Exam 1 Solutions (Form A) Spring 2005.pdf

Spring 2005 Math 152 Exam 1A: Solutions Mon, 21/Feb cİ2005, Art Belmonte For specificity, lengths are in centimeters unless stated otherwise. Graphs appear at the bottom of the right column to conserve space. 1. (b) We have f ave D 1 b?a R b a f .x/ dx D 1 p ?0 R p 0 x sin parenleftbig x 2 dx D ? 1 2 p cos parenleftbig x 2 p 0 D 1 2 1 p ? ? 1 2 1 p D 1 p . 2. (d) Via Hooke?s Law we have F.x/ D kx or 3 D k 1 6 , whence k D 18. Thus W D R 2 0 18xdxD9x 2 2 0 D36 ft-lb. 3. (a) The area is A D R 2 0 x 2 ? 1 ? 0 dx, computed thus: A D Z 1 0 1 ? x 2 dx C Z 2 1 x 2 ?1dx D x ? 1 3 x 3 1 0 C 1 3 x 3 ? x 2 1 D 2 3 ?.0/C 8 3 ?2 ? 1 3 ?1 D 4 3 C 8 3 ?2 D 2cm 2 : 4. (e) Use half-angle identities. R =2 0 sin 2 x cos 2 xdx D R =2 0 1 2 .1 ? cos 2x/ 1 2 .1 C cos 2x/ dx D 1 4 R =2 0 1 ? cos 2 2xdx D 1 4 R =2 0 1 ? 1 2 .1 C cos 4x/ dx D 1 4 1 2 R =2 0 1 ? cos 4xdx D 1 8 x? 1 4 sin 4x =2 0 D 16 ? 0 D 16 : 5. (c) We have dV D Adx D r 2 dx D y 2 dx D .e x / 2 dx D e 2x dx. Therefore V D R 3 0 e 2x dx D 1 2 e 2x 3 0 D 2 parenleftbig e 6 ?1 621:13 cm 3 . 6. (a) First compute an antiderivative, then apply the FTC. Let u D ln xdvDxdx du D 1 x dx v D 1 2 x 2 : Then R x ln xdxD 1 2 x 2 ln x ? R 1 2 xdxD 1 2 x 2 ln x ? 1 4 x 2 : Hence R 3 1 x ln xdxD 1 2 x 2 ln x ? 1 4 x 2 3 1 D 9 2 ln 3 ? 9 4 ? ? 1 4 D 9 2 ln 3 ? 2. 7. (b) Use the trig identity sin 2 x C cos 2 x D 1. Z sin 3 x cos 2 xdx D Z 1?cos 2 x cos 2 x sin xdx D Z cos 4 x ? cos 2 x .? sin x/ dx D 1 5 cos 5 x ? 1 3 cos 3 x C C 8. (e) We have R 2 1 x xC1 dx D R 2 1 1? 1 xC1 dx D .x ?ln jx C 1j/ 2 1 D .2 ? ln 3/ ? .1 ? ln 2/ D 1 C ln 2 ? ln 3 D 1 C ln 2 3 . 9. (d) Find where the curves intersect, draw a sketch, then set up and compute the integral for the volume. Now x 2 D 2x implies x 2 ? 2x D 0orx.x?2/D0 whence x D 0; 2. (See figure at bottom of column.) We have dV D Adx D parenleftbig r 2 o ? r 2 i dx D .2x/ 2 ? parenleftbig x 2 2 dx D parenleftbig 4x 2 ?x 4 dx. The volume is V D R 2 0 4x 2 ? x 4 dx D 4 3 x 3 ? 1 5 x 5 2 0 D 32 3 ? 32 5 D 32 5?3 15 D 64 15 13:40 cm 3 . 10. (c) First compute an antiderivative, then apply the FTC. Let u D xdvDf 00 .x/ dx du D dx v D f 0 .x/ : Then R xf 00 .x/ dx D xf 0 .x/? R f 0 .x/dx D xf 0 .x/? f.x/: Hence R 5 1 xf 00 .x/ dx D parenleftbig xf 0 .x/? f.x/ 5 1 D parenleftbig 5f 0 .5/? f.5/ ? parenleftbig f 0 .1/? f.1/ D parenleftbig 5.2/?3 ?.?1?4/D7C5D12. 