Introduction
 The process of finding functions from their gradient (derived) function is called integration
 Suppose we differentiate the function y=x^{2}. We obtain ^{dy}/_{dx}=2x
 Integration reverses this process and we say that the integral of 2x is x^{2}.
 From differentiation we know that the gradient is not always a constant. For example, if ^{dy}/_{dx} = 2x, then this comes from the function of the form y=x^{2 }+ c, Where c is a constant.
Example
Find y if ^{dy}/_{dx} is:
 3x^{2}
 4x^{3}
Solution
 dy = 3x^{2}
dx
Then, y =x^{3 }+ c  dy = 4x^{3}
dx
Then, y =x^{4 }+ c
Note;
 To integrate we reverse the rule for differentiation. In differentiation we multiply by the power of x and reduce the power by 1.
 In integration we increase the power of x by one and divide by the new power.
 If ^{dy}/_{dx} = x^{n},then y = X^{n+1} + c, where c is a constant and n≠ 1.since c can take any value we call it an arbitrary constant
n+1
Example
Integrate the following expression
 2x^{5}
 x^{1}
 5x^{3} 2x +4
Solution
 dy =2x^{5}
dx
Then, y = 2x^{5+1 }^{}+ C
5+1
= 2x^{6 }+ C
6
= x^{6 }+ C^{} 3  dy = x^{2}
dx
Then, y = x^{2+1}+ C
2+1
=x^{1}+ C
1
= x^{1 }+ C  dy = 5x^{3 } 2x +4
dx
Then, y =5x^{3+1 }−2x^{1+1} − 4x^{0+1}+ C
3+1 1+1 0+1
=^{5}/_{4}x^{4} − ^{2}/_{2}x^{2} + 4x + C
=^{5}/_{4}x^{4} − x^{2} + 4x + C
Example
Find the equation of a line whose gradient function is ^{dy}/_{dx} = 2x + 3 and passes through (0,1 )
Solution
Since ^{dy}/_{dx} = 2x + 3,the general equation is y =x^{2} + 3x + C. The curve passes through ( 0,1 ). Substituting these values in the general equation ,we get 1 = 0 + 0 + C
1 = C
Hence, the particular equation is y =x^{2 }+ 3x + 1
Example
Find v in terms of h if ^{dV}/d_{h} =3h^{2} + 4 and V =9 when h=1
Solution
The general solution is
V = ^{3}/_{3}h^{3 }+ 4h + C
= h^{3 }+ 4h + C
V = 9 when h = 1 .Therefore
9 = 3 + 4 + C
9 = 5 + C
4 = C
Hence the particular solution is ;
V = h^{3 }+ 4h + 1
Definite and Indefinite Integrals
 It deals with finding exact area.
 Estimate the area shaded beneath the curve shown below
 The area is divided into rectangular strips as follows.
 The shaded area in the figure above shows an underestimated and an overestimated area under the curve.
 The actual area lies between the underestimated and overestimated area. The accuracy of the area can be improved by increasing the number of rectangular strips between x = a and x = b.
 The exact area beneath the curve between x = a and b is given by
∫^{b}_{a}yδx  The symbol ∫ is an instruction to intergrate.
 Thus ∫ydx means integrate the expression for y with respect to x.
 The expression ∫^{b}_{a}yδx, where a and b are limits, is called a definite integral. ‘a’ is called the lower limit while 'b' is the upper limit. Without limits, the expression is called an indefinite integral.
Example
∫^{6}_{2}(2x^{2}+3 )dx
The following steps helps us to solve it
 Integrate 2x^{2 }+ 3 with respect to x , giving ^{2}/_{3}x^{3} + 3x + c.
 Place the integral in square brackets and insert the limits, thus
[^{2}/_{3}x^{3}+3 x+c]^{6}_{2}  Substitute the limits ;
X = 6 gives ^{2}/_{3} x 6^{3 }+ 3 x 6 + c = 162 + C
x = 2 gives ^{2}/_{3} x 2^{3 }+ 3 x 2 + c = ^{34}/_{3 }+ C  Subtract the results of the lower limit from that of upper limit, that is;
(162 + C) – (^{34}/_{3} + C ) = 150^{2}/_{3}
We can summarize the steps in short form as follows:
∫^{6}_{2}(2x^{2}+3 )dx = [^{2}/_{3}x^{3}+3 x+c]^{6}_{2}
=[2 x 6^{3}+(3 x 6)] – [2 x 2^{3} +3 x 2]
3 3
=150 ^{2}/_{3}
Example
 Find the indefinite integral
 ∫(x^{2} + 1) dx
 ∫( x^{2 }+ 4x) dx
 ∫(x^{2} + 1) dx
 Evaluate
 ∫^{1}_{0} ( x^{4 }− 5) dx
 ∫^{2}_{1}(− x^{3}+5x−2) dx
 ∫^{3}_{2}(3x^{2}−4x+5) dx
 ∫^{1}_{0} ( x^{4 }− 5) dx
Solution

 ∫(x^{2 }+ 1) dx = x^{3}+ x + c
3  ∫( x^{2 }+ 4x) dx = x^{4}+ 2x^{2 }+ c
4
 ∫(x^{2 }+ 1) dx = x^{3}+ x + c

 [x^{5} − 5x]^{1}_{0}
5
(^{1}/_{5 }−5) − (^{0}/_{5} − 0)
= 4^{4}/_{5}  x=[x^{4}+ 5x^{2 }−2x]^{2}_{1}
4 2
(–4 + 10–4 ) – (–^{1}/_{4} + ^{5}/_{2}+ 2)
= 2 – 4^{1}/_{4}= –2^{1}/_{4}  _{}[ 3x^{2}– 4x^{2}+5x]^{3}_{2}
3 2
= (27 – 18 +15) – (8 – 8 +10)
= 14
 [x^{5} − 5x]^{1}_{0}
Area Under the Curve
Find the exact area enclosed by the curve y = x2,the axis, the lines x = 2 and x = 4
Solution
The area is given by;
∫^{4}_{3}x^{2}dx=[^{1}/_{3}x^{3}]^{4}_{2}
^{64}/_{3}  ^{8}/_{3}= 18 square units
Example
Find the area of the region bounded by the curve = x^{3 } 3x^{2 }+ 2x, the x axis x = 1 and x = 2
Solution
The area is given by;
∫^{2}_{1}( x^{3 }– 3x^{2 }+ 2x)dx = [x^{4}–x^{3}+x^{2}]^{2}_{1}
4
= (4 – 8 + 4) – (¼ 1 + 1)
= 0  ¼ = ¼
Note;
The negative sign shows that the area is below the x – axis. We disregard the negative sign and give it as positive. The answer is ¼ square units.
