Quiz Solutions, Math 36, Sec 6, 7, 6. Spring, Warm-up Problems - Review. Find appropriate substitutions to evaluate the following indefinite integrals: a) b) t + 4) 3 dt. Set u t + 4. We have du dt t. The integral becomes t u 3 du u4 + c t + 4) 4 + c. + ) ) t t dt. Let u + /t and du dt. The integral becomes t u du u + c + t ) + c. c) e x + e xdx. Let u + ex and hence du e x dx. The integral becomes du u ln u + c ln + ex + c.. a) As suggested, u z + and du zdz, we have z 3 z + dz u 3 3 du u 3 3 + c z + ) 3 + c. b) u 3 z + du z 3 z + ) z 3 dz 3 u dz 3u du zdz. Hence our integral becomes u du 3 3 u u + c 3 z + ) 3 + c. c) u tanx du sec xdx sec xtan xdx udu u + c tan x + c. d) u sec x du secxtan xdx sec xtan xdx udu u + c sec x + c. e) The answers in parts c and d are the same as the solution for an indefinite integral is determined up to the addition of a constant. The answers in parts c and d differ by by the virtue of the identity sec x + tan x. 3. The average population is the following. 9 9 Wt)dt [ 3t 3 + t 6t + 86 ) 9 dt + 7 [ 3 7 4 t4 + 3 t3 8t + 86t [ ] 38 + 936 7 4 463 4. ) 4t 3 + 3t 8t + 8 ) ] dt + t 4 + t 3 9t + 8t ) 9 ]
4. The number produced in the period of our interest is 4 dx dt dt 4 ) dt t + ) ) 4 + t + + 4 8. Quiz Solutions, Given during the week of January 7 Section 6. points) Find the definite integral 3x + 3 x + x + 6) dx 3 x + )dx. We have 3x + 3 x + x + 6) dx. x + x + x + 6) dx. Setting u x + x + 6, we have du x + dx 3 9 du 6 u 3 9 u 6 6 + 4.. points) Find the ares of the region bounded by the graph of fx) x + 9 and the x-axis. ) The graph intersects the x-axis at x ±3. The area must be Section 7 3 3. points) Find the definite integral x + 9 ) dx ) 3 3 x3 + 9x 8 + 4 36. 3 x + x 3 + 6x + ) 4 dx. Setting u x 3 + 6x +, we have du 3x + 6)dx 3x + )dx. We have 7 3 u 4du 7 9 u 3 9 ) 7 3 393 447.. points) Find the ares of the region bounded by the graph of fx) x + 4 and the x-axis. The graph intersects the x-axis at x ±. The area must be x + 4 ) dx ) 3 x3 + 4x 6 3 + 6 3 3. Quiz Solutions, Given during the week of February 3 Section 6
. points) Find the area bounded by the curves y fx) x + 4 and y gx) x. We first need to find the points of intersection of the graphs. Setting fx) gx), we get x + 4 x x + x + x + )x ) x,. Hence the integral we must compute is x + x + ) dx 3 x3 + x + x 83 ) + + 4 3 + ) 9.. points) The Maplewood Economists Association determined that the Lorentz curve for dentists in Maplewood was given by 7 x + x while the Lorentz curve for clinical psychologists was given by 7 3 x3 + x. Find the Gini Indeicies for dentists and clinical psychologists in Maplewood. Which profession has a more equitable distribution of 3 income? For dentists, the Gini index is x 7 x 7 ) x dx 4 7 For psychologists, the Gini index is x 3 x3 ) 3 x dx 4 3 x + x ) dx 4 7 Clearly, the incomes of dentists are more equitably distributed. Section 7 3 x3 + ) x. x 3 + x ) dx 4 3 4 x4 + ) x 6 3.. Points) Find the area bounded by the curves y fx) x + 9 and y gx) x + 3. We first need to find the points of intersection of the graphs. Setting fx) gx), we get x + 9 x + 3 x x + 6 x + 3)x ) x, 3. Hence the integral we must compute is x x + 6 ) dx 3 x3 x + 6x 3 3 83 ) + 9 9 ) 8 6.. points) The Mayville Economists Association determined that the Lorentz curve for clinical psychologists was given by 3 7 x + 4 x while the Lorentz curve for dentists was given by 7 x3 + 3 x. Find the Gini Indicies for dentists and clinical psychologists in Mayville. Which profession has a more equitable distribution of income? For psychologists, the Gini index is x 37 x 47 ) x dx 6 7 For dentists, the Gini index is x x3 3 ) x dx 4 x + x ) dx 6 7 Clearly, the incomes of psychologists are more equitably distributed. 3 3 x3 + x ) 7. x 3 + x ) dx 4 4 x4 + ) x.
