−3x − 2+5x − 3«dx= [−3 ln |x − 2| + 5 ln |x − 3|]54= 3 ln(2/3) + 5 ln 2.Math 132 - Exam III - Spring 2008 22. Evaluate the integralZ21dxx(x2+ 1)dxA) ln2√5B) ln1√2C) ln3√2− ln5√2D) ln3√2E) ln7√3− ln1√5*F) ln2√5+ ln√2G) ln5√5H) ln7√2I) ln2√7+ ln3√2J) ln 3The partial fractions decomposition has the form1x(x2+ 1)=Ax+Bx + Cx2+ 1.Equating the numerators after setting both sides over the same denominator:A(x2+ 1) + (Bx + C)x = 1.Equivalently,(A + B)x2+ Cx + A = 1.This gives B = −A, C = 0 and A = 1. Therefore,Z211x(x2+ 1)dx =Z21„1x−xx2+ 1«dx=hln |x|− lnp1 + x2i21= ln(2/√5) + ln(√2).Math 132 - Exam III - Spring 2008 33. Determine whether the following improper integrals converge or diverge:(a)Z∞1dxx19/20, (b)Z50dxx20/19, (c)Z∞−3dx(x + 4)3/2, (d)Z10dxx.(Below, c stands for ‘converges’ and d for ‘diverges.’)A) c, c, c, cB) d, c, d, d*C) d, d, c, dD) c, c, c, dE) c, d, c, cF) d, d, d, dG) c, c, d, cH) c, c, d, dI) d, c, c, dJ) c, d, d, dRecall that if a is a finite positive number, the integralRa0dxxpconverges for p <and diverges for p ≥ 1. Therefore, both (b) and (d) diverge. The integralR∞adxxpconverges for p > 1 and diverges for p ≤ 1. Therefore, (a) diverges. The integral(c) is equal toR∞1duu3/2(this is seen by doing a substitution u = x + 4), whichconverges.Math 132 - Exam III - Spring 2008 44. Determine whether the integralI =Z10x ln x dxis convergent or not. If it is convergent, evaluate it.A) convergent, I = 1/8B) convergent, I = 1/2C) convergent, I = 1D) convergent, I = −1/2E) convergent, I = −1F) convergent, I = 2*G) convergent, I = −1/4H) divergentI) convergent, I = 4J) convergent, I = −4I = lima→0+R1ax ln x dx. Integration by parts gives:Z1ax ln x dx =»x22ln x −Zx2–1a=»x22ln x −x24–1a= −14−a2ln a2+a24.It can be shown using L’Hˆospital’s rule that lima→0+a ln a = 0. From this limitit immediately follows that lima→0+a2ln a = 0. Therefore, the limit as a → 0exists and is e qual to −1/4.Math 132 - Exam III - Spring 2008 55. Calculate the arc length of the graph of y = x3/2over the interval [1, 2].A) 2e +√7B)727h525/2−345/2i*C)827h1123/2−1343/2iD)83h1125/2−1145/2iE)83h921/2−741/2iF)14h1323/2−1541/2iG)13h723/2−1741/2iH)319h1523/2−1143/2iI)891343/2J)8271123/2The arclength I is given byI =Z21p1 + f0(x)2dx =Z21r1 +94x dx.The change of variables u = 1 + 9x/4 givesI =49Z1+921+94u1/2du =827hu3/2i11/213/4=827h(11/2)3/2− (13/4)3/2i.Math 132 - Exam III - Spring 2008 66. Compute the surface area of revolution defined by the function y = x + 1over the interval [0, 1].A) 2√2πB) 5√2πC) 7√2πD) 3√5πE) 7√5πF) 2√5πG) 8√2π*H) 3√2πI) 3√7πJ) 5√3πThe surface area is given by2πZ10f(x)p1 + f0(x)2dx = 2πZ10(x + 1)p1 + 12dx= 2√2πZ10(x + 1)dx= 2√2π»x22+ x–10= 2√2π3/2= 3√2π.Math 132 - Exam III - Spring 2008 77. Which of the following integrals correctly represents the surface area ofrevolution obtained by rotating the graph of y = sin x about the x-axisover the interval [0, π] ?A)Rπ0sin x√1 + cos2x dxB) πRπ0sin x√1 + cos2x dxC) 2πRπ0cos xp1 + sin2x dx*D) 2πRπ0sin x√1 + cos2x dxE) 2πRπ0√1 + cos2x dxF) πRπ0√1 + cos2x dxG)Rπ0√1 + cos2x dxH) 2πRπ0cos x√1 + cos2x dxI) 2πRπ0sin xp1 + sin2x dxJ) 4πRπ0cos xp2 + sin2x dxThe general integral expression for the area of a surface of revolution is I =2πRbaf(x)p1 + f0(x)2dx. Therefore, if f (x) = sin x, we haveI = 2πZπ0sin xp1 + cos2x dx.Math 132 - Exam III - Spring 2008 88. If w denotes the weight density of water, find the fluid force on a submergedvertical s quare plate of side 2 m eters having its top side at a depth of 1meter.A) wB) 2wC) 4w*D) 8wE) 3wF) 5wG) 7wH) 9wI) w/2J) w/4We fix the y-axis with origin at a depth of 1 meter pointing down. Therefore,the top side of the square is at y = 0 and the bottom side is at y = 2. Thewidth at level y is constant, equal to l(y) = 2, the depth of a point associatedto coordinate y is 1 + y, and the pressure at level y is p(y) = w (1 + y). So theforce is obtained byF =Z20p(y)l(y) dy = 2wZ20(1 + y)dy = 2w[2 + 22/2] = 8w.Math 132 - Exam III - Spring 2008 99. Find the area and the x-coordinate of the centroid of the region lyingbetween the graphs of y = x2and y =√x over the interval [0, 1].A) A = 1/6, xCM= 5/9*B) A = 1/3, xCM= 9/20C) A = 1/6, xCM= 5/11D) A = 1/3, xCM= 5/16E) A = 1/6, xCM= 5/13F) A = 1/6, xCM= 3/5G) A = 1/3, xCM= 9/10H) A = 1/6, xCM= 3/7I) A = 1/6, xCM= 3/10J) A = 1/3, xCM= 3/10The area isA =Z10`√x − x2´dx =»2x3/22−x33–10=13.The x-coordinate of the centroid isxCM=1AZ10x`√x − x2´dx = 3»2x5/25−x44–10= 3»25−14–=920.Math 132 - Exam III - Spring 2008 1010. Find the area and the x-coordinate of the centroid of the quarter of theunit disc centered at the origin (0, 0) and lying in the first quadrant.A) A = π/2, xCM=4πB) A = π/3, xCM=4πC) A = π/4, xCM=4πD) A = π/4, xCM=83π*E) A = π/4, xCM=43πF) A = π/2, xCM=2πG) A = π/2, xCM=3πH) A = π/4, xCM=8πI) A = π/4, xCM=73πJ) A = π, xCM=3πThe x-coordinate of the centroid equals, by symmetry, the y-coordinate, andthe latter is given by: of the region between y = 0 andxCM=12AZ10f(y)2dywhere in this case f(y) =p1 − y2. The area of the quarter disc of radius 1 isA = πr2/4 = π/4.


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WUSTL MATH 132 - m132_E3sSP08

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