DHS Applications of Integration (9649) (TR Solutions)

1 H2 Double Math Applications of Integration (Solutions) Qn Su ggested Solution 1(a)(i) Area R  π 4 π 4 1 π0 4 1tan d 1 d 1yy y y   Or Area R  π 4 0 1 πtan d 1 1 24yy    Or Area R 11 1 00 π 11 d t a n d4 xx x x    Or Area R  1 1 0 1 π11 t a n d24 xx    π 4 π 4 0 0 1tan d 1 1 24 1ln sec 1 24 1ln 2 1 24 14 πln 2 (shown)28 yy y                 OR  1 1 0 1 11 200 1 2 0 1 π11 t a n d24 1 π1t a n d24 1 1 π 11l n ( 1 )24 4 2 14ln 228 xx xxx x x x                              (ii) Volume generated by R abt y-axis  π 4 π 4 2 12 π0 4 1tan d d 1 0.89892 0.899 yyy y       Or Volume generated by R abt y-axis   π 4 22 0 1π tan d 1 1 34 0.89892 0.899 yy       Or Shell method : Volume

2 1 1 0 π21 1 t a n d4 0.899 x xx x       (b) Area of required region    ln 0 2 1 dd ln 1 d ln 1 dd sin ln 1 d 0 : 1; ln : 1.9786 1.98 xyx t x t t t tt t x t x t                   Qn Suggested Solution 2(a) ed e ed ee e( 1 ) xx x xx x x xx x x C xC               (b)(i) 2(2 ) , e2 dd 2, edd t t txy xy ttt       dd d dd d 1 e( 2 )t y yx x tt t     When t = 0, x = 2, y = 1 and d1 d2 y x  So, gradient of normal = 2 Hence, equation of normal at t = 0: 12 ( 2 ) 2 3y x y x    

3 (ii) Required Area  2 0 0 2 0 2 00 22 00 22 22 22 11 d( 1 )22 d1 dd4 1e2 d 4 1ed 2 ed 4 1e ( 1) 2 e using result in (a)4 11e ( 1 ) 2 1e 4 13e units4 t tt tt yx xyt t tt tt t t                                     Qn Su ggested Solution 3(a)  42 4 11 2 2 22 d d dx xxx xxxx x        24 12 24 12 22 1 d 1 d 2l n 2l n 22 l n 21 [ 42 l n 4( 22 l n 2 ) ] 1 xxxx xx xx                     If 2x , then |x  2| = (x  2) If x >2, then |x  2| = (x  2)

4 (b) Volume of solid generated   3 22 2 32 2 1 11 12 ( 2 ) d3 7 232 71 32 11 6 yy y y                  Alternative : Shell method   1 2 0 1 32 0 143 2 0 Volume 2 2 d 2 2 d 2 43 11 1 1 2 1 43 6 xx x x xx x x xx x                y x y = x y = x2 + 2 x = 1 2 1 3

5 Qn Su ggested Solution 4(i)   22 2 22 2 22 2 2 1 2 1 d (4 ) 1 2sec d (4 4 tan ) 2sec d 16(1 tan ) 11 d8 sec 1 cos d8 1 cos 2 1 d16 1s i n 2 16 2 1 sin cos16 12 tan16 2 4 x x C C xx C x                                  (ii) Required Volume  2 220 2 1 2 0 1 3 1π d (4 ) π 2 tan use (a)16 2 4 π 1 tan (1)16 2 π 1 π 16 2 4 π 2 π units64 y y yy y                Qn Su ggested Solution 5(a)   22 24 262 d d 14 14 xx xx xx xx      22 422 d d 14 14 xxx xx xx    22 24 2 d d 5( 2 ) 14 xxx xx x    12 22sin 2 1 4 5 x xx c     22 2 2 Let 2 tan 2sec 2sec tan 2 sin 4 2cos 4 x dx dx dd x x x x               x 2  y x y = 2 R

