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