(x/2) 2 +1 Add them together and throw in a constant c.

Transcription

1 Calculus, 5/6, Solutions Solutions Chapter Solutions to Eercises... (a) c (b) 5 + c (c) This integral can be evaluated using the substitution u +. Alternatively, we can first use polynomial division to get , and then integrate this to find / + +ln + + c. (d) /8( +) 6 + c (substitution u +) (e) /cos( ) + c (substitution u ) (f) /9( +7) 9 + c (substitution u +7) (g) /ep( )+c (substitution u ) (h) /8sin 8 ()+c (substitution u sin()) (i) ln cos() + c (substitution u cos()) (j) Note that + +.Integrateseparately d /ln( +) (subst. u +), + d / d /arctan(/) (/) + Add them together and throw in a constant c. (k) sin()+cos()+c (subst. u /). (integration by parts with u, v cos) (l) ( +) +) + c (integration by parts with u, v ( +) ) (m) / cos() +/sin() +c (integration by parts with u, v sin()) (n) e e + c (integration by parts with u, v e ) (o) e e e + c (use integration by parts twice: first time with u and v e,secondtimewithu and v e ) (p) We use the substitution tanu. Then d du cos u arctan d arctan(tan u) cos u du and hence u cos u du. Integrating this by parts gives u du u tan u tan u du u tan u +ln cos u + c. cos u

2 Calculus, 5/6, Solutions Now note that cos u sin u+cos u cos u tan u + and hence cosu / p +tan u.thus arctan d u tan u+ln cos u +c arctan()+ln / p + +c. (q) ln() + c (integration by parts with u ln and v ) p (r) arccos() + c (first substitution cos(u), then integration by parts). (a) R d R ln u (b) R du ln u + c ln ln + c d R u du ln u + ( ) u u + c ln + + c (c) R sec 5 tan d R u du 5 u5 + c 5 sec5 + c (d) R e +e d R +u du arctan(u)+c arctan(e )+c (e) R 5sin 7 d R 5( u ) du 5u +5u u u7 + c 5cos()+5cos () cos 5 ()+ 5 7 cos7 ()+c (f) R p + d R 8u (u ) du u u5 + 8 u + c 7/ 56 ( +) ( 5 +)5/ + 8( +)/ + c (g) R p d R cos u du u + sin u cos u + c arcsin()+ p + c (The integral of cos u in this computation can be found using integration by R parts: cos u du R R cos u cos u du cosusin u ( sin u)sinu du cosusin u + R R ( cos u)du cosusin u + u cos u du, hence R cos u du u +sinucos u + c.) Solutions to Eercises... (a) R e 5 d lim T! e 5e T 5 5 (b) undefined, because R pd lim [p ] T T! eist (c) R d lim + [arctan T! ]T / e (sin (d) R e cos d lim T!. If s 6 then T lim s T s+ d lim T! T! s + p and the limit lim T does not T! cos ) T s + which converges if and only if s <. If s thenweendupwith lim T! ln(t ), which diverges. Thus, the integral eists for s< anddoesn t eist for s.

3 Calculus, 5/6, Solutions. We define b a from the right. (a) R (b) R f() d lim T!a b pd lim / T! T d lim ( ) ( ) / / 9/ T! T Solutions to Eercises... R R y d dy R. R / T 8 y dy f() d, whereinthelimitt approaches a R / sin( + y) d dy R / ( cos( /+y)+cos(y)) dy sin( /) + sin( /) + p Solutions to Eercises... RR ( + y) d dy R R +y ( + y) d dy R y +y + 5 y dy 7 6. RR ( y) d dy R 9 R py ( y) d dy R 9 9 9y +y y/ y 5/ dy 9 5. If you plot the points you get a square rotated by 5 degrees. So, I split the integral into two. The first has limits apple y apple and y apple apple +y and the second integral is apple y apple andy apple apple y. +y y y d dy y d dy + y d dy y y dy + y (y y )dy. Another split. First region is apple y apple andapple apple and the second is apple y apple andappleapple y. y y d dy y d dy + y d dy P 9 y dy + y y + y dy 8 5. The integral RR d dy is the volume under the surface z andabovethe region. This volume is the area of times the height, i.e. it s just the same as the area of the region. Limits of integration: apple y apple and y apple apple y +. y+ d dy ddy + y y (y )dy 9

4 Calculus, 5/6, Solutions Solutions to Eercises.5.. (a) (i) (ii) y ( + y )d dy y dy 7.5 y (iii) new limits: apple apple and apple y apple (b) (i) y dy d (8 8 + )d 8 (ii) y.5.5 (iii) new limits: apple y apple andapple apple ( y)/ (c) (i) p y y y d dy y y dy

5 Calculus, 5/6, Solutions 5 (ii) y.5.5 (iii) new limits: apple apple and apple y apple. (a) It is di cult to integrate in the current order. Switching the order gives the new limits apple apple andapple y apple /. y 6y p + d dy / 6y p + dy d p + d (b) Again, it is worth switching on account of the integration. The new limits are / apple apple andapple y apple /. /y / cos( y)d dy / / / ( sin() sin( )) d cos( y)dy d cos(/) sin(/) + cos(/) cos() + sin(). If you draw the region, you will see that the region can be epressed more compactly as apple apple and/ apple y apple. / sin y dy d d 8

6 6 Calculus, 5/6, Solutions Solutions to Eercises.6.. y d dy y y dy 7. p y e y d dy Solutions to Eercises.7. ye y dy. Set u + y and v y.thentheregionin the (, y)-plane becomes the rectangle in the (u, v)-plane given by apple u apple 6andapple v apple. The Jacobian /( + ), which is always positive in the region interest. Thus, the integral y) ( + y)( + ) d dy ( + y)( + ) du v) u du dv 6 u du dv which is now easy to integrate. You should get.. Set u y and v y/ p. Then the region in the (, y)-plane becomes the rectangle in the (u, v)-plane given by apple u apple andapplevapple p 5. The Jacobian 7 y y) d dy du dv y v) p 5 du dv, 7 uv which can now be easily integrated to give /7ln( p 5) ln().. Use polar coordinates. The integral becomes a (r cos( )) p r r d dr a r cos ( ) d dr a 5.. Use polar coordinates. We have that apple r apple, so apple r apple. The integral becomes / r cos( )r sin( ) p r r d dr / 5. The Jacobian of the change of ) ( + r cos( ))( + r sin( ))r dr d. r sin( )cos( ) d dr. r, andtheintegralbecomes