Short Question 1


Question 1:
Find the integral of each of the following.(a)(3x252x3+2)dx(b)x2(x5+2x)dx(c)3x4+2xx3dx(d)(7+x)(7x)x4dx(e)(5x1)3dx(f)3(4x+7)8dx


Solution:
(a)
(3x252x3+2)dx=(3x25x32+2)dx=3x1+5x24+2x+c=3x+54x2+2x+c


(b)
x2(x5+2x)dx=(x7+2x3)dx=x88+2x44+c=x88+x42+c


(c)
3x4+2xx3dx=(3x4x3+2xx3)dx=(3x+2x2)dx=3x222x+c


(d)
(7+x)(7x)x4dx=(49x2x4)dx=(49x41x2)dx=(49x4x2)dx=49x33+1x+c=493x3+1x+c


(e)
(5x1)3dx=(5x1)4(4)(5)+c=120(5x1)4+c


(f)
3(4x+7)8dx=3(4x+7)8dx=3(4x+7)7(7)(4)+c=328(4x+7)7+c

Integration as the Inverse of Differentiation


3.4c Integration as the Inverse of Differentiation

Example:
Shows that ddx[2x+5x23]=2(x2+5x+3)(x23)2Hence, find the value of 20(x2+5x+3)(x23)2 dx


Solution:
ddx[2x+5x23]=(x23)(2)(2x+5)(2x)(x23)2                  =2x264x210x(x23)2                  =2x210x6(x23)2                  =2(x2+5x+3)(x23)2202(x2+5x+3)(x23)2 dx=[2x+5x23]20220(x2+5x+3)(x23)2 dx=[2x+5x23]20    20(x2+5x+3)(x23)2 dx=12[(2(2)+5223)(2(0)+5023)]                                 =12[9(53)]                                 =12×323                                 =163                                 =513

3.6 Integration as the Summation of Volumes


3.6 Integration as the Summation of Volumes

(1).


The volume of the solid generated when the region enclosed by the curve y = f(x), the x-axis, the line x = and the line x = b is revolved through 360° about the x-axis is given by

Vx=πbay2dx



(2).


The volume of the solid generated when the region enclosed by the curve x = f(y), the y-axis, the line y = a and the line y = b is revolved through 360° about the y-axis is given by
Vy=πbax2dy

Long Question 1 & 2


Question 1:
A curve with gradient function 5x5x2  has a turning point at (m, 9).
(a) Find the value of m.
(b) Determine whether the turning point is a maximum or a minimum point.
(c) Find the equation of the curve.

Solution:
(a)
dydx=5x5x2At turning point(m,9),dydx=0.5m5m2=05m2=5mm3=1m=1

(b)
dydx=5x5x2=5x5x2d2ydx2=5+10x3When x=1, d2ydx2=15 (> 0)Thus, (1,9) is a minimum point.

(c)
y=(5x5x2) dxy=5x22+5x+cAt turning point (1,9), x=1 and y=9.9=5(1)22+51+cc=32Equation of the curve: y=5x22+5x+32




Question 2:
A curve has a gradient function kx2– 7x, where k is a constant. The tangent to the curve at the point (1, 3 ) is parallel to the straight line  y + x– 4 = 0.
Find
(a) the value of k,
(b) the equation of the curve.

Solution:
(a)
y + x – 4 = 0
y = – x + 4
m = –1

f ’(x) = kx² – 7x
Given tangent to the curve at the point (1, 3) parallel to the straight line
k (1)² – 7 (1) = –1
k – 7 = –1
k = 6

(b)
  f'

Short Question 5 – 7


Question 5:
Given  ( 6 x 2 +1 )dx=m x 3 +x +c,  where m and c are constants, find (a) the value of m. (b) the value of c if  ( 6 x 2 +1 )dx=13 when x=1.

Solution:
(a)
( 6 x 2 +1 )dx=m x 3 +x +c 6 x 3 3 +x+c=m x 3 +x+c 2 x 3 +x+c=m x 3 +x+c Compare the both sides,  m=2

(b)

( 6 x 2 +1 )dx=13 when x=1. 2 ( 1 ) 3 +1+c=13            3+c=13                 c=10



Question 6:
It is given that  5 k g(x)dx=6 , and  5 k [ g( x )+2 ]dx =14, find the value of k.

Solution:
5 k [ g( x )+2 ]dx =14 5 k g( x )dx + 5 k 2dx =14                6+ [ 2x ] 5 k =14                 2( k5 )=8                      k5=4                           k=9



Question 7:
Given  k 2 (4x+7)dx=28 , calculate the possible value of k.

Solution:
k 2 (4x+7)dx=28 [ 2 x 2 +7x ] k 2 =28 8+14( 2 k 2 +7k )=28 222 k 2 7k=28 2 k 2 +7k+6=0 ( 2k+3 )( k+2 )=0 k= 3 2  or k=2

Short Question 8 – 10


Question 8:
Given y= 5x x 2 +1  and  dy dx =g( x ), find the value of  0 3 2g( x )dx.

Solution:
Since dy dx =g( x ), thus y= g( x ) dx 0 3 2g( x )dx=2 0 3 g( x )dx   =2 [ y ] 0 3   =2 [ 5x x 2 +1 ] 0 3   =2[ 5( 3 ) 3 2 +1 0 ]   =2( 15 10 )   =3



Question 9:
Find  5 k ( x+1 )dx, in terms of k.

