CHAPTER 6: INTEGRATION 6.1 INTRODUCTION
Integration is the reverse process of differentiation or called as antidifferentiation.For example, since the derivative of the distance travelled by a car is its velocity, the distance travelled is the antiderivative of the velocity of the car.

* Types of integral: 1. Indefinite Integral: 2. Definite Integral: whereis the function, is antiderivative and is an arbitrary constant

* Properties of Indefinite Integral

Example

Integrate the following function

* The Integral of Trigonometric Functions Standard form | General form Where : | | | | | | | | | | | | | Remark: must be a LINEAR FUNCTION which means Example

Evaluate the following:

6.2 METHOD OF INTEGRATION

* Integration By Substitution * For product with the same type of functions

* Steps: i) Substitute and to obtain the integral ii) Evaluate by integrating with respect to iii) Example
Replace by in the resulting expression

* Integration By Parts * For product with the different types of functions

* Where are both functions of Example

NOTE: 1. The choice must be such that the u part becomes a constant after successive differentiation 2. The dv part can be integrated from standard integrals 3. Normally, we give priority to the following functions as u part by following LIATE order: | Functions | Example | a) | Logarithmic function | | b) | Inverse trigonometric functions | | c) | Algebraic functions | | d) | Trigonometric functions | | e) | Exponential functions | |

6.3 INTEGRATION OF TRIGONOMETRIC EXPRESSIONS

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* *

6.4 THE DEFINITE INTEGRALS * The Area Between A Curve CASE 1: Area of the region above x-axis Let a function f be continuous on an interval [a,b] and let for any xϵ [a,b], f(x) ≥ 0, then is equal to the area of the region Ω bounded above by the graph of the function f(x), below by the x-axis and lines x = a, x = b from the sides. CASE 2: Area of a region below x-axis Let a function f be continuous on an interval [a,b] and let for any xϵ [a,b], f(x) ≤ 0, then is equal to minus the area of the region Ω bounded above by the x-axis , graph of the function f(x) below and lines x = a, x = b from the sides.

CASE 3: Area of a region above and below the x-axis Let a function f be continuous on an interval [a,b] and let function f changes its sign on the interval [a,b], then is equal to the area of the region above [a,b] minus the area below [a,b]

Example

Determine the area of the region bounded by the curve and x-axis for x = -1 and x = 1

* The Area Between Two Curves Example

Example

Example

Example

* The Area Between Two Curves With Respect To Y

Example

* Volume

* Volume of The Solid Bounded by a Curve

Example

* Volume Of A Solid Bounded By Two Curves

6.5 APPLICATION Distance Travelled, Velocity, Acceleration If function x = x(t)describes the position of an object moving along the coordinate line as a function of time t, then its instantaneous velocity and acceleration are given respectively by the formulas When the object moves from left to right, the velocity is positive, when the object stops, velocity is zero and when it moves from right to left, velocity is negative.

Displacement From the graph below, the object moves over time interval [t0,t1]. During this time interval the position of the object changes from x(t0) to x(t1). The displacement over this interval is x(t1)-x(t0). Since v(t) = x’(t), we obtain x(t1)-x(t0) = So we have displacement over where v(t) is the velocity of the moving object. Caution: Displacement over time interval from t0 to t1 is not equal to the distance travelled during this time interval.

Speed, Distance Travelled There is a difference between velocity and speed of a moving object. A velocity describes how fast an object is moving and in which direction. As speed described how fast an object is moving without regard to direction. Mathematically the instantaneous speed s(t) is the absolute value of instantaneous velocity v(t). That is, The displacement of an object over time interval [t0,t1] is, x(t1)-x(t0) = and, the distance travelled during the time interval [t0,t1] is

Acceleration In a motion along straight line, the rate at which the velocity of an object changes with time is called its acceleration. If x(t) is the position of a moving object on the x-axis, then the instantaneous acceleration is defined by The graph below shows the position, velocity and acceleration of the moving object. | | An object is speeding up (accelerating) when its instantaneous speed is increasing and is slowing down (decelerating) when its instantaneous speed is decreasing. Observing the graphs of x(t), v(t), s(t) and a(t), we can see that an object in motion along the straight line is speeding up when its velocity and acceleration have the same sign and slowing down when they have opposite signs | | Example

An object initially placed at x = 1, moves along the x-axis. If the velocity of the object is

Determine: a) Position function x = x(t) b) Acceleration function a = a(t) c) Displacement over the time interval [0.5,4] d) Distance travelled during time interval [0.5,4]
Solution:
a) The position function x = x(t) is an antiderivative of the velocity function. Therefore,

Since the initial position of the object is x(0) = 1, we have c = 1.
Thus,

b) Acceleration is a derivative of velocity. Therefore,

c) The displacement over [1,4] equals

d) The distance travelled during time interval [0.5,4] equals

Factoring the velocity function v(t), we obtain

Within the interval [0.5,4], v(t) is negative on [0.5,0) U (2,3) and positive on (1,2) U (3,4]. Therefore,

Example

A footballer kicks a ball directly upwards with an initial velocity of 24.5m/s at a point is 1 m above the ground.

By placing the coordinate x-axis vertically with origin on the ground, determine: a) The velocity function v = v(t), where t is time in seconds b) The position function x = x(t) c) The height to which the ball will rise d) The time when the ball hits the ground e) The displacement upon time interval from 1 sec to 3 sec f) The distance travelled during time interval [2,4]
Solution:
a) It is a fact from physics that a body moving on a vertical line near the earth’s surface moves with a constant accelerating g = -9.8 m/s2. Air resistance is ignored.
Since velocity is an antiderivative of the acceleration we have v = -9.8t + c. The initial velocity v0 at t = 0 is 24.5. v0 = -9.8t + c = 24.5 Therefore c = 24.5. So the velocity of the function is v = -9.8t + 24.5

b) Similarly the position function x(t) is an antiderivative of the velocity function. That is, x(t) = -4.9t2 + 24.5t +c
The ball was stuck at 1m above the ground, therefore x(0) = c =1
Thus, the position function is x(t) = -4.9t2 +24.5t +1

c) At the highest point, v(t) = 0
Therefore, -9.8t + 24.5 = 0 t = 2.5 sec
The position function at t = 2.5 is = -4.9(2.5)2 + 24.5(2.5) + 1 = 30.625
So, the ball will rise as high as 30.625m

d) The ball is on the ground when x(t) = 0
Solving the equation -4.9t2 + 24.5t + 1 = 0 t1 = -0.04 and t2 = 5.04
Ignoring the first root, we conclude that the ball hit the ground 5.04 sec after the kick.

e) The displacement over the time interval [1,3] equals

f) The displacement traveled during the interval [1,3] equals

Exercise