Midterm Exam #2. Please staple this cover and honor pledge atop your solutions.

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1 Math 50B Itegral Calculus April, 07 Midterm Exam # Name: Aswer Key David Arold Istructios. (00 poits) This exam is ope otes, ope book. This icludes ay supplemetary texts or olie documets. You are ot allowed to work i groups or pairs o the quiz. You are ot allowed to elist the aid of a tutor or fried to help with the quiz. You must aswer all of the exercises o your ow. You are also ot allowed to share your work with your fellow studets. Hoor Pledge: I promise that the solutios submitted were doe etirely by me. I received o help from colleagues, frieds, or tutors. I also did ot share ay parts of my solutios with ay of my classmates. Sigature: Please staple this cover ad hoor pledge atop your solutios. Neatess Requiremet: Each problem o this exam must be doe o a separate sheet of paper. All pecil ad paper calculatios must be doe usig pecils. Deductios will be made for ay work doe i pe. Ay mistakes made must be carefully erased with a eraser. If there is ay scratchig out of work or ay disorgaized presetatio, poits will be deducted. (0 pts ). Use pecil ad paper calculatios to determie the legth of the curve x = (y )/ from y = to y = 5. Use Mathematica s CotourPlot commad to sketch the curve x = (y )/ from y = to y = 5. Obtai a pritout of your code ad curve ad attach it behid your had-calculatios for this problem. Note: To obtai full credit for this problem, you must set up the problem with the correct formula, the show all of the steps required to verify your solutio. Solutio: Substitute y = ad y = 5 ito x = (y )/. x = ( )/ x = (5 )/ x = (0)/ x = (4)/ x = 0 x = 6 With this iformatio, we ca draw the cotour plot. CotourPlot[x == / (y - )^(/), {x, 0, 6/}, {y,, 5}, ImageSize -> 00] This produces the followig image.

2 Math 50B Itegral Calculus/Midterm Exam # Page of 6 Name: Aswer Key For our first step, let s simplify ds = + (dx/dy), give that x = (y )/. ds = + (dx/dy) ( ) = + D y (y )/ = + ((y ) / ) = + y = y Fixig a y-value betwee y = ad y = 5, we ca itegrate to fid the legth. L = = = = (dx/dy) dy y dy y / dy [ y/ ] 5 = (5)/ ()/ = (5/ ) = (5 5 ) Thus, the legth is L = (5 5 ).

3 Math 50B Itegral Calculus/Midterm Exam # Page of 6 Name: Aswer Key (0 pts ). Use pecil ad paper calculatios to determie the surface area of the surface obtaied by rotatig the curve y = 5 x5 + x from x = to x = about the x-axis. Use Mathematica s Plot commad to draw the curve ad Mathematica s RevolutioPlotD commad to draw the surface of revolutio. Obtai a pritout of your code, curve, ad surface ad attach it behid your hadcalculatios for this problem. Note: To obtai full credit for this problem, you must set up the problem with the correct formula, the show all of the steps required to verify your solutio. Solutio: We start by drawig the curve. Plot[/5 x^5 + /( x^), {x,, 5}, ImageSize -> 00] This produces the followig image Now we ca use Mathematica s RevolutioPlotD to draw the surface. RevolutioPlotD[{x, /5 x^5 + /( x^), 0}, {x,, }, RevolutioAxis -> {, 0, 0}, ImageSize -> 50] Which produces the followig image.

