SPM Additional Mathematics- by KBJIM - jimmigmig@yahoo.com
Integration – Indefinite integral
1. The reverse of differentiation is called integration. The sign
∫
means to integrate.
∫2 dx
2. If we differentiate y = 2x, we get dy/dx = 2. Therefore
= 2x
3. If we differentiate y = 2x + 3 , we also get dy/dx = 2. Therefore ∫2 dx = 2x, which is not correct.
We are supposed to get back 2x + 3.
4. This means that we must add a constant c to the result to ‘take care’ of the answer
and when given certain other data, we can always get back the actual equation.
5.Therefore we always add a constant c to the result of the integration.
The integration that has a constant c is called the indefinite integral.
Hence the indefinite integral of 2 with respect to x is 2x + c.
We write it as ∫2 dx = 2x + c.
The general formula is ∫ x = x
but n = –1 is not allowed.
n+1
n
n+1
Example D-1: Integrate the following functions with respect to x.
(i) x
(ii) √x (iii) 3x + 2x +3 (iv) x – 1 (v) 1/x (vi) (x – 2)
8
2
4
2
3
(vii) (x – 1)/x
Solution:
(i) ∫x dx = x
+ c = x +c
8 +1
9
8
8+1
9
(iii) )
∫ 3x3 + 2x2 + 3
(iv)
∫ x– 1
2
2
∫ x–1
x
3
∫3x3 dx
dx = x – x + c
2
(vi) ) ∫(x – 2) dx
(vii)
dx =
dx =
∫√x dx
(ii) )
= ) ∫x
1/2
2
4
∫1
x
dx =
4
– 1 dx
3
x
x
2
3
=
3/2
+c
∫ x –2 – x –3
3
∫ x– 4 dx
= ) ∫x – 4x + 4 dx = x – 4x + 4x + c
3
2
∫1
2x
3
+ ∫2x dx + ∫3 dx = 3x + 2x + 3x + c (only write one c
4
3
for all the constants)
(v) )
2
dx =
2
dx = – x
–1
–3
= x
–3
3
+c = – 1 + c
3
3x
2
= x – 2x + 4x + c
3
+ x
2
–2
+c = –1 +1 + c
2
x 2x
You should now try the exercises in your text. Don’t forget to write the constant sign c in your answer.
1|Page
Example D-2: A curve passes through the point
P(2, 4).The tangent to the curve at any point is
3
given by dy/dx = kx . If the tangent at P is 2,
find the equation of the curve.
Example D-3: The gradient of a curve which
passes through the point P(2,2) is given by.
x + 4 . Find the equation of the curve.
Solution:
Solution:
3
dy/dx = kx = 2.
When x = 2, k = 2 = 1
3
4
2
3
dy/dx = x + 4
∫ dy = ∫ x + 4 dx
3
Therefore dy = kx dx = x dx
4
∫ dy = ∫ x3 dx
4
4
y= x +c
16
4
= (1) x +c
4 4
y
2
= x + 4x + c
2
2
At P(2,2), we have 2 = 2 + 4(2) + c
2
c= –8
4
At P(2, 4), we have 4 = 2 + c
16
c=3
2
Therefore equation of curve is y = x
2
4
+ 4x – 8
Therefore equation f curve is y = x + 3
16
By Substitution
Example D-4: Integrate with respect to x (i) ∫(x – 2) dx (ii) )
5
∫
dx
5
(x – 2)
(iii)) ∫4(x – 2) dx
5
Solution:
(i) let u = x – 2
du/dx = 1
du = dx
(ii) let u = x – 2
du/dx = 1
du = dx
therefore ∫(x – 2) dx = ∫u du
6
= u +c
6
6
= ((x – 2) + c
6
5
5
therefore
∫ dx
(x – 2)
5
= ∫u
–5
–4
du
+ c = –1 + c
4
4u
= u
–4
= –1
+ c
4
4(x – 2)
(iii) let u = x – 2
du/dx = 1
du = dx
therefore ∫4(x – 2) dx = 4 ∫u du = 4( u + c )
6
6
= 2u + c
3
6
= 2 (x – 2) + c
3
5
5
6
6
note we should get 2u + 4c
3
But 4c is a also constant, so we use c instead.
2|Page
Definite integral
∫
The definite integral has the formula
b
f(x) dx where a is the lower limit and b is the upper limit.
a
Let’s look at some examples of definite integrals.
