# 2011_Solution_4 - MA 1505 Mathematics I Tutorial 4...

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Unformatted text preview: MA 1505 Mathematics I Tutorial 4 Solutions 1. (a) Observe that sec 2 x > 0 and- 4sin 2 x ≤ 0 on [- π/ 3 , π/ 3]. Area = Z π/ 3- π/ 3 h 1 2 sec 2 x- (- 4sin 2 x ) i dx = h 1 2 tan x + Z (2- 2cos2 x ) dx i π/ 3- π/ 3 = tan π 3 + (2 x- sin2 x ) fl fl fl π/ 3- π/ 3 = √ 3 + 4 3 π- 2sin π 3 = 4 3 π. (b) The points of intersection: x = x 2 / 4 implies x = 0 or x = 4. Hence the points of intersection are (0 , 0) and (4 , 4). Note that y = x 2 / 4 ⇔ x = 2 √ y . The required area = Z 1 £ 2 √ y- ( y ) / dy = h 4 3 y 3 / 2- 1 2 y 2 i 1 = 4 3- 1 2 = 5 6 . (c) We have that (2- x )- (4- x 2 ) = x 2- x- 2 = ( x + 1)( x- 2) is negative if and only if x ∈ (- 1 , 2). Hence Area = Z 3- 2 fl fl (2- x )- (4- x 2 ) fl fl dx = h Z- 1- 2 + Z 3 2 i ( x 2- x- 2) dx + Z 2- 1- ( x 2- x- 2) dx = h Z 3- 2- 2 Z 2- 1 i ( x 2- x- 2) dx = h 1 3 x 3- 1 2 x 2- 2 x i 3- 2- 2 h 1 3 x 3- 1 2 x 2- 2 x i 2- 1 = 1 3 £ (27 + 8)- 2(8 + 1) /- 1 2 £ (9- 4)- 2(4- 1) /- 2 £ 5- 2(3) / = 1 3 17 + 1 2 + 2 = 49 6 . 2. (a) The parabola and the line meet at ( x,y ) with 3 = y 2 + 1 , i.e. at (3 , ± √ 2). By formula, Volume = Z √ 2- √ 2 π £ ( y 2 + 1)- 3 / 2 dy = π Z √ 2- √ 2 £ y 4- 4 y 2 + 4 / dy = π h 1 5 y 5- 4 3 y 3 + 4 y i √ 2- √ 2 = π 2 h 1 5 4 √ 2- 4 3 2 √ 2 + 4 √ 2 i = 64 15 √ 2 π. MA1505 Tutorial 4 Solutions...
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2011_Solution_4 - MA 1505 Mathematics I Tutorial 4...

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