Section 6. Points) Find the area bounded by the curves y fx) x + 9 and y gx) x + 3. We first need to find the points of intersection of the graphs. Setting fx) gx), we get x + 9 x + 3 x x + 6 x + 3)x ) x, 3. Hence the integral we must compute is 3 x x + 6 ) dx 3 x3 x + 6x 3 83 ) + 9 9 ) 8 6.. Points) Find the consumers surplus if the demand function is p.x.x + 8 and the market price is $ per unit. We will need to find the unit demanded at p. We have.x.x + 8 x + x 3 x + 3)x ) x 3,. We can clearly discard the negative solution. Hence the integral we must compute is.x.x + 8 ) dx. ) 3 x3.x + 8x + 8 3 3. Quiz 3 Solutions, Given during the week of February Note: I try to sketch the solids of interest in this quiz with Maple 8 and import the graphics into the solutions. But the graphs are all ugly and hardly shows anything worth seeing. Moreover, all solids in question are simple to draw. I am positive you can picture them in your head. Hence, imagination is the best painter in the world and I shall not include the sketches of the solids in the solutions. I will be more than happy to show you how to generate the graphs using Maple if you want. Section 6. points) Find the volume of the solid of revolution obtained by revolving the region in the first quadrant bounded by the curves y fx) x and y gx) x about the x-axis. Draw a sketch. Show all work. First, the curves intersect when x x x x x,. Hence, the area of the cross-section perpendicular to the x-axis is π x x4 4, and hence the volume of the solid of our interest is π ) ) x x4 x 3 dx π 4 3 x 8 π 3 3 ) 6π.. points) Find the volume of the solid of revolution obtained by revolving the region bounded by the x-axis, the curve y x and the lines x and x 4 about the x-axis. Draw a sketch. Show all work. The area of the cross-section perpendicular to the x-axis is πx and hence the volume of the solid of our interest is 4 π xdx π 4 x π. Sections 7 and 6 4
. Points) Find the volume of the solid of revolution obtained by revolving the region in the first quadrant bounded by the curves y fx) x and y gx) x about the x-axis. Draw a sketch. Show all work. First, the curves intersect when x x 4x x x, 4. Hence, the area of the cross-section perpendicular to the x-axis is π x x 4, and hence the volume of the solid of our interest is π 4 ) ) x x x 4 dx π 4 x3 π 8 6 ) 8π 3 3.. Points) Find the volume of the solid of revolution obtained by revolving the region bounded by the x-axis, the curve y fx) x and the line x and x about the x-axis. Draw a sketch. Show all work. The area of the cross-section perpendicular to the x-axis is π x4 and hence the volume of the solid of our 4 interest is x 4 π 4 dx π x 3π. Quiz 4 Solutions, Given during the week of February 7 Section 6 x. points) Evaluate the indefinite integral lnxdx. Show all work. We shall integrate by parts. Let f x and hence f 3 x3 ; let g lnx and hence g. We have x. points) Evaluate the definite integral xlnxdx 4 3 x3 lnx 3 x dx 3 x3 lnx 9 x3 + c. xe x dx. Show all work. Let f e x and hence f e x ; set g x and hence g. By partial integration, we have Sections 7. Points) Evaluate the indefinite integral xe x dx e x x) e x dx e x x e x ) + 3e. lnx x dx. Show all work. We shall integrate by parts. Let f and hence f x; let g lnx and hence g. We have x x lnx dx xlnx x x dx xlnx 4 x + c.