6 (b) Volume =  2 2 e 1 ln 1 de 2 3e x xx      e 221 ln (e 2)d 3e x xx    e 2 1 11 1( e 2 )(ln ) d 3exx xx x          e 2 1 ln 1 (e 2) 3e x xx     2 2( e 2 )1 e3 e    2 22 e3 e 3 e     2 721 3e 3e  Qn Su ggested Solution 6 area of shaded rectangle < Area under the curve from x= n to x = n+1 hence 1 x 40n < 1 40 d n n xx   Drawing similar triangles as shown above, Sum of area of rectangles < actual area under curve 0 1 1 1 2 1 .... 80 1     < 41 40 40 dxx   Hence 41 40 1 2 3 ... 8 0 4 0 d xx      --- (1) 40 -38 y x -40 41 0 -39 -37 39 0 n -40 41n+1 x y 1 2 3 4 −0.4 −0.2 0.2 0.4 0.6 0.8 1 x y R 2 L C 1

7 Draw rectangle as shown below: area of shaded rectangle > Area under the curve from x= n to x = n+1 hence 1 x 41n > 1 40 d n n xx   By drawing rectangles as shown above, Sum of area of rectangles > actual area under curve 1 1 2 1 .... 81 1    > 41 40 40 dxx   Hence 41 40 1 2 3 . . . 81 40 d xx      --- (2)  4141 3 2 4040 240 d 40 3xx x      413 2 40 2 40 4863 x    (or use GC) From (1) and (2), 41 41 40 40 4 0 d 8 1 1 2 3 ... 8 0 < 4 0 dxx xx          477 1 2 3 . . 80 486    95 3 1 2 3 . . 8 0 95 4   53a  x y -40 41 -39 -38 -37 40 39 0 0 n 4yx -40 41n+1 x 41n

8 Qn Su ggested Solution 7(i) To obtain x -intercepts, let 0y   2 ln 1x  ln 1x 11e o r ex  To obtain the turning point, find d 2lnd y xx  . Let d 02 l n 0d y xx   1x Thus coordinates of turning point is 1, 1 . (ii) Area of region R 1 e 2 e ln 1 dxx      11 1 ee e2 ee e 2l nln d xxx x x x x         1 1 ee11 e e ee 2 l n 1 d eexx x                11 1 1ee 2ee ee ee          14e (iii) Make x the subject:  2 ln 1yx ln 1xy  1 e y x   Thus the volume obtained   0 2211 1 e e d 12.2 (to 3 s.f.) yy y      Alternative : Shell method Volume 1 e 2 e 2[ 1 ( l n ) ] d 12.2 xx x   

9 Qn Suggested Solution 8(i) (ii) Since the figure is symmetrical,   1 0 0 2 2 0 22 0 Area d4 d sin d sin dd 4 sin 2 sin d 0, ; 1, 0 2 4s i n s i n 2 d 8s i nc o s d xy xt x t t t tt t x t x t tt t tt t                     3 2 0 sin8 3 8 3 t     (iii) Volume 1 2 0 0 2 2 322 0 222 0 35 2 0 2 d 2 sin 2 ( sin ) d refer to previous part on area 8s i n c o s d 8s i n ( 1 c o s ) c o s d cos cos8 35 11 1 68 35 1 5 yx tt t tt t tt t t tt                      

10 Qn Su ggested Solution 9  2 3 0 2113 22 0 3 0 3 0 2113 22 0 113 22 0 313 22 0 d1d d 11 = 1 d 22 11 11 d 42 4 11 1 d42 4 11 d22 11 d22 1 3 133 3 2 3 ( s h o w n )3 ysx x x xx xxx xxx xx x xx x xx                             2 3 0 13 1 13 22 2 2 0 113 2 22 0 313 2322 0 d21 d d 11 1 2d 32 2 11 1d 33 21 1 =2 33 9 23 3 23 3 3 43 3 ySy x x x xx x x xx x x x xx x x x x                                          