0 1 2 0 1 2 3 x y X1A/3 0 1 2 0 1 2 3 4 x y X1A/9 thickness dx 0 1 2 3 0 5 10 15 20 25 x y X1A/5 thickness dx 1 11. A diagram of the tank, its semicircular end, and a rectangular differential layer of water appears at bottom right. Clearly the layer is 3 m long, but how wide is it? Look at a diagram of the semicircular end of the tank. The width of the layer is 2y D 2 p 1 2 ? z 2 D 2 p 1 ? z 2 . The rectangular layer of water has an area of A D LW D 3 2 p 1?z 2 D 6 p 1?z 2 : Its thickness is dz. Here are the volume of the layer, its weight, and the work required to lift it to the top of the tank. Recall that D g D 9800 N=m 3 . dV D Adz D 6 p 1?z 2 dz dF D dV D 9800 .6/ p 1 ? z 2 dz dW D dF .0?z/ D?9800 .6/ z p 1 ? z 2 dz The work required to pump the water out of the tank is W D R dW D R 0 ?1 ?9800 .6/ z parenleftbig 1 ? z 2 1=2 dz D .9800/.6/ 1 2 2 3 parenleftbig 1?z 2 3=2 0 ?1 D .9800/.2/?0D19;600 J. 12. The volume of a differential cross-section is dV D AdyDs 2 dy D .2x/ 2 dy D 4x 2 dy D 4.1? 1 4 y 2 /dy: Hence the total volume is R 2 ?2 4 ? y 2 dy D 2 R 2 0 4? y 2 dy D 2 4y ? 1 3 y 3 2 0 D 2 8? 8 3 ?0 D 32 3 10:67 cm 3 : 13. Find where the curves intersect, draw a sketch, then set up and compute the integral for the volume. Now p x D 2 ? x implies x D 4 ? 4x C x 2 or x 2 ? 5x C 4 D .x ? 1/.x ?4/D0, whence x D 1; 4. Toss out x D 4. See the figure in the next column. Via cylindrical shells, we have V D R 2 rhdy. V D R 1 0 2 y parenleftbig .2? y/?.y 2 / dy D 2 R 1 0 2y ? y 2 ? y 3 dy D 2 y 2 ? 1 3 y 3 ? 1 4 y 4 1 0 : D 2 1? 1 3 ? 1 4 ?0 D 5 6 2:62 cm 3 : 14. Let x D 5sin .Thendx D 5cos d . Hence (see figure at bottom of the next column) R dx x 2 p 25?x 2 D R 5cos d 25 sin 2 5cos D 1 25 R csc 2 d D? 1 25 cot C C D? p 25?x 2 25x C C 15. We?ll integrate the rational function via partial fractions. Split the integrand into a sum of partial fractions. 10 .x ? 1/ parenleftbig x 2 C 9 D A x ? 1 C Bx CC x 2 C9 10 D A x 2 C 9 C .Bx CC/.x?1/ 0x 2 C0x C10 D .A C B/ x 2 C .C ? B/ x C .9A ? C/ Thus A C B D 0, C ? B D 0, and 9A ? C D 10. Adding the first two equations we obtain A C C D 0, whence C D?A. Substituting this into the third equation gives 10A D 10 whence A D 1, C D?1, and B D C D?1. Therefore, 10 .x ? 1/ parenleftbig x 2 C 9 D 1 x ? 1 ? x x 2 C 9 ? 1 x 2 C 9 : Now integrate term-by-term. Z 10 .x ? 1/ parenleftbig x 2 C 9 dx D Z 1 x ?1 ? x x 2 C9 ? 1 x 2 C9 dx D ln jx ? 1j ? 1 2 ln parenleftbig x 2 C 9 ? 1 3 tan ?1 parenleftbig x 3 C C x z y X1A/11: Tank ?1 ?0.5 0 0.5 1 ?1 ?0.5 0 1 y z X1A/11: Semicircular back end of tank y z ?1 0 1 ?2 ?1 0 1 2 x y X1A/12 thickness dy 0 1 2 0 1 x y X1A/13 thickness dy x 5 (25 ? x 2 ) 1/2 ? X1A/14 2