. Points) Evaluate the definite integral xe x dx. Show all work. Let f e x and hence f e x ; set g x and hence g. By partial integration, we have xe x dx e x x ) + e x dx e x x e x) 3e +. Section 6 Section 6 did not have recitation this week due to class cancellations. Quiz Solutions, Given during the week of February 4 Section 6. Points) Use the Trapezoidal Rule with n 4 to approximate + lnx)dx. [ ] + lnx)dx + ln + + ln /4 + + ln 3/ + + ln 7/4 + + ln 4.77337.. Points) Use Simpson s Rule with n 4 to approximate e x dx. e x dx [e + 4 e /4) + e /) + 4 e 3/4) + e ] 3 4.463776. Section 7. Points) Use the Trapezoidal Rule with n 4 to approximate + ex dx. + ex dx [ + e + + e 4 /4 + + e / + + e 3/4 + ] + e.9843.. Points) Use Simpson s Rule with n 4 to approximate lnx 3 dx. lnx 3 dx [ ln + 4 ln/4) 3 + ln3/) 3 + 4 ln7/4) 3 + ln 3] 3 4.8778688. Section 6 6
. Points) Evaluate the following integral x ln xdx. Show all work. With g x and f lnx, we have g x and f. Partial integration gives x ) xlnxdx x lnx xdx ln ) ln 3 4 4.6369436.. Points) Use Simpson s rule with n 4 to approximate the integral in problem. Show all work. xlnxdx ln + 4 4 3 4 ln 4 + 3 ln 3 + 4 74 ln 74 ) + ln.63639898. As expected, our approximation using Simpson s rule is not too bad at all. The Week of March 3 During the week of March 3, all sections had their first mid-term examination and no quiz was given during that week. However, students in section 6 was given the extra credit of naming their favorite cast member of Saturday Night Live. The result of the survey is given in the solutions of Quiz 6 for section 6. Quiz 6 Solutions, Given during the week of March Section 6. Points) Evaluate or show divergence: dx. Show all work. x + ) 3 a dx lim x + ) 3 a 3 lim a. Points) Solve the differential equation: work. dx lim x + ) 3 a a + ) ) a+ 3. ) a+ 6 du 3 lim u3 a u dy dx y + 4x 3 ) subject to the condition y). Show all dy dx y + 4x 3 ) dy y dx dx y x + x 4 + c y Moreover, y) /c c. Hence, we have y Section 7 + 4x 3 dx y dy x + x 4 + c x 4 + x + c. x 4 + x. 7
6. Points) Evaluate or show divergence: dx. Show all work. 3 x + 3) 4 3 6 a dx lim x + 3) 4 a 3 6 3 lim a. Points) Solve the differential equation: work. dy dx y + 3x ) a+3 6 dx lim x + 3) 4 a 6 a + 3) 3 ) 6 8. ) a+3 6 u3du 6 3 lim a u 3 6 dy dx y + 3x ) subject to the condition y). Show all dy y dx dx y x + x 3 + c y + 3x dx y dy x + x 3 + c x 3 + x + c. Moreover, y) /c c. Hence, we have y x 4 + x. Section 6. Points) Evaluate or show divergence: xe x dx lim a a Therefore, the integral of our interest converges.. Points) Solve the differential equation: work. xe x dx. Show all work. xe x dx lim ) a a e x lim e a e 4) a e 4. dy dx y + 4x 3 ) subject to the condition y). Show all dy dx y + 4x 3 ) dy y dx dx y x + x 4 + c y Moreover, y) /c c. Hence, we have y + 4x 3 dx y dy x + x 4 + c x 4 + x + c. x 4 + x. 3. Extra Credit) Name your favorite cast member of Saturday Night Live, past or present. The following is the result of this survey. Name Votes Name Votes Name Votes Chris Farly 7 Will Farrell 4 Chevy Chase John Belushi John Candy Colin Quinn Adam Sandler Surprisingly, this is the first time someone in my classes picked Colin Quinn. 8
Quiz 7 Solutions, Given during the week of March 4 Section 6 The police in the township of Junebury keep statistics on the number of traffic accidents. Let X be the number of accident in the intersection of Main Street and Broad Avenue on the Fourth of July. The police reports indicate that the following probabilities apply: PX )., PX )., PX ).4, PX 3).3.. Points) Draw a histogram to represent this probability data. Label your axes.) I haven t learned how to draw the histogram on the computer yet. But I am sure that you all know what it looks like.. Points) What is the expected number of accidents in the intersection of Main Street and Broad avenue in Junebury on the Fourth