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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 01/05/2012 for the course MATHEMATIC MA1505 taught by Professor Freudleong during the Spring '10 term at National University of Singapore.

Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online