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3.8.2 Integration, Long Questions (Question 3 & 4)


Question 3:
The gradient function of a curve which passes through P(2, –14) is 6x² – 12x. 
Find
(a) the equation of the curve,
(b) the coordinates of the turning points of the curve and determine whether each of the turning points is a maximum or a minimum.

Solution:
(a)
Given gradient function of a curve  dy dx =6 x 2 −12x The equation of the curve, y= ∫ ( 6 x 2 −12x )  dx y= 6 x 3 3 − 12 x 2 2 +c y=2 x 3 −6 x 2 +c −14=2 ( 2 ) 3 −6 ( 2 ) 2 +c, at point P ( 2,−14 ) −14=−8+c c=−6 y=2 x 3 −6 x 2 −6


(b)
dy dx =6 x 2 −12x At turning points,  dy dx =0 6 x 2 −12x=0 6x( x−2 )=0 x=0, x=2 x=0,  y=2 ( 0 ) 3 −6 ( 0 ) 2 −6=−6 x=2,  y=2 ( 2 ) 3 −6 ( 2 ) 2 −6=−14 d 2 y d x 2 =12x−12 When x=0,  d 2 y d x 2 =12( 0 )−12=−12 <0 ( 0,−6 ) is a maximum point. When x=2,  d 2 y d x 2 =12( 2 )−12=12 >0 ( 2,−14 ) is a minimum point.


Question 4:
Diagram below shows a curve x = y2 – 1 which intersects the straight line 3y = 2x at point Q.
Calculate the volume generated when the shaded region is revolved 360o about the y-axis.


Solution:

 x= y 2 −1→( 1 ) 3y=2x x= 3 2 y→( 2 ) Substitute (2) into (1), 3 2 y= y 2 −1 2 y 2 −3y−2=0 ( 2y+1 )( y−2 )=0 y=− 1 2    or   y=2


When y=2,x= 3 2 ( 2 )=3, Q=( 3, 2 ) I 1 ( Volume of cone ) = 1 3 π r 2 h= 1 3 π ( 3 ) 2 ( 2 ) =6π  unit 3 I 2 ( Volume of the curve ) = π ∫ 1 2 x 2 dy = π ∫ 1 2 ( y 2 −1 ) 2 dy = π ∫ 1 2 ( y 4 −2 y 2 +1 )dy =π [ y 5 5 − 2 y 3 3 +y ] 1 2 =π[ ( 2 5 5 − 2 ( 2 ) 3 3 +2 )−( 1 5 5 − 2 ( 1 ) 3 3 +1 ) ] =π( 46 15 − 8 15 ) = 38 15 π  unit 3 ∴ Volume generated = I 1 − I 2  =6π− 38 15 π  = 52 15 π  unit 3

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