4 Math 50B Itegral Calculus/Midterm Exam # Page 4 of 6 Name: Aswer Key Now, if we fix a value betwee x = ad x = 5, the followig image y (x,y) y x leads to the formula for the surface area. Area = πy ds We will make a choice of: ds = + ( ) dy dx dx

5 Math 50B Itegral Calculus/Midterm Exam # Page 5 of 6 Name: Aswer Key Give that y = 5 x5 +, we ca write: x ( dy dx = D x 5 x5 + ) x dy dx = x4 4 x 4 Therefore, + ( ) dy = + (x dx 4 4 ) x 4 = + x x 8 = x x 8 ( = x 4 + ) 4 x 4 = x x 4 Absolute values are ot eeded because this expressio is positive for all values betwee x = ad x =, the bouds of our itegral. Therefore, we ca cotiue with: ( ) dy Area = πy + dx dx ( = π 5 x5 + ) (x x ) x 4 dx ( = π 5 x9 + 0 x + x + ) 48 x 7 dx ( = π 5 x9 + 5 x + ) 48 x 7 dx

6 Math 50B Itegral Calculus/Midterm Exam # Page 6 of 6 Name: Aswer Key Now we ca itegrate. [ Area = π 5 x0 + 5 x 88 x 6 [( = π ) ( )] 88 [( 05 = π ) ( )] 88 [( 05 = π ) ( )] 84 [ 0 = π ] 84 [ 0 = π ] 048 ] Now we eed a commo deomiator of 5,00. [ Area = π ] 500 = π = π 5600 (0 pts ). Show all the required steps to evaluate the limit ( + lim No credit will be awarded for a correct aswer uless there is clear evidece i your pecil ad paper work that you uderstad the processes that your are usig. Solutio: If we take the limit, we arrive at: lim ( + ) ) ( = lim + ) =

7 Math 50B Itegral Calculus/Midterm Exam # Page 7 of 6 Name: Aswer Key This is a l Hopital s idetermiate power situatio. So, let: ( y = + ) ( l y = l + ) ( l y = l + ) Now we take the limit. ( lim l y = lim l + ) This is a ( )(0) situatio, which ca be writte as a divisio. l ( ) + lim l y = lim Now we have a 0 0 situatio, so we ca use l Hopital s rule. lim l y = lim lim l y = lim lim l y = lim lim l y = lim lim l y = + D D ( ) Now, recall that we started with y = ( + /), so ow we ca write: ( ) ( + lim = lim + ) = lim y = lim e l y lim l y = e = e

8 Math 50B Itegral Calculus/Midterm Exam # Page 8 of 6 Name: Aswer Key (0 pts ) 4. Cosider the recursively defied sequece a =, a + = (a + 9) ad perform each of the followig steps: (a) Usig pecil ad paper calculatios, use mathematical iductio to prove that the sequece is icreasig. Solutio: We will use mathematical iductio to show that a < a + for all. Step Oe: Because a = ad a + = (a + 9), we ca write: a = (a + 9) = ( + 9) = () = 4 Therefore, a < a, ad we ve proved that a < a + for =. Step Two: Next, we assume that the statemet is true for = k. a k < a k+ Now we must show that the statemet is true for = k +. a k < a k+ a k + 9 < a k+ + 9 (a k + 9) < (a k+ + 9) a k+ < a (k+)+ Therefore, we ve proved that a < a + for = k + Fial Step: Because we ve proved that a < a + is true for = ad wheever it is true for = k it is also true for = k +, by mathematical iductio we ve show that a < a + for all. (b) Use Mathematica s RecurreceTable commad to geerate umerical approximatios of the first 0 terms of the sequece. Based o the data, make a guess at a upper boud for the sequece. Usig pecil ad paper calculatios, use mathematical iductio to prove that the sequece is bouded above by your guess. Obtai a pritout of your code ad umerical output ad attach it behid your had-calculatios for this problem. Solutio: We use Mathematica s RecurreceTable to geerate approximatios of the first 0 terms of the sequece. Clear[a, ] RecurreceTable[{a[ + ] == / (a[] + 9), a[] == }, a[], {, 0}] // N