of July. Show all work. The expected value is µ. +. +.4 + 3.3.9. Section 7 The police in the township of Junebury keep statistics on the number of traffic accidents. Let X be the number of accident in the intersection of Main Street and Broad Avenue on the Fourth of July. The police reports indicate that the following probabilities apply: PX )., PX ).4, PX ).3, PX 3)... Points) Draw a histogram to represent this probability data. Label your axes.) I haven t learned how to draw the histogram on the computer yet. But I am sure that you all know what it looks like.. Points) What is the expected number of accidents in the intersection of Main Street and Broad avenue in Junebury on the Fourth of July. Show all work. The expected value is µ. +.4 +.3 + 3..3. Section 6. Points) Suppose that X is a normal random variable with µ and σ 7. Find P46 < X < 8). 46 µ P46 < X < 8) P < Z < 8 µ ) σ σ ) 46 8 P < Z < 7 7 P.33 < Z <.67).67 ) exp t dt π.33 [.67 π [.438 ) exp t dt.33.438 +.99).637. 9 exp t.33 ) dt )] ) ] exp t dt
. Points) Let Z be the standard normal variable. Find the values of z if z satisfies PZ > z).7486. Applying a change of variables, we have.7486 PZ > z) z ) exp t dt From the table, we see that z.67 and hence z.67. z Quiz 8 Solutions, Given during the week of March 3 Section 6. Points) Use the Taylor Polynomial of order 3 to approximate 3 7.3. ) exp t dt. Clear for convenience, we shall use the Taylor polynomial centered at a 7 and our function is fx) 3 x x 3. WE have Therefore, f x) 3 x 3 f x) 9 x 3 f 3) x) 7 x 8 3. f7) 3, f 7) 7, f 7) 87, f3) 7) 7747. Hence the Taylor Polynomial that we must use is 3 + x 7) 7 87 x 7) + 344 x 7)3. Inserting x 7.3 into the above polynomial, we have 3 7.3 3 + 7.3 7) 7 87 7.3 7) + 7.3 7)3 344 3 + 9 43 + 39366 83379 39366 3.73. If you remember anything from Calculus I in which you might have approximated some number using so-called differentials, you will realize that the you were in fact using the first degree Taylor Polynomial to approximate some numbers.. Points) Use the Taylor Polynomial of order at a to approximate.4 ln3x + )dx. We have fx) ln3x + ) and hence f 3 x) 3x + and f 9 x). Therefore, f), 3x + ) f ) 3 and f ) 9. Therefore, the Taylor Polynomial of our interest is 3x 9 x. Hence, we have Section 7.4 ln3x + )dx.4 3x 9 x dx 3 x x 3).4 8.. Points) Use the Taylor Polynomial of order 3 to approximate 3 8.3. Clear for convenience, we shall use the Taylor polynomial centered at a 8 and our function is fx) 3 x x 3. We have f x) 3 x 3 f x) 9 x 3 f 3) x) 7 x 8 3.
Therefore, f8), f 8), f 8) 44, f3) 8) 346. Hence the Taylor Polynomial that we must use is 3 + x 8) 88 x 8) + 736 x 8)3. Inserting x 7.3 into the above polynomial, we have 3 8.3 + 8.3 8) 88 8.3 8) + 8.3 8)3 736 + 9 3 + 36 9679 468.8. If you remember anything from Calculus I in which you might have approximated some number using so-called differentials, you will realize that the you were in fact using the first degree Taylor Polynomial to approximate some numbers.. Points) Use the Taylor Polynomial of order at a to approximate.3 lnx + )dx. We have fx) lnx + ) and hence f x) x + and f 4 x). Therefore, f), x + ) f ) and f ) 4. Therefore, the Taylor Polynomial of our interest is x x. Hence, we have Section 6.3 ln3x + )dx.3 x x dx x 3 ).3 x3 9.. Points) Find the fourth degree Taylor polynomial of fx) ln + x) centered at x. fx) ln + x) f x) + x f x) + x) f x) Hence we have f), f ), f ), f ), f 4) ) 6. Therefore, the fourth degree Taylor polynomial of fx) ln + x) centered at x is x x + 3 x3 4 x4. + x) 3 f4) x) 6 + x) 4. { } n +. Points) Determine whether the sequence converges or no. Find its limit if it does. n n n + lim n n lim n + n n n n + n lim n Hence the sequence converges as n tends to infinity. Also note that we could have applying L Hôpital s rule in the evaluation of the limit and arrive at the same conclusion..
