tutorial 3 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA 1505 Mathematics I Tutorial 3 (Solution Notes) 1. Find the area of the following region. (a) The region bounded between y = 1 2 sec 2 x, y = 4sin 2 x, x = π 3 and x = π 3 . (b) The region in the ±rst quadrant bounded by y = x, y = 1 4 x 2 and below y =1 . (c) The region bounded by y =4 x 2 ,y =2 x, x = 2and x =3 . Preliminaries 0 x y a b y=ƒ y=© S™ The area between the curves y = f ( x )and y = g ( x ) and between x = a and x = b with a<b is S = ± b a | f ( x ) g ( x ) | dx. (1) Remarks: The formula is easy to remember, but difficult to use. Before integrating | f ( x ) g ( x ) | , you must remove the absolute value, then you need to know all the intersection points of f ( x g ( x ), and the sign of f ( x ) g ( x ). If we let g ( x ) = 0, then the physical meaning of integral is quite obvious. Symmetrically, the area between the curves x = f ( y x = g ( y ) and between y = c and y = d with c<d is S = ± d c | f ( y ) g ( y ) | dy. (2) In the following solutions, we ±rst draw the graphs of the function, but it is not necessary to draw the graphs. Actually, you can solve these questions directly. 35
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Solution (a) Qn: The region bounded between y = 1 2 sec 2 x, y = 4sin 2 x, x = π 3 and x = π 3 . 2 -2 x y 0 Ans: The boundaries are x = π 3 and x = π 3 , and so we use formula (1): S= ± b a | f ( x ) g ( x ) | dx = ± π/ 3 3 ² ² ² 1 2 sec 2 x ( 2 x ) ² ² ² dx = ± 3 3 ² ² ² 1 2 sec 2 x +4sin 2 x ² ² ² dx = ± 3 3 ³ 1 2 sec 2 x 2 x ´ dx (Remove absolute value) = ± 3 3 ³ 1 2 sec 2 x +(2 2cos2 x ) ´ dx (s in 2 x = 1 cos 2 x 2 ) = ³ 1 2 tan x +2 x sin 2 x ´ 3 3 = ³ 1 2 tan π 3 · π 3 sin π 3 ´ ³ 1 2 tan µ π 3 · µ π 3 sin µ π 3 ¶´ = 4 3 π. 36
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(b) Qn: The region in the frst quadrant bounded by y = x, y = 1 4 x 2 and below y =1 . 2 -2 x y 0 Ans: The boundaries are y = 0 (in the frst quadrant) and y = 1, and so we use Formula (2): S = ± d c | f ( y ) g ( y ) | dy. Note that y = x 2 / 4= x = ± 2 y = x = f ( y )=2 y (in the frst quadrant) , y = x = x = y = x = g ( y )= y. ±or 0 <y< 1, we know that y< 2 y = g ( y ) <f ( y ). S = ± d c | f ( y ) g ( y ) | dy = ± 1 0 ² ² 2 y y ² ² dy (Note that g ( y ) ( y )) = ± 1 0 2 y y dy = ³ 4 3 y 3 / 2 1 2 y 2 ´ 1 0 = 5 6 . 37
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(c) Qn: The region bounded by y =4 x 2 ,y =2 x, x = 2and x =3 . 4 2 -2 -4 -6 x y 0 ± Ans: The boundaries are x = x = 3, and so we use formula (1): S = ± b a | f ( x ) g ( x ) | dx. DeFne f ( x )=2 x and g ( x )=4 x 2 ,thenwenot icethat f ( x ) g ( x )=(2 x ) (4 x 2 )= x 2 x 2=( x +1)( x 2) < 0 ⇐⇒ x ( 1 , 2) . Thus, we have: f ( x ) g ( x ) 0i f 2 x ≤− 1 , f ( x ) g ( x ) < f 1 <x< 2 , f ( x ) g ( x ) f 2 x 3 . Hence S = ± b a | f ( x ) g ( x ) | dx = ± 3 2 ² ² (2 x ) (4 x 2 ) ² ² dx = ± 3 2 ² ² x 2 x 2 ² ² dx = ± 1 2 ( x 2 x 2) dx + ± 2 1 ( x 2 x 2) dx + ± 3 2 ( x 2 x 2) dx = ³ 1 3 x 3 1 2 x 2 2 x ´ 1 2 ³ 1 3 x 3 1 2 x 2 2 x ´ 2 1 + ³ 1 3 x 3 1 2 x 2 2 x ´ 3 2 = 49 6 .
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tutorial 3 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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