9 Math 50B Itegral Calculus/Midterm Exam # Page 9 of 6 Name: Aswer Key This yields the followig umbers.., 4., 4., , , 4.498, , 4.499, , It appears that the umbers are bouded above by 5. Now, let s use mathematical iductio to prove that a < 5 for all. Step Oe: a = < 5, so the statemet a < 5 is true for =. Step Two: Next, we assume that the statmet a < 5 is true for = k, ad we ca write: a k < 5 a k + 9 < (a k + 9) < (5 + 9) (a k + 9) < 4 (a k + 9) < 4 a k+ < 4 Thus, we ve show that a k+ is also less tha 5. Fial Step: Fial Step: Because we ve proved that a < 5 is true for = ad wheever it is true for = k it is also true for = k +, by mathematical iductio we ve show that a < 5 for all. (c) State whether the sequece coverges ad diverges based o the evidece obtaied i parts (a) ad (b). If the sequece coverges, use pecil ad paper calculatios to determie the limit of the sequece. Solutio: Because the sequece is icreasig ad it is boud above by 5, by the mootoic sequece theorem, the sequece coverges. Recall that if a sequece coverges, ay subsequece of the sequece also coverges to the same limit. That is, if lim a = L, the lim a + = L. Thus, we ca write:

10 Math 50B Itegral Calculus/Midterm Exam # Page 0 of 6 Name: Aswer Key a + = (a + 9) lim a + = lim (a + 9) lim a + = ( lim a + 9) L = (L + 9) L = L + 9 L = 9 L = 9 L = 4.5 Therefore, lim a = 9/, or equivaletly, lim a = 4.5. Note that this agrees with the evidece produced i our RecurreceTable output above. e (0 pts ) 5. Fid the sum of the ifiite series. No credit will be awarded uless there = is sufficiet pecil ad paper calculatios that demostrate that you uderstad the processes your are usig to obtai the aswer. Solutio: Start by multiplyig umerator ad deomiator by, the pull the i the umerator outside the summatio symbol. = e = e = e = = = ( e ) = Now, write out the first few terms of the series. ( e ) ( e ( e ) ( ) e = ) = = e + e + e 9 + Note that the series is geometric with the first term a = ad the commo ratio r = e/,

11 Math 50B Itegral Calculus/Midterm Exam # Page of 6 Name: Aswer Key which lies betwee ad. Thus, the sum is give by the formula: a r = e e = e e (0 pts ) 6. Use the itegral test to determie whether the series Full credit will be awarded if ad oly if: e coverges or diverges. = i) A fuctio f(x) is selected that is cotiuous ad positive. ii) Usig pecil ad paper calculatios, apply the first derivative test to fid the iterval [a, ) where the fuctio is decreasig. iii) Usig pecil ad paper calculatios, evaluate the itegral whether it coverges or diverges. iv) Make a fial argumet as to why the series a f(x) dx ad state e coverges or diverges. Note: Somethig must be said here how addig a fiite umber of terms does ot affect the covergece or divergece. Solutio: Use the fuctio f(x) = xe x. Note that f() = e agrees with our series. Moreover, as f(x) = x/e x, the deomiator is ever zero, so there are o poits of discotiuity ad both umerator ad deomiator are positive for all value of x > 0. Thus we have a positive cotiuous fuctio. To show where the fuctio is decreasig, we take the first derivative. = f (x) = x( e x ) + e x = e x ( x) = x e x Note that the deomiator is positive for all values of x, so the derivative will be egative whe the umerator is egative: x < 0 x < x > Hece, f is decreasig o (, ). Next, we use the itegral test ad write: f(x) dx = = lim T xe x dx T xe x dx

12 Math 50B Itegral Calculus/Midterm Exam # Page of 6 Name: Aswer Key Usig itegratio by parts, we get: Usig l Hopital s Rule, we ca write: [ = lim xe x e x] T T {[ = lim T e T e ] T [ e e ]} T { = lim T T e T e + T e + } e { = lim T e T e + T e + } e = e Hece, the itegral xe x dx coverges, so the series = e also coverges. Because addig a fiite umber of terms does ot affect the covergece of a coverget series, the series also coverges. (0 pts ) 7. Use the compariso test to determie whether the series coverges or diverges. + 4 = Note: I order to receive full credit, you must show that the hypotheses of the compariso test are satisfied. You caot just claim that the terms of the series you ve selected for compariso are less tha or greater tha the terms of the series. You must + 4 = provide a proof of less tha or greater tha. You must also state a clear reaso based o the series you selected for compariso why the the series coverges or + 4 = diverges. = e Solutio: First, ote that a = /( + 4) is positive for all. Secod, ote that + 4 > for all values of. Ivertig both sides requires that we reverse the iequality sig. The, multiplyig both sides by, + 4 < + 4 <