Example
Find the area enclosed by the curve x^{2 } 10x + 9,the x – axis and the lines x = 4 and x =1 0.
Solution
The required area is shaded below.
Area = ∫^{9}_{4}(x^{2}10x+9)dx+∫^{10}_{9}(x^{2}10x+9)dx
= [x^{3}−5x^{2}+9x]^{9}_{4 }+ [x^{3}−5x^{2}+9x]^{10}_{9}
3 3
= [(243−405+81)−(^{64}/_{3}−80+36)] + [(^{1000}/_{3} − 500+90 ) − (243 − 405+81)]
= 58^{1}/_{3} + 4^{1}/_{3} (Drop negative sign for area under x  axis)
= 62^{2}/_{3} square units
Example
Find the area enclosed by the curve y = 9x−x^{2} , and the line y =x.
Solution
The required area is
To find the limits of integration, we must find the x coordinates of the points of intersection when;
x = 9x  x^{2}
→0 = 8x  x^{2}
0 = x (8  x)
x = 0 or x = 8
The required area is found by subtracting area under y = x from area under y = = 9x  x^{2}
The required area =∫^{8}_{0}(9xx^{2})dx −∫^{8}_{0}xdx
[9x^{2} x^{3}]^{8}_{0 −}[x^{3}]^{8}_{0}
2 3 2
= 117^{1}/_{3} − 32
= 85^{1}/_{3} square units
Application in Kinematics
The derivative of displacement S with respect to time t gives velocity v, while the derivative of velocity with respect to time gives acceleration, a
Differentiation. Integration
Displacement. Displacement
↓↑
Velocity. Velocity.
↓↑
Acceleration. Acceleration.
Note;
Integration is the reverse of differentiation. If we integrate velocity with respect to time we get displacement while if velocity with respect to time we get acceleration.
Example
A particle moves in a straight line through a fixed point O with velocity ( 4 – 1 )m/s.Find an expression for its displacement S from this point, given that S = when t = 0.
Solution
Since ^{dS}/_{dt} = 4  t
S = 4t  t^{2 }+ C
2
Substituting S = 4, t = 0 to get C;
4 = 4 x 0  0^{2 }+ C
2
4 = C
Therefore S = 4t  t^{2 }+ 4.
2
Example
A ball is thrown upwards with a velocity of 40 m s
 Determine an expression in terms of t for
 Its velocity
 Its height above the point of projection
 Find the velocity and height after:
 2 seconds
 5 seconds
 8 seconds
 Find the maximum height attained by the ball. (Take acceleration due to gravity to be 1 0 m/s2.
Solution
 ^{dv}/_{dt} = 10 (since the ball is projected upwards)
dt
Therefore, v =10t + C
When t = 0, v = 40 m/s
Therefore, 40 = 0 + C
40 = C
The expression for velocity is v = 40 – 1 0t
Since ^{dS}/_{dt} = v = 40  10t;
S = 40t  5t^{3 }+ c
When t = 0 , S = 0 C = 0
The expression for displacement is ;
S = 40t  5t^{2}  ^{}Since v = 40 – 1 0t
 When t = 2
v = 40 – 10(2)
= 40 – 20
= 20 m/s
S = 40t  5t^{2}= 40(2) – 5(2)^{2}
= 80 – 20
= 60 m  When t = 5
V = 40 – 10 (5)
= 10 m/s
S = 40 (5)  4 (5)^{2}
= 200  125
= 75 m  When t = 8
V = 40 – 10(8)
= 40 m/s
S= 40(8)  5 (8)^{2}
= 320 – 320
= 0
 When t = 2
 Maximum height is attained when v = 0.
Thus , 40 – 10t = 0
t= 4
Maximum height S = 160 – 80
= 80 m
Example
The velocity v of a particle is 4 m/s. Given that S = 5 when t =2 seconds:
 Find the expression of the displacement in terms of time.
 Find the :
 Distance moved by the particle during the fifth second.
 Distance moved by the particle between t =1 and t =3.
Solution
 ^{dS}/_{dt} = 4t + c
S=4t + c
Since S = 5 m when t =2;
5 = 4 (2) + C
5 – 8 = C
3 = C
Thus, S =4t – 3 
 [4t3] 5 4 = [(203 )  (163 )]
= 17  13
= 4 m  [4t3] 3 1 = [(123 )  (43 )]
= 9  1
 [4t3] 5 4 = [(203 )  (163 )]
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