Math 1A: introduction to functions and calculus Oliver Knill, Spring 20114/5/2011: Second midterm practice IYour Name:• Start by writing your name in the above b ox.• Try to answer each question on t he same page as the question is asked. If needed, use theback or the next empty page for work. If you need additional paper, write your name on it.• Do not detach pages from this exam packet or unstaple the packet.• Please write neatly. Answers which are illegible for the grader can not be given credit.• Except for multiple choice problems, give computations.• No notes, books, calculators, computers, or other electronic aids are allowed.• You have 90 minutes time to complete your work.1 202 103 104 105 106 107 108 109 1010 10Total: 110Problem 1) TF questions (20 points) No justifications a re needed.1)T FThe formulaRx0f′′(x) dx = f′(x) − f′(0) holds.Solution:Apply the fundamental theorem t o the derivative.2)T FThe area of the upper half disc is the integralR1−1√1 − x2dxSolution:The circle has the equation x2+ y2= 1. Solving for y gives y =√1 − x2.3)T FIf the graph of the function f(x) = x2is r otated around the interval [0, 1]we obtain a solid with volumeR10πx4dx.Solution:Yes, the cross section area is A(x) = πx4. Integrate this from 0 to 1.4)T FThe function f(x) = exis the only anti derivative of ex.Solution:We can scale this function too. The function 2exfor example has also the same antiderivative 2ex.5)T FIf f has a critical point at 1, then F (x) =Rx0f(t) dt has an inflection pointat 1.Solution:By the fundamental theorem of calculus, F′= f and F′′= f′. If f has the critical point1, then f′(1) = 0 and so F′′(1) = 0 which by definition means that F has an inflectionpoint. An inflection point for F is a point, where F′′= f′changes sign. That meansf′= 0.6)T FCatastrophes are parameter values c for a family of functions fc(x) , forwhich a local minimum of fcdisappears.Solution:This is the definition of a catastrophe.7)T FThe volume of a cylinder of height and radius 1 minus the volume of a coneof height and radius 1 is half the volume of a sphere of radius 1.Solution:This was Archimedes insight and you have discovered yourself in a homework.8)T FRolle’s theorem tells that if 0 < c < 1 is a critical point of f and f(0) = f (1),then the critical point is in the interval [0, 1].Solution:There could also be critical points outside the interval.9)T FRolle also introduced the notation |x|1/3for roots.Solution:Yes, this was mentioned in class and on the handout.10)T FIntegrals are linear:Rx0f(t) + g(t) dt =Rx0f(t) dt +Rx0g(t) dt.Solution:You have verified this in a homework.11)T FThe function Li(x) =Rx2dt/ log(t) has an anti-derivative which is a finiteconstruct of trig functions.Solution:No, it is known that this logarithmic integral has no elementary anti-derivative.12)T FThere is a region enclosed by the graphs of x5and x6which is finite andpositive.Solution:Yes, there is a finite region enclosed. It is between x = 0 and x = 1.13)T FThe integralR1−11/x4dx = −1/(5x5)|1−1= −1/5 −1/5 = −2/5 is negative.Solution:It is not defined at 0 and can not even be saved using the Cauchy principal value.14)T FGabriel’s trumpet has finite volume but infinite surface area.Solution:Yes, we have seen this in class.15)T FA function f(x) is a probability density function, if f(x) ≥ 0 andR∞−∞f(x) dx = 1.Solution:This is the definition of a probability density function.16)T FThe mean of a function on an interval [a, b] isRbaf(x) dx.Solution:For the mean, we would have to divide by (b − a).17)T FThe cumulative probability density f unction is an antideriative of the pr ob-ability density function.Solution:Yes, by definition. It is a particular antiderivative which has the property thatlimx→−∞F (x) = 0.18)T FThe integralR∞−∞(x2− 1) dx is finite.Solution:No way. The function does not even go to zero at infinity.RR−R(x2−1) dx = (x3/3−x)|R−R=2R3/3 − 2R does not converge for R → ∞.19)T FThe total prize is the derivative of the marginal prize.Solution:Its reversed. The ma r ginal prize is the derivative of the total prize.20)T FThe acceleration is the anti-derivative of the velocity.Solution:It is reversed. The acceleration is the derivative of the velocity.Problem 2) Matching problem (10 points) No justifications are needed.Match the following functions with their anti derivat ives. Of course only 6 of the 30 functions willappear.Function Antiderivative Enter 1-30cos(3x)sin(3x)3xFunction Antiderivative Enter 1-301/(3x)tan(3x)1/(1 + 9x2)1) sin(3x)2) −sin(3x)/33) sin(3x)/34) −3 sin(3x)5) 3 sin(3x)6) cos(3x)7) −cos(3x)/38) cos(3x)/39) −3 cos(3x)10) 3 cos(3x)11) log(x)/312) 1/(3 − x)13) 1/(3x)14) log(x/3)15) −1/(3x2)16) 3x217) x2/218) 3x2/219) 320) x221) arctan(3x)/322) 3 arctan(3x)23) 1/(1 + 9x2)24) 3/(1 + 9x2)25) −3/(1 + x2)26) 1/ cos2(3x)27) log(cos(3x))28) −log(cos(3x))/329) log(cos(3x))/330) 3/ cos3(3x)Solution:The magic numbers are 3, 7, 18, 11, 28, 21.Problem 3) Matching problem (10 points) No justifications are needed.Which of the following formulations is a Riemann sum approximating the integralR30f(x) dx off(x) = x2over the interval 0, 3.Sum Check if t his is the R iemann sum.nPn−1k=0(3k/n)21nPn−1k=0(3k/n)2nP3n−1k=0(k/n)21nP3n−1k=0(k/n)2Solution:It is the fourth entry.Problem 4) Area computation (10 points)Find the area of the region enclosed by the three curves y = 6 − x2, y = −x and y = x.Solution:Make a picture:-2-112-2246We can look at the region on the rig ht of the y-axis and have therefore the area betweenthe graph of f (x) = x and f(x) = 6 − x2. These graphs intersect at x = 2. The integralisZ20(6 − x2− x) x dx = 2 2/3 .Since we have only computed half, we have to multiply this with 2. The area is 44/3.Problem 5) Volume computation (10 points)Jeanine grows some magical plants in a pot which is a rotationally symmetric solid for which theradius at position x is 5 + sin(x) and 0 ≤ x ≤ 2π. Find the volume of the pot.Solution:The area of the cross section at hight x is A(x) = π(5 + sin(x))2. The volume isZ2π0π(5 + sin(x))2dx = πZ2π025 + 10 sin(x) + sin2(x) dx = (50π + π)π = 51π2.To find the anti-derivative of sin2(x) we use the double angle formula 1 − 2 sin2(x) =cos(2x) leading to sin2(x) = [1 − cos(2x)]/2 so thatRsin2(x) dx = x/2 − sin (2x)/4.Problem 6) Definite integrals (10 points)Find the following definite integralsa) (5