2
∫ (x – 2)
Example D-5: Evaluate (i)
5
dx
(ii)
1
∫
3
1
2
dx
2
(x – 2)
(iii)
∫ 4 (x – 2)5 dx
0
Solution:
(i) let u = x – 2
du/dx = 1
du = dx
(ii) let u = x – 2
du/dx = 1
du = dx
therefore ∫(x – 2) dx = ∫u du
6
= u +c
6
6
= ((x – 2) + c
6
5
5
2
∫(x – 2)5 dx
therefore
∫ dx
(x – 2)
1
6
]
∫
1
6
1
Substitute x = 1
= 1 [ (2 – 2) – (1 – 2)
6
6
6
–5
du
–4
= u
+ c = –1 + c
4
–4
4u
+ c
= –1
4
4(x – 2)
2
1 [(x – 2 )
=
5
= ∫u
]
3
3
dx
=
5
(x – 2)
[
–1
4
4(x – 2)
= 1 [ – 1 – ( –1 )
4
4
4 (3 – 2)
(1– 2)
Substitute x = 2 (upper limit)
Substitute x = 3
(upper limit)
therefore ∫4(x – 2) dx = 4 ∫u du = 4( u + c )
6
6
= 2(x – 2) + c
substitute x = 2
3
5
5
2
0
]
substitute x = 1
(lower limit)
= 1 ( –1 + 1 ) = 0
4
let u = x – 2
du/dx = 1
du = dx
∫ 4(x – 2)5 dx
1
(lower limit)
= –1
6
(iii)
]
6
substitute x = 0
2
=
[ 2(x – 2)6 ]
3
=
0
2 [ (2 – 2) – (0 – 2 )
3
6
6
]
6
= 2 ( 0 – 2 ) = – 128
3
3
Notice that all the constants ‘ c ‘ are not used at all.
The answers do not contain ‘c’.
Therefore if you come across integrals with upper and lower limits, then you are dealing with
definite integrals. The answers will not contain ‘c’
3|Page
2
Example D-6: Evaluate (i)
∫ 5x
3
2
0
dx (ii) ∫ 2x + x dx
2
0
∫4 (x + 2)2 dx
(iii)
1
1
Solution:
2
3
∫ 5x2 dx
(i)
0
∫
(ii)
0
2
1
1
2
3
= 5 ∫ x dx
2
3
1
= – 4 ∫ (x + 4x + 4 ) dx
2
2
= [ 2x + x ]
1
3
2
0
2
3
(iii) ∫4 (x + 2) dx
2
2x + x dx
3
= 5[ x ]
0
3
0
( when you interchange the limits
2
= [ 2(3) + 3 – ( 2(1) + 1 ) ]
3
2
3
2
3
= 5 [2 –0]
3
= 18 + 9 – 7
2
6
= 40
3
= 64
3
the integral changes sign )
3
1
2
= – 4 [ x + 4x + 4x ]
0
3
2
3
= – 4 [ 1 + 2(1)
3
2
+ 4(1) ]
= – 76
3
2
∫ f(x) dx
Example D-7: If
6
= 3, and
0
6
∫ f(x) dx
(i)
0
=
∫
2
(ii)
0
0
6
∫ (3f(x) + 5 ) dx
0
2
f(x) dx = 9 , find the followings: (i) ∫f(x) dx (ii) ∫ (3f(x) + 5) dx
2
∫ f(x) dx
f(x) +
0
2
6
∫
= 3 + 9 = 12
2
2
= 3 ∫ f(x) +
0
2
∫ 5 dx
2
= 3(3) + [5x]
0
= 9 + (10) = 19
0
Area under a curve
The area under a curve, y = f(x), bounded by the x-axis, x = a and x = b is given by using the formula
Y
y = f(x)
b
A = ∫ y dx
a
0
a
b
X
Y
For area bounded by the Y-axis, y = a and y = b,
the formula is
b
b
A = ∫ x dy
a
y = f(x)
a
0
X
4|Page
2
Example D-8: Find the area bounded by the curve y = 6 – 2x , the x-axis ,the lines x = – 1 and x = 1.
Solution:
Y
1
1
1
Area A =
∫ f(x) dx
∫
=
–1
= [ 6x – 2x
3
2
6 – 2x dx
–1
= [ ( 6 – 2 ) – (– 6 + 2 ) ] = 32 units
3
3
3
3
]
y = f(x)
–1
2
X
–1
0
1
2
Example D-9: Find the area bounded by the curve y = 8 – x , and the line y = x + 2 .
Solution:
Y
P
y=x+2
y = x + 2 .........(1)
B (2, 4)
2
y = 8 – x ........(2)
2
y=8–x
A (– 3, 1)
Solving eq (1) and (2),
X
–3
0
2
2
x+2 =8–x
2
x +x–6=0
x = 2 and y = 4 or x = – 3 and y = 1
Area required = Area under graph APB – Area under line AB
=
∫
2
2
2
8–x
dx – ½ (1 + 4) 5 =
–3
[ 8x – x ]
3
–3
3
= [ (16 – 8 ) – (–24 – 9) ] – 25 = 203
3
2
6
– 25
2
(area under line AB is a trapezium)
= 33.83 units
3
2
2
Example D-10: Find the area bounded by the curve y = x – 3x + 2x and the x –axis.
