Review Problem for Midterm #1

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1 Review Problem for Midterm # Midterm I: - :5.m Fridy (Sep. ), Topics: 5.8 nd Office Hours before the midterm I: W - pm (Sep 8), Th -5 pm (Sep 9) t UH 8B Solution to quiz cn be found t mtsui/ mth76/hw.html No clcultor is llowed in the em. You should know how to solve these problems without clcultor.. Evlute the following indefinite integrls: () + d (b) Solution. We cn first simplify + = + = +. So +d = ( + )d = d d + d = ln + + C ( ) = ln + C. We hve used the formul d = + + C + if nd d = d = ln + C. + d (c) Solution. We cn first simplify + = + = 5 +. So + d = ( 5 + )d = 5 d d + d = 7 7 = C 7 ( )( + )d + + C Solution: First we simplify ( )( + ) = ( + ) ( + ) = + 6 = + 6. Thus ( )( + )d = ( + 6)d = C. (d) sec ( ) cos() sin( )d Solution: sec ( ) cos() sin( )d = sec ( )d cos()d sin( )d = tn( ) sin() ( cos( )) MATH 76: pge of 8

2 Review Problem for Midterm # MATH 76: pge of 8 (e) (f) (g) = 8 tn( ) sin() + cos( ) + C. We hve used sec ()d = tn() + C, cos()d = sin() + C, sin()d = cos() + C. sec( ) tn( ) e d Solution: sec( ) tn( ) e d = sec( ) tn( )d e d = sec( ) e + C = 8 sec( ) e + C. We hve used sec() tn()d = sec() + d + C nd e d = e + C. Solution: First, we simplify = = + = Thus + d = ( )d + d = d d rcsin() + C = rctn() rcsin() + C. + We hve used d = rctn()+c nd + d = rcsin()+c. d Solution: d d = + C. ln ln() + C. We hve used d =. Evlute the following definite integrls: Note tht the functions in problem ()-(c) re continuous on the given intevl. So we cn use FTC to evlute definite integrls. () d Solution: First, we simplify the epression + + = + = + So + d = ( 5 + )d = 7 7 This implies tht + = + = C = C. d = ( ) = Note tht = =, 7 = ( ) 7 = 7 = = 8, 5 = nd =. ( ) ( ) So = = = 58 + = 58+ = Thus + d =

3 MATH 76: pge of 8 Review Problem for Midterm # (b) π π cos( ) d (c) (d) (e) Solution: We hve cos( ) d = Thus π π cos( ) π d = sin( ) π sin( π 6 ) = ( ) = + =. π 9 π 8 sin()d Solution: π π 9 π sin()d = cos() 9 8 π 8 cos( π) + cos( π) = + 6 cos( π) = nd cos( π) =. 6 + d cos( )d = sin( ) = sin( ) + C. = sin( π ) sin( π ) = sin( π 6 ) = cos( π 9 ) ( cos( π 8 ) ) = = +.. We hve used Note tht the function f() = is not continuous t =. + So f() = is not continuous on the intervl [, ]. We cn + not use the FTC to evlute + d d for now. + Solution:Note tht the function f() = is continuous on [, ] + ln + d = + = ln 5 ln = ln 5. We hve used d = +b ln +b + C nd ln =.. Suppose the vlue of the function f is shown in the following tble - -/ -/ -/ / / / f() () Approimte f()d using equl subintervls nd left endpoints. Solution: The length of ech subintervl is = tion is {,,,. So the prti- + f() = + = =., }. The Riemnn sum is f() + f( ) + f() = + + ( ) +