Quiz 9 Solutions, Given during the week of April 7 Sections 6 and 7 Note that Sections 6 and 7 had the same quiz this time.. Points) The weekly wages of construction works in Boston are normally distributed with a mean of $ and a standard deviation of $4. Find the probability that a construction worker selected at random in Boston has weekly wage of more than $ 66. Use the table provided to give your answer. You may use your calculator only to check your answer. We first must standardize the variable. 66 µ σ 66 4.6. Therefore, we have PX > 66) Pz >.6) Pz <.6).9.49.. Points) Give and use the Taylor Polynomial of order to approximate 9.. Show all work to get credit. fx) x f x) x f x) 4 x 3. Therefore, we have f9) 3, f 9) 6 and f 9). Therefore, approximating fx) by its second 8 degree Taylor polynomial around x 9, we see that Section 6 fx) 3 + x 9) 6 6 x 9) 9. f9.) 3 +. 6.4 6 6379 4.. Points) Express the following decimal as the quotient of two integers. 6.3333... 6.3333 6. +.3 +.3 +.3 + 6. + 3 3. Points) Is the following statement true? If Justify your answer. 3 + 3 3 + 3 99 669 99. a n + b n ) converges, then so do both n ) n n a n and n b n. The above statement is false. The counter example is as follows. Take a n and b n for all n. We see that a n + b n ) and converges, but a n b n diverge. n The Week of April 4 No quiz was given this week. All sections had a review for the second mid-term exam. Quiz Solutions, Given during the week of April Section 6 n n n
. Points) Determine whether the series Justify your answer. n ) 3 n 4 n+ is convergent. If it is convergent, find the sum. ) 3 n 4 n+ 4 ) n 3 is a geometric series whose common ratio is 3 n n strictly less than. Therefore, it is convergent and the sum is 4 3. which has absolute value n. Points) Determine whether the series n + ) 3 n is convergent. Justify your answer. n n n + ) 3 n The last series above is known to converge. Hence we see that the series of our interest is majorized by a convergent series. Consequently, we infer that the series of our interest also converges. Section 7. Points) Determine whether the series Justify your answer. ) n+ 3 6 n n n n n ) 3 n n. ) n+ 3 is convergent. If it is convergent, find the sum. n n ) n is a geometric series whose common ratio is strictly less than. Therefore, it is convergent and the sum is 6. which has absolute value n. Points) Determine whether the series n 3 + ) 4 3 n is convergent. Justify your answer. n n n 3 + ) 4 3 n The last series above is known to converge. Hence we see that the series of our interest is majorized by a convergent series. Consequently, we infer that the series of our interest also converges. Section 6 n n 3 ) 4 3 n n.. Points) Find the power series expansion for the function fx) convergence for the series. x. What is the radius of + x fx) x + x x x ) x x ) n ) n x n+. Transforming the fraction into series is only valid if x < or < x <. Therefore, the radius of convergence is. n n 3
. Points) Find the power series expansion for the function gx) ln + x ). What is the radius of convergence for the series. Hint: g x) fx), where fx) is as in problem.) gx) x + x dx + c ) n x n+ dx + c n n ) n xn+ n + + c. Toward the end of determining the constant c in the above expression, we set x. g) ln and all but the constant c vanishes in the right-hand side of the above expression as all terms in the series involve a positive power of x. Therefore c and gx) n ) n xn+ n +. Of course, the power series of gx) inherits that of the series for fx) which is. Quiz Solutions, Given during the week of 8 April Post hic tantament, in manu Dei est. Translation: After this test, you are in God s hand. Sections 6 and 7 Sections 6 and 7 had a review session and did not get a quiz. Section 6. Points) Find the domain of the function gs, t) s + t. It is known that a square of a real number is always non-negative and sums of non-negative numbers are always non-negative. Hence, s + t is always non-negative. Therefore, the domain of gs, t) is {s, t) s, t R}. R denotes the set of real numbers.. Points) Let fx, y) x 3y. Find the critical points) of the function and then use the second derivative test to classify the naturemin or max) of the point, if possible. f x 4x, f y 6y. Hence the critical point is, ). Moreover, f xx 4 <, f yy 6 and f xy. Therefore, Dx, y) 4 >. Now we may infer the the critical point, ) is a relative maximum. 4