x 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx

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1 . Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute u to get: e d e u du e u + C e + C d) We hve: + d d + d rcsin + d After substituting u we obtin: d u du u + C Hence: + d rcsin() + C. Compute the following integrls: ) b) c) 4 (t + ) 5 dt 4 d d

2 d) π/ sin() cos() d Solution: ) Substitute u t + to get: (t + ) 5 dt 5 ] 5 u 5 du u b) Note tht 4 < for nd 4 > for 4. Hence we cn write: 4 4 d ( ( 4)d + 8 ) + (4 c) We perform the substitution u : u / du 4 [ ( 4)d 4 ] 6 ( 8)) ( u) d du u u / du + d) After substituting u cos() we get [ u u / / du / ] u + u du u / [ u / / ( / / ) 4 (/ / ) + 5 (5/ 5/ ) π/ sin() d cos() u ] du u u du ] [ ] [ u 5/ + 5/ ]. Find n ntiderivtive F () of f() Solution: The ntiderivtives of f() re of the form: such tht F (). F () rcsin() + C for some constnt C. We hve F () rcsin() + C C, hence the nswer is F () rcsin() + 4. Use the definition of derivtive to show tht the derivtive of f() + is f () +. Solution: We hve: f f( + h) f() ( + h) + ( + h) ( + ) () lim lim h h h h + h + h + + h h + h + h lim lim lim + + h + h h h h h

3 5. Clculte the following limit: Solution: We hve: cos() cos() Since lim +cos(), we hve: lim cos() + cos() + cos() cos() + cos() sin() + cos() lim cos() sin() lim, sin() so tht we only hve to compute lim. Let us compute the side limits by L Hopitl s rule. We hve: sin() sin() cos() lim lim lim sin() sin() cos() lim lim lim sin() Since the two side limits re different, it follows tht lim does not eist. 6. Consider the function f() e. does not eist nd hence lim cos() ) Where is f continuous? Where is it differentible? b) Find ll the criticl points of f. Find ll the locl mim nd minim nd the intervls where f is incresing or decresing. c) Find the inflection points of f nd the intervls where f is concve down/up. d) Does f hve ny horizontl symptotes for f? e) Sketch the grph of f. Solution: ) The function g(), being polynomil, is continuous nd differentible everywhere. The function e is composition of two functions tht re differentible for every, hence it is lso differentible everywhere (hence continuous everywhere). b) We compute the first derivtive of f() using the chin rule: f () ( + )e Since f() is differentible everywhere, the criticl points re the points c stisfying f (c). We hve: f (c) (c + )e c c c + c To find out whether c is locl mimum or locl minimum, we need to compute f (): Hence: f () ( + ) e e (( + ) )e f ( ) (( + ) )e ( ) e 4 < so we conclude tht c is locl mimum, f() is incresing for < (since f () is positive there) nd f() is decresing for > (since f () is negtive there). c) To find the inflection points of f(), we need to solve f (). We hve: f () (( + ) )e ( + ) +,

4 Thus there re two inflection points. We hve: < < < + > + f () > f () < f () > hence f() is concve upwrd in ], [ ] +, + [ nd concve downwrd in [, + ]. d) Since lim + ( ) lim ( ), we hve: lim f() lim + + (e ) e lim + ( ) 4

5 7. Find the volume of pyrmid with squre bse of length l nd height h. Solution: Plce the pyrmid on coordinte system so tht the its verte coincides with the origin nd the bse is perpendiculr to the -is. Then its volume cn be obtined by the formul: V h A()d where A() is the re of the intersection of the pyrmid with plne prllel to its bse t distnce from the verte (the re of the red squre in the picture). We hve A() l(), where l() is the side of the red squre in the picture. We need n epression for l() in terms of. To find it, consider the following side view: Note tht there re two similr tringles which give: l()/ l/ h l() l h 5

6 Hence we hve: h ( V l ) ( ) l h d d h h ( ) [ ] l h ( ) l h h h l h 8. Compute the volume of the solid obtined by considering the region bounded by y nd nd y nd rotting it long the line y. Solution: To compute this volume, we cn shift the whole region up by units nd then rotte long the line y. The new region is bounded by y +, nd y. Applying the formul for the volume of the solid obtined by rotting region bounded between two curves, we obtin: V π(( + ) )d π( )d π[ ] 8π 7 9. A rectngulr poster is to be produced with design in the middle nd blnk spce round the edges. The blnk spce will etend 4cm from the top nd bottom edges nd cm from the left nd right edges. The re of the design in the middle will be cm. Find the dimensions of the poster with the smllest totl re. Solution: The centrl design is rectngle of sides (horizontl) nd b (verticl). Then the totl re of the poster is: A ( + 4)(b + 8) Using the reltion b, we get b A() ( + 4)( nd hence: + 8) To minimize A, we compute its derivtive with respect to : A () Hence A () Since A ( 5) 4 >, we obtin tht 5 minimizes 5 A. Hence 5, b 5 5 re the dimensions tht give the poster the smllest totl re.. Find the re enclosed between the two curves y nd y 4 +. Solution: First we hve to find the intersection points of the two curves: 4 + 4, If is in the intervl [, ], then < 4 + nd hence the re cn be found s the following integrl: A (4 + )d (4 )d Since this is the integrl of n even function on the intervl [, ], it equls: (4 )d ] [4 (4 ) 6

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