Solutions to Math 152 Review Problems for Exam 1

Transcription

1 Soltions to Math 5 Review Problems for Eam () If A() is the area of the rectangle formed when the solid is sliced at perpendiclar to the -ais, then A() = ( ), becase the height of the rectangle is and the base of the rectangle has length. Therefore, the volme of the solid is ( ) d = = ( ) d + ( )( ) d + ( ) d ( ) d. If we se the sbstittion = then we see that the integral over [, ] is eqal to /, and that the integral over [, ] is also eqal to /. Therefore, the volme of the solid is / + / = /. () The average of f() = over [, ] is ( d = π ) = π. We sed the following fact: d is half of the area of a circle of radis. Now we have to solve c = f(c) = π. The soltions are c = ± π /6. The eistence of at least one c is garanteed by the Mean Vale Theorem for Integrals. ()(a) This type of cone is obtained when the region bonded by y =, = H, y = (R/H) is rotated abot the -ais. The volme is π H ((R/H)) d = πr H/. ()(b) This type of cone is obtained when the region bonded by =, y = H, y = (H/R) is rotated abot the y-ais. The volme is R π(h (H/R)) d = πr H/. ()(a) The eqation y = ( ) is eqivalent to = + y/. The method of washers gives a volme of π ( + y/ ) ( ) dy = π y + y / dy = π( + /) = π/. ()(b) The method of shells gives a volme of π( )( ( )) d = π + 5 d = π(7/6) = π/. cos d d (5)(a) = sin, cot d = sin = = ln + C = ln sin + C. sin d d (5)(b) = cos, tan d = cos = = ln + C = ln cos + C, where the last epression eqals ln sec + C.

2 tan sec d d (5)(c) = sec, tan d = = = ln sec + C. sec sinh d d (5)(d) = cosh, tanh d = cosh = = ln + C = ln cosh + C, which eqals ln(cosh ) + C becase cosh is always positive for real vales of. cosh d d (5)(e) = sinh, coth d = sinh = = ln + C = ln sinh + C. (6)(A) When = csc + cot we get csc d = = csc d csc + cot csc csc + cot d = + csc cot d csc + cot = ln + C = ln csc + cot + C. (6)(B) When = csc cot we get csc cot csc csc d = csc csc cot d = csc cot d csc cot d = = ln + C = ln csc cot + C. (6)(C) The comptation below ses csc cot = at the end: ln csc + cot = ln csc + cot = ln csc cot (csc + cot )(csc cot ) = ln csc cot csc cot = ln csc cot. (7) For the first integral, we do an integration by parts and get sin d = sin sin d = ( cos )(sin ) ( cos )(cos ) d = cos sin + cos d = cos sin + ( sin ) d = cos sin + sin d.

3 After adding sin d to both sides, we discover sin d = cos sin + + C, and this is eqivalent to sin d = ( cos sin ) + C. The second integral is very similar. Integration by parts prodces cos d = cos cos d = (sin )(cos ) (sin )( sin ) d = sin cos + sin d = sin cos + ( cos ) d = sin cos + cos d. We add cos d to both sides and get cos d = sin cos + + C, and this is eqivalent to cos d = ( + sin cos ) + C. The method in this problem can be sed to derive the redction formlas for cos n d and sin n d. In order to get the cos redction formla, we proceed as follows: cos n d = cos cos n d = (sin )(cos n ) (sin ) ( (n ) cos n ( sin ) ) d = sin cos n + (n ) sin cos n d = sin cos n + (n ) ( cos ) cos n d = sin cos n + (n ) cos n d (n ) cos n d.

4 After adding (n ) cos n d to both sides, we get n cos n d = sin cos n + (n ) and this is eqivalent to cos n d = n sin cosn + n n cos n d, cos n d. (8) If we se the sbstittion = tan then we find tan sec d = tan sec sec d = tan ( + tan ) sec d = ( + ) d = + d = + + C = tan + tan + C. On the other hand, we can se the sbstittion = sec and obtain tan sec d = sec (tan sec ) d = d = + C = sec + C. To see that these two answers are really the same, note sec + C = (sec ) = tan + C = ( + tan ) ( ) + + C + tan becase + C is an arbitrary constant. + C = + tan + tan = tan + tan (9) We se integration by parts and long division (applied to + ) to conclde tan d = tan + d = tan d = + tan ( tan ) + C. () There are three integrations by parts: (ln ) d = (ln ) (ln ) d = (ln ) (ln ) d [ = (ln ) (ln ) ln = (ln ) (ln ) + 6 ln d [ = (ln ) (ln ) + 6 ln + C ] d ] d = (ln ) (ln ) + 6 ln 6 + C. + C