Solution:
Y
3
2
when y = 0, x – 3x + 2x = 0
x(x-1)(x – 2) = 0
X
0
1
2
Graph cuts x-axis at x = 0, x = 1 and x = 2
1
Area above x-axis =
∫ (x3 – 3x2 + 2x) dx
0
1
=
[ x4 – 3x3
4
3
+ 2x
2
2
]
1
=[x –x +x
4
2
=[x –x +x
4
2
4
0
3
]
= ( ¼ –1 + 1) = ¼ unit
2
0
2
Area above x-axis =
∫ (x3 – 3x2 + 2x) dx
1
2
=
[x
4
3
– 3x + 2x
4 3
2
2
]
1
2
Therefore required area = 2 x ¼ = ½ units
4
2
3
]
= (4 – 8 + 4) – (¼ – 1 +1) = – ¼ unit
2
1
( When the areas are below the x-axis you need to add them
separately ignoring the minus sign )
5|Page
2
Example D-11: Find the area bounded by the curve y = 2x, the y-axis and the line y = 3.
Solution:
3
∫x
Area AOB =
3
dy =
0
∫
3
y dy = ½ ∫ y dy
0
2
2
0
Y
3
2
3
= ½[ y
3
3
]
=
1 ( 27 – 0 ) = 9 units
6
2
0
2
0
X
2
Example D-12: Find the area bounded by the curve 4 – y = x and the y-axis.
Solution:
2
∫x
Area AOB =
2
dy =
–2
∫
2
2
4 – y dy =
–2
Y
2
[ 4y – y3 ]
3
–2
2
= [ 4(2) – 2
3
]
= 16 – 16
3
= 10.667 units
3
–2
= ( 8 – 8 ) – (– 8 + 8 )
3
3
0
X
2
–2
Solids of Revolution
The volume of a solid generated by rotating about the x-axis, the area of a curve, y = f(x),
from x = a and x = b is given by the formula
y = f(x)
b
V=
∫y
2
dx
a
b
a
Y
For the volume of a solid generated about the Y-axis, from
y = a and y = b, and the area of the curve, y = f(x),
b
b
V=
∫ x2
dy
a
y = f(x)
a
0
X
6|Page
Example D-13: Find the volume of the solid generated about the x-axis when the area of the
2
curve, y = 3x , bounded by the lines x = 0 and x = 1, is rotated through 360°.
Y
y = f(x)
Solution:
y = 3x
2
2
y = 9x
4
X
1
Volume generated =
1
∫y
2
V=
0
∫x
4
dx = 9
0
1
dx
0
1
= 9
=
[
9
5
x ] = 9
0
5
[( 1 – 0]
5
units
5
3
Example D-14: Find the volume of the solid generated about the x-axis when the area of the
2
curve, y = 2x – x , bounded by the lines x = 0 and x = 1, is rotated through 360°.
Y
Solution:
y = 2x – x
2
2
2
= x(2 – x)
y = x (2 – x)
2
y = f(x)
2
2
4
3
= x (4 – 4x + x ) = x – 4x + 4x
1
Volume generated =
∫y
2
V=
∫
dx =
0
2
X
0
1
4
3
1
2
x – 4x + 4x dx
0
1
5
4
[x – x
=
5
=
8
units
15
+ 4x
3
3
]
[( 1 – 1 + 4 ) – 0]
=
0
5
3
3
Example D-15: Find the volume of the solid generated about the y-axis when the area of the
2
curve, y = 3x , bounded by the lines y = 0 and y = 1, is rotated through 360°.
Y
y = f(x)
Solution:
y = 3x
2
2
x = y
3
X
1
Volume generated =
∫x
2
V=
0
1
∫ y dy
dy =
0
3
0
1
[
=
3
2
y
2
]
[( 1 – 0]
=
0
6
=
units
3
6
7|Page
2
Example D-16: The area of the segment of the curve y = x + 2 is bounded by the lines
y = 2, y = 3 and the y –axis. It is rotated through 360° about the y- axis..
Find the volume generated.
Solution:
2
Points of intersection of the line y = 3 and the curve y = x + 2 are given by
2
y=x +2
Y
2
x = y–2
3
∫x
2
Volume generated = V =
3
dy =
2
∫ (y – 2) dy
3
2
y= f(x)
3
=
[y
2
– 2x
]
2
2
2
X
=
=
[( 9 – 6) – ( 2 – 4 ) ]
2
5
2
units
0
3
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