INTEGRATION 2

Mathematics SSS 3 Second Term

WEEK 2

INTEGRATION 2

Performance Objectives

Student should be able to:

1. Solve problems of integration using: integration by part method.
2. Solve problems on integration using: integration by partial fraction method.
3. Solve problems involving definite integral.
4. Apply integration to find the area under the curve or between a curve.

Content

Integration by Part Method

This method has been introduced in week 1 hence we would treat a few more examples to further deepen our understanding of the method.

Example 1: Evaluate the following using the integration by part method

1.            ii.    

Solution

1. 

Let u=ex following the order of priority and since it can be differentiated easily ∴       du=exdx

And  dv=cosxdx      ∴     v=cosxdx      ⟹     v=sinx

By integration by parts

By evaluating  using u=ex, dv=sinxdx we have

                        

x

                                 

 

2. 

Put  and

 and v=cosxdx

By integration by parts

But

Therefore

 

ASSESSMENT

Evaluate

Answer =

 

Integration by Partial Fraction

This method has also been introduced in week 1 hence we would treat few more examples to further deepen our understand of the method.

Example 2: Evaluate the following

1.  

           

2.  

Solution

1. 

                    

If

 ⟹   -3=B       ∴    B=-3

If

 ⟹   -23=-5A-3   ⟹5A=20    ∴    A=4

Hence

Therefore

                             

2. 

                   

If 

If

Hence

                             

Therefore

                 

              

ASSESSMENT

Evaluate

Answer = 3lnx+1+lnx-2+K

 

Definite Integrals

So far all the integrals we have been evaluating are indefinite integrals because of the presence of the arbitrary constants C or K. The integral

Is called the definite integral of the function f(x) with a and b the lower and upper limits of the integral respectively. The integral

Geometrically represents the area bounded by the curve , hence and the x=axis. So in general, to evaluate a definite integral

1. We first find the indefinite integral) omitting the constant
2. We substitute the upper limit  in  to get  
3. We also substitute the lower limit  in to get
4. We then subtract  from ) i.e. to get the final result. Thus

Example 3: Evaluate the following definite integrals.

1.                ii.               

iii.                       (WAEC)                                                                                        

Solution

1.  

 

2. 

                                        

                                   

                                  

 

3. 

               

        

ASSESSMENT

Evaluate the integral

241x2dx

Answer = 14

 

Area Under a Curve or Between Curves

One of the most important applications of integration is in evaluating the area under a curve or between or bounded by two or more curves

Geometrically, the area bounded by the curve , the lines  and the x-axis (i.e. the line ) as shown in the figure above is given by the definite integral

 

If the curve is below the x-axis as shown in the figure below, the area  is given by

 

In general, if  and  are two curves bounded by the lines x=a and x=b as shown in the figure below, then the area enclosed by the curves and the lines is given by the definite integral

Provided  in the interval .

 

Example 4: Find the area under the curves

1. 
    and between the lines  and the x axis
2.  from  to

Solution

Let A be the required area

x

1. 
   If , we have

               4 sq. units

 

2. If , we have

  sq. units

Example 3: Find the area between the curves

1.  and the lines x=-1,  x=1
2.  and the line .

Solution

The area A is given by

Provided  in

1.  and  in ]

3

=3313 sq. units

 

2.  and  intersect when

 or x=2 

Also  in the interval (0, 2)

 sq. units

 

Note: when  and when . Also  and when , hence the curve  and the line  intersect at