5 Of corse, we cold have derived a redction formla for (ln ) n d and sed it three times. () The first integration by parts gets s as far as cos(a) cos(b) d = ( a sin(a)) cos(b) a sin(a)( b sin(b) ) d = a sin(a) cos(b) + b sin(a) sin(b) d. a Now we focs on sin(a) sin(b) d and do another integration by parts: sin(a) sin(b) d = ( ( a cos(a)) sin(b) a cos(a)) b cos(b) d = a cos(a) sin(b) + b cos(a) cos(b) d. a When this eqation is sbstitted into the previos eqation, we get cos(a) cos(b) d = a sin(a) cos(b) + b a which can be rewritten as ) ( b a cos(a) cos(b) d = a This is eqivalent to cos(a) cos(b) d = [ a cos(a) sin(b) + b a ] cos(a) cos(b) d, b sin(a) cos(b) cos(a) sin(b) + C. a a sin(a) cos(b) b cos(a) sin(b) a b + C. One can show that this answer is eqivalent to eqation 5 on page of the tetbook. () Integration by parts and the trigonometric identity + tan = sec allow s to write sec d = sec sec d = tan sec tan (tan sec ) d = tan sec tan sec d = tan sec (sec ) sec d = tan sec sec d + sec d = tan sec sec d + ln tan + sec. 5

6 After adding sec d to both sides, we get sec d = tan sec + ln tan + sec + C. After dividing by, we get sec d = tan sec + ln tan + sec + C. The method in this problem can be sed to derive the redction formlas for sec m d and csc m d. In order to get the sec redction formla, we proceed as follows: sec m d = sec sec m d = (tan )(sec m ) (tan ) ( (m ) sec m tan sec ) d = tan sec m (m ) tan sec m d = tan sec m (m ) (sec ) sec m d = tan sec m (m ) sec m d + (m ) sec m d. After adding (m ) sec m d to both sides, we get (m ) sec m d = tan sec m + (m ) sec m d, and this is eqivalent to sec m d = m tan secm + m m sec m d. () We se the sbstittion = tan θ and the reslt of problem (): 9 + d = ( sec θ)( sec θ) dθ = 9 sec θ dθ = 9 tan θ sec θ + 9 ln tan θ + sec θ + C = ln C = ln( ) + C, 6

7 where we sed 9 ln C = 9 ln ln + C, the fact that 9 ln + C is an arbitrary constant C, and the positivity of for any. () Here the sbstittion = sin θ is appropriate and we se problem (7): 9 d = cos θ cos θ dθ = 9 cos θ dθ ( = 9 (θ + sin θ cos θ) + C = 9 ( sin ) + ) 9 + C. (5) Here the sbstittion = tan θ is appropriate and we also se problem (7). The trick sin θ cos θ = tan θ sec θ is helpfl at the end when we rewrite everything in terms of. d ( + ) = sec θ dθ sec ( + tan θ) = θ dθ (sec θ) = dθ sec θ = cos θ dθ = (θ + sin θ cos θ) + C = ( θ + tan θ ) sec + C θ = ( tan + ) + + C. (6) The correct partial fractions setp is ( )( ) d = ( + )( ) d = A + + B + C ( ) d. Now we have to solve for A, B, C in the eqation Since this can be rewritten as ( + )( ) = A + + B + C ( ). ( + )( ) = A( ) + B( )( + ) + C( + ) ( + )( ), we have to solve for A, B, C in the eqation = A( ) + B( )( + ) + C( + ). 7

8 If we plg in = and = into this last eqation, then we get C = and A =, respectively. If we compare the coefficients of in this same last eqation, then we get = A + B. Now we conclde B =. Finally, ( )( ) d = ( ) d = ln + + ln + + C. (7) The correct partial fractions setp is + A + B ( + )( + ) d = + + Now we have to solve for A, B, C in the eqation + ( + )( + ) = A + B + + This is the same as solving for A, B, C in the eqation C + d. C + = (A + B)( + ) + C( + ) (. + )( + ) ( ) + = (A + B)( + ) + C( + ) = (A + C) + (B + A) + (B + C). Eqating coefficients, we get ( ) A + C =, B + A =, B + C =. There is a short ct for solving these three eqations in three nknowns: If we go back to the eqation + = (A + B)( + ) + C( + ) in ( ) and plg in =, then we obtain = C, hence C =. Plgging C = into part A + C = of ( ) we get A =. Plgging A = into B + A = of ( ) we get B =. Or reslts are consistent with B + C = of ( ). This is a way of checking or arithmetic. Plgging these A, B, C into or partial fractions setp, we get + + ( + )( + ) d = d = ln( + ) + tan ln + + C. (8) We know < < e for < and we know that implies that e d diverges. Trning to the other improper integral, we obtain se the sbstittion =. If we can show that 8 e d = d diverges. All this e d when we e d converges then we will be

9 able to conclde that can show that e d converges. It is clear that e d is finite then it will be clear that e d is finite. If we e d = e d + e d is finite, and hence convergent. If then < e e. The last two sentences imply that e d is finite. e. In addition, e d = (9) We solve ( + )( + ) = A + + B. This is eqivalent to A( + ) + + B( + ) =. The sbstittion = / gives B =. The sbstittion = / gives A =. Now we get d ( + )( + ) = + + d = ln + ln + +C = ln + + +C. The comptation lim ln R + R R + = ln = ln(/) d leads to ( + )( + ) = lim ln R + R R + ln + + = ln(/) ln(/9). () We know cos for. Since cos d converges. d converges, we conclde that 9