Solution:
5 k ( x+1 )dx=[ x 2 2 +x ] 5 k   =( k 2 2 +k )( 5 2 2 +5 )   = k 2 +2k 2 35 2   = k 2 +2k35 2



Question 10:
Given that= 2 5 g(x)dx=2 . Find (a) the value of  5 2 g(x)dx, (b) the value of m if  2 5 [ g(x)+m( x ) ]dx=19

Solution:
(a)  5 2 g(x)dx= 2 5 g(x)dx                      =( 2 )                      =2

(b)  2 5 [ g(x)+m( x ) ]dx=19       2 5 g(x)dx+m 2 5 xdx=19               2+m [ x 2 2 ] 2 5 =19                        m 2 [ x 2 ] 2 5 =21                     m 2 [ 254 ]=21                             21m=42                                 m=2

Short Question 11 – 13


Question 11:
Given  2 3 g(x)dx=4 , and  2 3 h(x)dx=9 , find the value of (a)  2 3 5g(x)dx, (b) m if  2 3 [ g(x)+3h( x )+4m ]dx=12

Solution:
(a)
2 3 5g(x)dx=5 2 3 g(x)dx                  =5×4                  =20

(b)
2 3 [ g(x)+3h( x )+4m ]dx=12 2 3 g(x)dx+3 2 3 h( x )dx+ 2 3 4mdx=12 4+3( 9 )+4m [ x ] 2 3 =12        4m[ 3( 2 ) ]=19                       20m=19                           m= 19 20



Question 12:
(a) Find the value of  1 1 ( 3x+1 ) 3 dx. (b) Evaluate  3 4 1 2x4  dx.

Solution:
a)  1 1 ( 3x+1 ) 3 dx=[ ( 3x+1 ) 4 4( 3 ) ] 1 1                            = [ ( 3x+1 ) 4 12 ] 1 1                            = 1 12 [ 4 4 ( 2 ) 4 ]                            = 1 12 ( 25616 )                            =20

(b)  3 4 1 2x4  dx= 3 4 1 ( 2x4 ) 1 2  dx                             = 3 4 ( 2x4 ) 1 2  dx                             = [ ( 2x4 ) 1 2 +1 1 2 ( 2 ) ] 3 4                             = [ 2x4 ] 3 4                             =[ 2( 4 )4 2( 3 )4 ]                             =2 2



Question 13:
Given that y= x 2 2x1 , show that dy dx = 2x( x1 ) ( 2x1 ) 2 . Hence, evaluate  2 2 x( x1 ) 4 ( 2x1 ) 2  dx .

Solution:
y= x 2 2x1 dy dx = ( 2x1 )( 2x )x( 2 ) ( 2x1 ) 2     = 4 x 2 2x2 x 2 ( 2x1 ) 2     = 2 x 2 2x ( 2x1 ) 2     = 2x( x1 ) ( 2x1 ) 2  ( shown ) 2 2 2x( x1 ) ( 2x1 ) 2  dx = [ x 2 2x1 ] 2 2 1 8 2 2 2x( x1 ) ( 2x1 ) 2  dx = 1 8 [ x 2 2x1 ] 2 2 1 4 2 2 x( x1 ) ( 2x1 ) 2  dx = 1 8 [ ( 2 2 2( 2 )1 )( ( 2 ) 2 2( 2 )1 ) ]                            = 1 8 [ ( 4 3 )( 4 5 ) ]                            = 1 8 ( 32 15 )                            = 4 15

Long Question 3


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( x2 )=0 x=0x=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 =12x12 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.

Long Question 6


Question 6:
Diagram below shows part of the curve   y = 2 ( 3 x 2 ) 2  which passes through (1, 2).


(a) Find the equation of the tangent to the curve at the point B.
(b) A region is bounded by the curve, the x-axis and the straight lines x = 2 and x = 3.
(i) Find the area of the region.
(ii) The region is revolved through 360° about the x–axis. Find the volume generated, in terms of p.

Solution:
(a)
y = 2 ( 3 x 2 ) 2 = 2 ( 3 x 2 ) 2 d y d x = 4 ( 3 x 2 ) 3 ( 3 ) d y d x = 12 ( 3 x 2 ) 3 d y d x = 12 ( 3 ( 1 ) 2 ) 3 , x = 1 d y d x = 12 y 2 = 12 ( x 1 ) y 2 = 12 x + 12 y = 12 x + 14


(b)(i)

Area = 2 3 y d x = 2 3 2 ( 3 x 2 ) 2 d x = 2 3 2 ( 3 x 2 ) 2 d x = [ 2 ( 3 x 2 ) 1 1 ( 3 ) ] 2 3 = [ 2 3 ( 3 x 2 ) ] 2 3 = [ 2 3 [ 3 ( 3 ) 2 ] ] [ 2 3 [ 3 ( 2 ) 2 ] ] = 2 21 + 1 6 = 1 14 unit 2


(b)(ii)
Volume generated = π y 2 d x = π 2 3 4 ( 3 x 2 ) 4 d x = π 2 3 4 ( 3 x 2 ) 4 d x = π [ 4 ( 3 x 2 ) 3 3 ( 3 ) ] 2 3 = π [ 4 9 ( 3 x 2 ) 3 ] 2 3 = π [ 4 9 [ 3 ( 3 ) 2 ] 3 ] [ 4 9 [ 3 ( 2 ) 2 ] 3 ] = π ( 4 3087 + 4 576 ) = 31 5488 π unit 3

3.5 Integration as the Summation of Areas

3.5 Integration as the Summation of Areas

(A) Area of the region between a Curve and the x-axis.



Area of the shaded region;  A = a b y d x


(B) Area of the region between a curve and the y-axis.


Area of the shaded region;  A = a b x d y


(C) Area of the region between a curve and a straight line.


Area of the shaded region;  A = a b f ( x ) d x a b g ( x ) d x