INTEGRATION
PURE MATHEMATICS 1
Objectives
Integration as the reverse process of differentiation
Integration is also known as anti-derivatives.
𝑥𝑛
Indefinite integrals
where
𝑛 ≠ −1
and
c
𝑑𝑥 =
+𝑐
is the constant of integration.
Therefore,
(i)
𝑘 𝑑𝑥 = 𝑘𝑥 + 𝑐
(ii)
𝑘𝑥 𝑛
𝑑𝑥 =
𝑥 𝑛+1
𝑛+1
𝑘𝑥 𝑛+1
𝑛+1
+𝑐
Example - Find the following indefinite integral
(i)
5 𝑑𝑥 = 5𝑥 0+1 + 𝑐 = 5𝑥 + 𝑐
(ii)
3𝑥 2
(iii)
2
(iv)
𝑑𝑥 =
3𝑥 2+1
2+1
𝑥 + 5 𝑑𝑥 =
(2𝑥 − 3𝑥
−4
+ 𝑐 = 𝑥3 + 𝑐
𝑥 2+1
2+1
) 𝑑𝑥 =
+
5𝑥 1
1
2𝑥 2
2
+𝑐 =
3𝑥 −3
−
−3
𝑥3
3
+ 5𝑥 + 𝑐
2
+𝑐 =𝑥 +
1
𝑥3
+𝑐
Example – Find the following indefinite integrals
(v)
(2𝑥 4
−
4
𝑥2
) 𝑑𝑥 = (2𝑥 4 − 4𝑥 −2 ) 𝑑𝑥
=
(vi)
(vii)
(viii)
𝑥 𝑑𝑥 =
(2
+ 𝑥3
1
4
𝑥
5
4
+𝑐 =
𝑑𝑥 = 3 + 𝑐 =
2
+ 5𝑥 −3
5𝑥 𝑑𝑥 =
−
4𝑥 −1
−1
3
𝑥2
1
2
5
5𝑥 4
2𝑥 5
5
) 𝑑𝑥 = 2𝑥
5
4
2 𝑥3
3
𝑥4
+
4
+ 𝑐 = 4𝑥 + 𝑐
2 5
𝑥
5
4
+
𝑥
+𝑐
+𝑐
5𝑥 −2
+
−2
+ 𝑐 = 2𝑥 +
𝑥4
4
−
5
2𝑥 2
+𝑐
Example – Find the following indefinite integrals
(ix)
11𝑥 10
+
10𝑥 9
𝑑𝑥 =
11𝑥 11
11
+
10𝑥 10
10
+𝑐
= 𝑥 11 + 𝑥 10 + 𝑐
(x)
𝑥4
+ 3𝑥 2
+ 2𝑥 + 1 𝑑𝑥 =
𝑥5
5
3𝑥 3
3
2𝑥 2
+
2
+
=
𝑥5
5
+ 𝑥3 + 𝑥2 + 𝑥 + 𝑐
+
𝑥1
1
+𝑐
Integrating a linear expression
𝑛+1
𝑛+1
(𝑎𝑥
+
𝑏)
(𝑎𝑥
+
𝑏)
(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 =
+𝑐 =
+ 𝑐,
𝑑
𝑎(𝑛 + 1)
(𝑛 + 1) ×
(𝑎𝑥 + 𝑏)
𝑑𝑥
for all 𝑛 ≠ −1.
Example
4
𝑑𝑥 =
7
(1 − 2𝑥)
−7+1
−6
4(1
−
2𝑥)
(1
−
2𝑥)
4(1 − 2𝑥)−7 𝑑𝑥 =
+c=
+𝑐
−7 + 1 (−2)
3
Equation of a curve
In differentiation,
𝑑𝑦
𝑑𝑥
can be found if the equation of curve 𝑦 = 𝑓(𝑥) is given.
Therefore, if the gradient function is given, then the general equation of the
curve can be found by integration.