13 Math 50B Itegral Calculus/Midterm Exam # Page of 6 Name: Aswer Key for all values of. Therefore, we have two sequeces; a = /( + 4) ad b = /. Both are sequeces of positive terms ad a < b for all. However, the series is a p-series with p >, so it coverges. Therefore, = so the series = also coverges. (0 pts ) 8. Cosider the alteratig series = = = also coverges. Thus, by the compariso theorem, = /( +4) ( ). = i) State why the first hypotheses of the alteratig series test is satisfied. ii) State why the secod hypotheses of the alteratig series test is satisfied. That is, prove that the sequece b = / ultimately decreases. For what values of is the sequece decreasig? Note: Select a fuctio f ad show all of the steps of the first derivative test. iii) Perform the third ad fial part of the alteratig series test to determie whether the series coverges. Note: Somethig must be said here how addig a fiite umber of terms does ot affect the covergece or divergece. Solutio: Defie: f(x) = x x Note that both umerator ad deomiator are positive umbers for x >. Thus, f(x) is positive (, ). Moreover, there are o discotiuities, so f is cotiuous o (, ). Now, to fid where is f decreasig, use the first derivative test. f (x) = x x x l ( x ) = x x x l x = x ( x l ) x = x l x Now, because the deomiator is positive o (, ), the derivative will be egative where the umerator is egative. x l < 0 x l < x l > x > l

14 Math 50B Itegral Calculus/Midterm Exam # Page 4 of 6 Name: Aswer Key I the last step, l > 0, so we did ot reverse the iequality sig. Thus, f (x) is egative whe x > / l, which is equivalet to x >.447. Now we ca say that f is positive, cotiuous, ad decreasig o [, ). Hece, the sequece {/ } represeted by the fuctio f(x) = x/ x is positive ad decreasig for =,, 4, 5,... Now we take the limit. Because both umerator ad deomiator approach ifiity as, we ca apply l Hopital s rule ad differetiate both umerator ad deomiator to obtai: lim = lim = 0 Hece, by the alteratig series test, the series ( ) = l coverges. Now, because addig a fiite umber of terms does ot affect the covergece, the series also coverges. = ( ) (0 pts ) 9. Use the ratio test to determie whether the series coverges or diverges. To! = receive full credit for your solutio, you must show a sufficiet umber of steps that make it clear that you uderstad the process. Solutio: First, simplify: a + a = + ( + )!( + )! We ca remove the absolute value bars because all terms are positive for. We ll also reorder. = +! ( + )! +! = ( + )! + = + + = ( + )

15 Math 50B Itegral Calculus/Midterm Exam # Page 5 of 6 Name: Aswer Key Now we ca take the limit. lim a + a = lim = lim = lim ( + ) [ ( + )] ) ( + = lim ( ) + = lim ( ) ( + 0) = 0 The limit is zero, which is less tha. Hece, by the Ratio Test, the series! coverges. That is, the series = =! coverges absolutely. Therefore, the series! also coverges. ( ) + (0 pts ) 0. Use the root test to determie whether the series coverges or diverges. + 5 = To receive full credit for your solutio, you must show a sufficiet umber of steps that make it clear that you uderstad the process. Solutio: Note that: = ( + a = + 5) Because (( + )/( + 5)) is positive for all, we ca remove the absolute value bars. = ( ) = + + 5

16 Math 50B Itegral Calculus/Midterm Exam # Page 6 of 6 Name: Aswer Key Now we ca take the limit. lim a = lim = However, this limit is larger tha, so by the Root Test, the series diverges. = ( ) + + 5