Integration Techniques: Definite Integrals, Substitution, and Chain Rule, Summaries of Computer Science

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Introduction to

Integration

By Ronald Golooba

Integration is the Reverse of

Differentiation

The process of integration reverses the process of differentiation. In differentiation, if f( x ) =

2 x

2

, then 𝑓

′

( x ) = 4 x. Thus the integral of 4 x is 2 x

2

. We can represent this process pictorially

as follows:

The situation gets a bit more complicated, because there are an infinite number of functions

we can differentiate to give 4 x. Here are some of those functions:

f ( x ) = 2 x

2

• 7 ; g ( x ) = 2 x

2

• 8 ; h ( x ) = 2 x

2

1

2

.

In differentiation, the differential coefficient indicates that a

function of x is being differentiated with respect to x , the dx

indicating that it is “with respect to x ”.

In integration, the variable of integration is shown by adding

d ( the variable ) after the function to be integrated. When we

want to integrate a function, we use a special notation :.

Thus, to integrate 4 x , we will write it as : + C

Note that along with the integral sign (, there is a term of the form dx , which

must always be written, and which indicates the variable involved, in our

example x.

We say that 4 x is integrated with respect x, i.e:

The function being integrated is called the integrand.

Technically, integrals of this type are called indefinite integrals , to

distinguish them from definite integrals.

When you are required to evaluate an indefinite integral, your answer must

always include a constant of integration .; i.e:

= 2 + c; where c

1. Find

Solution

2. Find
3. Find

Adding 1 to the power and then dividing would lead to undefined term

Recall where C is a constant of integration

Integrating with respect to y

Riemann

Sums

• If the width of the rectangles is

not so obvious, the interval [a,b]

is split up into n different

rectangles, and each will have

width The process of using

rectangles to approximate area is

called Riemann sums.

• Approximate the area beneath

the curve on the interval [0,3]

using a left Riemann sum with

four rectangles.

• Solution:
• Width then we use the left the

heights reached by he curve i.e

Therefore area of a rectangle is

wxh

This area is much smaller than the actual area

Area of rectangles =

• We realize that as we make the width of a rectangle smaller the area

left out either for left or right Riemann's Approximation becomes very

small and hence we become more accurate. So we make approach zero

making

• Therefore the area
• The it's a sum over an infinite number of terms.
• Problem: Approximate the area under the curve and above the x-axis

using n rectangles. Let the height of each rectangle be given by the

value of the function at the right side of the rectangle.

13

SUMMAR Y

the curve y  f ( x ),

the lines x = a and x = b

the x -axis and

PROVIDED that

the curve lies on, or above, the x -axis between the values x = a and x = b



The definite integral or gives the area between



b

a

f ( x ) dx



b

a

y dx

y  f ( x ),



b

a

f ( x ) dx



b

a

y dx

14

y x 2 x

2

 

y x 2 x

2

Finding an area





 

0

1

2

a r e a A x 2 x dx

A B



  

1

0

2

a r e a B x 2 x dx

For parts of the curve below the x -axis, the definite

integral is negative, so

y x 2 x

2

 

y x 2 x

2





 

0

1

2

a r e a A x 2 x dx



  

1

0

2

a r e a B x 2 x dx

16

Harder Areas

e.g.1 Find the coordinates of the points of intersection of the curve and line shown. Find the area

enclosed by the curve and line.

2

x  2 x  x

Solution: The points of intersection are given by

0

2

 x  x   x ( x  1 )  0

x  0 o r x  1

2

y  2 x  x

y  x

2

x  2 x  x

0

2

 x  x   x ( x  1 )  0

x  0 o r x  1

2

y  2 x  x

y  x

17

2

y  2 x  x

y  x

x  0  y  0

Substitute in y  x

x  1  y  1

The area required is the area under the curve

between 0 and 1...

... minus the area under the line (a triangle )

1

0

3

2

1

0

2



x

x x dx x

Area of the triangle

Area under the curve

Required area

   

2

y  2 x  x

y  x

x  0  y  0

y  x

x  1  y  1

1

0

3

2

1

0

2



x

x x dx x

2

1

( 1 )( 1 )

2

1

 

   

Fundamental Theorem of Calculus

Definite integrals involving

It is particularly important to remember the modulus sign when evaluating definite integrals

of functions involving.

1

x

Find the area under the curve y = – between x = –3,

x = –1 and the x -axis, writing your answer in the form ln a.

1

x

0

x

y

• 3 – 1

1

y

x



The area is given by.

1

3

1

x







1 1

3

3

1

= ln x

x

 





   

=  ln  1   ln  3

=  ln1+ ln

Remember

that ln 1 = 0

1

x

= ln

units squared

1

x

1

x

1

y

x



1

3

1

x







1 1

3 3

1

= ln x

x

 

 

   

=  ln  1   ln  3

=  ln1+ ln

1

x

= ln