4 Review Problem for Midterm # MATH 76: pge of 8 (b) Approimte f()d using equl subintervls nd left endpoints. Solution: The length of ech subintervl is ( ) = =. So the prtition is {,,,, }. The Riemnn sum is f( ) + f( ) + f() + f() = ( ) = + + =.. Approimte ( 5)d using equl subintervls nd left endpoints. Solution:The length of ech subintervl is = =. So the prtition is {,,, 5, }. The Riemnn sum is f() +f() +f() +f(5). We hve f() = 5 = 5 =, f( ) = ( ) 5 = 9 5 = 9 5 =, f() = () 5 = 5 = 6 5 =, f( 5) = ( 5 ) 5 = 5 5 = 5 5 =. So f() + f() + f() + f(5) == ( ) = ( ) = = Evlute the following limits () n (c k ) k (b) lim P k= where P = { =,,, k,, n = }, c k [ k, k ], k = k k. Solution: We hve f(c k ) = c k, f() =, = nd b =. So lim P n k= (c k ) k = ( )d = ( ) ( ) = = = 8 + = 7 =. lim P n sin(c k ) k k= where P = { =,,, k,, n = π}, c k [ k, k ], k = k k. Solution: We hve f(c k ) = sin(c k ), f() = sin(), = nd b = π. So lim n P k= sin(c k) k = π ( ) = cos( π ) cos( ) used cos( π ) = nd cos() =. sin()d = cos() = cos( π ) + cos() = + = π. We hve

5 MATH 76: pge 5 of 8 Review Problem for Midterm # 6. Epress the re of the region enclosed by y = nd y = s n definite integrl (Do not evlute the definite integrl). FIGURE. Grph for problem 6 Solution: We first find the points of intersection by solving = This implies tht 5 5 = nd 5 6 =. We get ( 6)( + ) = nd = or = 6. From the grph, we know tht on the intervl [, 6]. So the re of the region enclosed by y = nd y = is ( )d = 6 ( )d = 6 ( )d. 7. () Epress the re of the region enclosed by y = +, y = +, is nd y is s n definite integrl. Solution: We need to find the coordintes of these points on the grph. The coordinte of point is =. The coordinte of point is given by + = which gives = nd we get =. The coordinte of point is given by solving + = + which gives ( + ) = ( + ) nd we get = +, = nd 7+5 =. Fctoring 7+5 =, we get ( )( 5) = nd = or = 5. Plugging = into +, we get 6 + =. Plugging = into +, we get + =. So = is the solution of + = +. Plugging = 5 5 into +, we get + = 5 + =. Plugging = 5 into 5 +, we get + = 9 =.

6 Review Problem for Midterm # MATH 76: pge 6 of 8 FIGURE. Grph for problem 7 So = 5 is not the solution of + = +. So the coordinte of point is =. On the intervl [, ], we hve + nd the re of the region bounded by y =, y = +, = nd = is ( + )d = + d. On the intervl [, ], we hve + + nd the re of the region bounded by y = +, y = +, = nd = is + ( + )d = ( + + ) + d. So the re of the shded region is ( + d ) + d. (b) Epress the volume of the solid obtined by rotting the bove enclosed region bout is s n definite integrl. Solution: On the intervl [, ], the cross section is circle of rdius + nd A() = π( + ) = π( + ). The volume of the first prt is π( + )d. On the intervl [, ], the cross section is n nnulus where the bigger circle hs rdius ( + ) = + nd the smller circle hs rdius ( + ) =. So the re of

7 MATH 76: pge 7 of 8 Review Problem for Midterm # the cross section is A() = π( + ) π( ) = π( + ( 6 + 6)) = π( + 7 5), The volume of the second prt is π( + 7 5)d. Thus the volume we re interested in finding is π( + )d + π( + 7 5)d. 8. Epress the re of the region enclosed by y =, y = nd = s n definite integrl. Solution: From the picture, we see tht FIGURE. Grph for problem 8 on the intervl [, ]. So the re of the region enclosed by y =, y = nd = is ( )d = ( + )d. 9. () Epress the re of the region enclosed by y =, y = s n definite integrl(do not evlute the definite integrl). Solution: First, we find the points where y = nd y = intersect by solving =. We get = ( ) = nd = or =. On the intervl, we hve. So the re of the region enclosed by y =, y = is ( )d. (b) Epress the volume of the solid obtined by rotting the bove enclosed region bout is s n definite integrl(do not evlute the definite integrl). Solution: The cross section is n nnulus where the rdius of the bigger circle is nd the rdius of the smller circle is. So the re of the cross section is π () π ( ) = π 6 π = π(6 ).

8 Review Problem for Midterm # MATH 76: pge 8 of 8 FIGURE. Grph for problem 9 So the volume of the solid obtined by rotting the bove enclosed region bout is is π(6 )d

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