𝒅𝒚
= 𝑓′ 𝑥
𝒅𝒙
𝑑𝑦 =
𝑓′(𝑥)
𝑦 =𝑓 𝑥 +𝑐
𝑑𝑥
Example
1)
Solution (i)
𝑦=
(6𝑥 2 + 5) 𝑑𝑥 = 2𝑥 3 + 5𝑥 + 𝑐
(ii) 9 = 2(1)3 +5 1 + 𝑐
∴ 𝑦 = 2𝑥 3 + 5𝑥 + 2
𝑐=2
3
(iii) 𝑦 = 2(−1) +5 −1 + 2 = −2 − 5 + 2 = −5
(shown)
Example
2)
Solution
𝑦=
1
𝑥2
+ 𝑥 −2 𝑑𝑥 =
3
𝑥2
3
2
𝑥 −1
2 𝑥3 1
+
+𝑐 =
− +𝑐
−1
3
𝑥
At (1, 5),
2 (1)3
1
5=
−
+𝑐
3
1
2 𝑥 3 1 16
∴𝑦=
− +
3
𝑥 3
16
𝑐=
3
Definite Integrals
Integration of a function within limits a and b are definite integrals, where a is
the lower limit and b is the upper limit. Solutions to these integrals are
numerical constants.
𝑏
𝑓 ′ 𝑥 𝑑𝑥 = 𝑓 𝑥
𝑎
𝑏
𝑎
= 𝑓 𝑏 − 𝑓(𝑎)
Example
=
4
3𝑥 −1
−1 1
=
−3 4
𝑥 1
=
−3
4
− −3
= 2.25
=
4
8𝑥 −2
−2 2
=
−4 4
𝑥2 2
=
−4
(−4)
−
16
4
=
3
4
4
12𝑥 3 2
3 2 1
=
−1
6𝑥 −2
−2 −3
4
3
2
8𝑥
1
=
−3 −1
𝑥 2 −3
= 8(8 − 1)
=
−3
(−1)2
=
−8
3
=
=
= 56
−
(−3)
(−3)2
=
2
−2
𝑥
0.5
=
𝑥 −1
−1
=
−1
𝑥
=
141
8
+
+ 3𝑥 −3 + 4𝑥 −4 𝑑𝑥
3𝑥 −2
−2
3
− 2
2𝑥
+
2
4𝑥 −3
−3 0.5
4 2
− 3
3𝑥 0.5
=
9 1
𝑥2
4
1
−𝑥
=
𝑥3 2
3 2
9
𝑥1 2
−
1 2 4
=
2
3
=
32
3
−2
𝑑𝑥
27 − 2(3) −
2
3
8 − 2(2)
An improper integral is a definite integral that has either or both limits
infinite or an integrand that approaches infinity at one or more points in
the range of integration.
Example
=
=
1
𝑥1 2
1 2 0
2 𝑥 10
=2
=
=
=
∞
𝑥 −2
−2 1
−1 ∞
2𝑥 2 1
−1
0−
2
=
1
2
Area below the curve
Area between the curve 𝑦 = 𝑓 𝑥 and the x-axis, bounded by
the line 𝑥 = 𝑎 and the line 𝑥 = 𝑏 is given by
𝑏
𝐴𝑟𝑒𝑎 =
𝑦 𝑑𝑥
𝑎
𝑏
=
𝑓 𝑥 𝑑𝑥
𝑎
Area below the curve
Area between the curve 𝑥 = 𝑓 𝑦 and the y-axis, bounded by the line 𝑦 = 𝑎
and the line 𝑦 = 𝑏 is given by
𝑏
𝐴𝑟𝑒𝑎 =
𝑥 𝑑𝑦
𝑎
𝑏
=
𝑓 𝑦 𝑑𝑦
𝑎
Area below the curve
Area above the x-axis (yellow) will give
a positive value and the area below the
x-axis (green) will give a negative
value. Hence Integrating the function
directly between limits a and b will not
give the total area below the graph.
To find total area, integration must be
carried out piecewise.
