MTH 142 Solution Practice for Exam 2

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1 MTH 4 Solution Pratie for Eam Updated /7/4, 8: a.m.. (a) = 4/, hene MID() = ( + + ) +/ +6/ +/ ( 4 ) =. ( LEFT = ( 4..). =.7 and RIGHT = (.. ). =.7. Hene TRAP =.7.. (a) MID = ( ). = ( TRAP = (.9 +.4)/ = 4.6 SIMP = ( )/ = (a) RIGHT, MID, TRAP, LEFT. Reason: sine the funtion is dereasing, we have RIGHT < LEFT. Also, TRAP is the average of RIGHT and LEFT. In addition, the funtion is onave up, whih implies MID < TRAP. ( Errors: eat - right() =., eat - mid() =., eat - trap()= -., eat - left () = -.4. () LEFT and RIGHT: the first deimals will be orret sine the error improves by deimal plae. MID: the first 4 deimals will be orret sine the error is improved by deimal plaes. TRAP: the first deimals will be orret sine the error is improved by deimal plaes. b 4. (a) Improper. = lim ( + ) / d = lim 6( + ) / b = + b b = lim 6 + b 6 =. Hene the integral diverges. b ( Improper. + d = lim + + d = lim ln + + = lim ln 6 ln + =. Hene the integral diverges. + () Improper. t + t = lim + ln t ln t + the integral diverges. (d) Improper. e t dt = + = lim + dt lim t(t + ) = + ln t t + = lim + e t dt + / t e t dt = (I) + (II). / t + dt = ln 8 ln + = Hene We now analyze (I) and (II) separately: (I) = lim e t et = lim e =. Hene (I) onverges. b (II) = lim e t b b b et = lim b eb =. Hene (II) diverges. We onlude that e t dt also diverges.. (a) Behaves-like analysis: = when is large (p=). Hene we suspet onvergene. We now ompare the integrand with a larger funtion whose integral onverges. We note that = for, whih implies that the following inequality is valid: + 6 =, for <. We onlude from the omparison test that d onverges.

2 t + ( Behaves-like analysis: t t = for large t (p=). Hene we suspet divergene. We now ompare the integrand with a smaller funtion whose integral diverges; t For this we note that t < t + and that t > t, whih imply that the following inequality is valid: = t t t + for t <. We onlude from the omparison test t that dt diverges. t + t 6. By taking setions perpendiular to the ais of rotation, we get washers. At the tikmark j the washer has inner radius r j = j, outer radius R j =, and thikness. The sum that approimates the volume is V (πrj πr j ) = n (π π( j ) ) j= j= The eat volume is obtained by taking limit as. We have, / (π π4 4 )d = π By taking setions perpendiular to the ais of rotation, we get disks. At the tikmark y j the radius is r j = j = yj / and the thikness is y The sum that approimates the volume is πrj y = π( y j /) y = πy j / y j= j= j=! The volume is obtained by taking limit as y. We have, π ydy = π 4 8. By taking setions perpendiular to the ais of rotation, we get washers. At the tikmark y j the washer has inner radius r j =, outer radius R j = + y j /, and thikness y. The sum that approimates the volume is V (πrj πrj ) y = (π( + j= j= The volume is obtained by taking limit as. We have, (π( + y j /) π() ) y y/) π)dy = π( 4 + ).747

3 9. a). a). ) = j= 7 A plot of y = ( )/( + ) produed with a graphing alulator shows that the plate has the shape shown in the figure. It is lear that the urve meets the X and Y aes at = and y = respetively. Sine the plate has onstant density, the formulas in page 6 of the tet apply. The total mass of the plate is Mass = is given by =.4 πr r + rj (.4)πr + r r ( +.) j= ( +.)d =.7 ( +.)d. = ( +.)d.7 =.666. d + Mass d The enter of mass (, y) + = By symmetry we have that y = = Hene (, y) = (.7686,.7686). Note: to obtain y with a alulation, it may be done as follows: Solve for in y = ( )/( + ) to obtain = ( y)/( + y). Then, y = y. y +y dy Mass = Slie the drum with horizontal, irular setions. Eah setion orresponds to a tikmark y j on the vertial ais. The number of bateria in a setion at tikmark y j is Number(Setion j ) density volume = (..y j ) π(4) y The total number of bateria is approimated by the Riemann Sum N umber(container) = (..y j ) π(4) y j=

4 The eat number is obtained by passing to the limit as y. It is, 6 (..y) π(4) dy = 697 million bateria. A ross-setion of the one (shown in the figure) is bounded by the lines y = ± and y =. Introdue tik marks in the y-ais. - The slab S j at height y j is a disk with radius R j = j = y j / and thikness y, so its volume is π(y j /) y, and its weight is 6.4 π(y j /) y. The work involved in raising the slab a distane of ( y j ) to the top of the one is The total work is approimated by The eat work is given by w j = ( y j ) 6.4 π(y j /) y W ( y j ) 6.4 π(y j /) y j= ( y) 6.4 π(y/) y = A sketh of the dam is shown in the figure below. 4 6 Note that the equation of the right hand, non-horizontal side is y =. Introdue tik marks y, y,..., y n, in the y ais. At height y j, the slab has area (y j + ) y, and the pressure at this height is 6.4( y j ). Therefore the fore on the slab is The total fore is approimated by F j = 6.4( y j )(y j + ) y F 6.4(y j + )( y j ) y j=

5 . a) The eat value of the total fore is obtained by taking the limit as y :.7.. T d =.69 d = F = 6.4(y j + )( y j )dy =,, 6. a) d =. = T =. = T + =. = T = 4 = d = d = ln() = T e d =. = e T =. = e T + =. = T = b e = b e d = lim e d = lim b b e + = lim ( b b e + b e ) ( b ) = ln a) The umulative distribution funtion is an antiderivative of the density, so P = d = +. We now find. We also know that P = when = (first value of ). Substituting we find =, so P () = +. Another way to solve it: P () = dt = t = +. t P = e d = e + C. We also know that P = when =. Substituting into P we have = + C, that is, C =. We onlude that P = e +. Another way to solve it: P () = e t dt = e t = e The inreasing funtion is the Cumulative Distribution Funtion, whih we know has range from to. This gives the vertial range, so the tik marks on the y ais are at {,.,.4,.6,.8,.}. Also, we know that the region under the density funtion (appro.. retangles) has area. Then eah retangle has area /. =.4. But we also know the height of the retangle is.. Then the base of the retangle is approimately.4/. =. Therefore the tik marks on the -ais are at {,, 4, 6, 8, }.