Example
1)
𝐴=
=
5
3 𝑑𝑥
(𝑥
−
3)
3
5
(𝑥−3)4
4
3
=4
𝐴=
=
=
4
2 𝑑𝑥
(𝑥
−
4)
2
4
(𝑥−4)3
3
2
8
3
Area between a curve and a line
Area between the curve 𝑦 = 𝑓 𝑥 and
the line 𝑦 = 𝑔(𝑥) , bounded by the limits
𝑥 = 𝑎 and 𝑥 = 𝑏 is given by
𝑏
𝐴𝑟𝑒𝑎 =
𝑓 𝑥 − 𝑔 𝑥 𝑑𝑥
𝑎
Area between curves
Area between the curve 𝑦 = 𝑓 𝑥 and 𝑦 = 𝑔(𝑥) , bounded by the limits
𝑥 = 𝑎 and 𝑥 = 𝑏 is given by
𝑏
𝐴𝑟𝑒𝑎 =
𝑓 𝑥 − 𝑔 𝑥 𝑑𝑥
𝑎
Area between curves
Area between the curve 𝑥 = 𝑓 𝑦 and 𝑥 = 𝑔(𝑦) , bounded by the limits 𝑦 = 𝑐
and 𝑦 = 𝑑 is given by
𝑑
𝐴𝑟𝑒𝑎 =
𝑓 𝑦 − 𝑔 𝑦 𝑑𝑦
𝑐
Examples
Example
1) The diagram shows the curve
𝑦 = (𝑥 − 2)2
and the line 𝑦 + 2𝑥 = 7, which
intersect at points A and B.
Find the area of the shaded region.
Solution:
𝐴=
3
7
−1
= 7𝑥
=
Line = curve
𝑥 2 − 4𝑥 + 4 = −2𝑥 + 7
𝑥 2 − 2𝑥 − 3 = 0
𝑥 = −1 𝑜𝑟 3
32
3
− 2𝑥 − (𝑥 − 2)2 𝑑𝑥
− 𝑥2
3
(𝑥−2)3
−
3
−1
2) The diagram shows the graph of 𝑦 = 4𝑥 − 𝑥 3 . The point A has
coordinates (2, 0).
(i) Find
𝑑𝑦
.
𝑑𝑥
Solution :
Then find the equation of tangent to the curve at A.
𝑑𝑦
𝑑𝑥
= 4 − 3𝑥 2
At 2 , 0 ;
𝑑𝑦
𝑑𝑥
= −8
Equation of tangent at A;
𝑦 = −8𝑥 + 16
(ii) The tangent at A meets the curve again at the curve again at the point B.
Find the coordinates of B.
Equation of tangent = curve
𝐵(−4 , 48)
(iii) Calculate the area of the shaded region between the straight line AB and the
curve.
𝐴=
2
16 −
−4
= 16𝑥 −
8𝑥 − (4𝑥 − 𝑥 3 ) 𝑑𝑥
6𝑥 2
= 108 𝑢𝑛𝑖𝑡 2
2
𝑥4
+
4 −4
Volume of revolution
If we rotate the area between the curve 𝑦 = 𝑓 𝑥 and the x-axis, bounded by
the line 𝑥 = 𝑎 and the line = 𝑏 , a solid of revolution is formed.
The volume of this solid is given by
𝑏
𝑦 2 𝑑𝑥
𝑉= 𝜋
𝑎
𝑏
=𝜋
𝑓 𝑥
𝑎
2
𝑑𝑥
Volume of revolution
If we rotate the area between the curve 𝑥 = 𝑓 𝑦 and the y-axis, bounded
by the line 𝑦 = 𝑐 and the line = 𝑑 , a solid of revolution is formed.
The volume of this solid is given by
𝑑
𝑥 2 𝑑𝑦
𝑉= 𝜋
𝑐
𝑑
=𝜋
𝑓 𝑦
𝑐
2
𝑑𝑦
Example
Solution :
(i) & (ii)
(iii)
(a) Volume = 𝜋
1
3
4 3𝑥 2
𝑑𝑥
0 4
(b) Volume = 𝜋 3
2
=
4
9𝑥 3
(3)16 0
4 = 12𝜋
= 12𝜋