[Maths Class Notes] on Definite Integral Pdf for Exam

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Introduction

Differentiation and integration are the two important concepts of calculus. Calculus is a branch of mathematics that deals with the study of problems involving a continuous change in the values of quantities. Differentiation refers to simplifying a complex function into simpler functions. Integration generally refers to summing up the smaller function to form a bigger unit. Integration and differentiation are entirely contradictory to each other and hence they are inverse Mathematical operations. There are two types of integrals. They are definite integrals and indefinite integrals.

Indefinite and Definite Integrals:

Indefinite integrals are those integrals that do not have any limit of integration. It has an arbitrary constant. Definite integrals are those integrals that have an upper and lower limit. Definite integral has two different values for the upper limit and lowers limit when they are evaluated. The final value of a definite integral is the value of integral to the upper limit minus the value of the definite integral for the lower limit.

[int_a^b {fleft( x right).dx} = Fleft( b right) – Fleft( a right)]

In the above equation,

F (b) and F (a) are the integral functions for the upper and lower limit respectively.

f (x) is the integrand

dx is the integrating agent

‘b’ is the upper limit of the definite integral

‘a’ is the lower limit of the definite integral

Properties of Definite Integrals:

Property 1:

Definite integrals between the same limits of the same function with different variables are equal.

[int_a^b {fleft( p right),dp} = int_b^a {fleft( q right)dp} ]

Property 2:

The value of a definite integral is equal to its negative when the upper and lower limits of the integral function are interchanged.

[int_a^b {fleft( p right),dp} = – int_b^a {fleft( p right)dp} ]

Property 3:

The definite integral of a function is zero when the upper and lower limits are the same.

[int_a^a {fleft( p right).dp = 0} ]

Property 4:

A definite integral can be written as the sum of two definite integrals. However, the following conditions must be considered.

The lower limit of the first addend should be equal to the lower limit of the original definite integral.
The upper limit of the second addend should be equal to the upper limit of the original definite integral.
The upper limit of the first addend should be equal to the lower limit of the second addend integral.

[int_a^b {fleft( b right).dp = int_a^c {fleft( p right).dp + int_c^b {fleft( p right).dp} } } ]

Other Properties of Definite Integrals:

Property 5:

[int_a^b {fleft( p right).dp = int_a^b {fleft( {a + b – p} right).dp} } ]

Property 6:

[int_0^a {fleft( p right).dp = int_a^b {fleft( {a – p} right).dp} } ]

Property 7:

[int_0^{2a} {fleft( p right).dp} = int_0^a {fleft( p right).dp + int_0^a {fleft( {2a – p} right).dp{text{ }}if{text{ }}fleft( {2a – p} right) = fleft( p right)} } ]

Important Integration Formula:

A few important integrals of various MathematicalMathematical functions (integrands) defined in terms of a variable ‘x’ are listed below.

[int 1dx = x+c]
[int adx = ax+c]
[int x^{n}dx=frac{x^{n+1}}{n+1}+C;nneq-1]
[int text{sinx dx}=text{-cox x+C}]
[int text{cosx dx}= text{sin x+C}]
[int sec^{2} x dx= text{tan x+C}]
[int csc^{2} x dx= text{-cot x+C}]
[int secxleft ( tanx right ) dx= text{sec x+C}]
[int cscxleft ( cotx right ) x dx= text{-csc x+C}]
[int frac{1}{x}dx=lnmid x mid +C]
[int e^{x}dx=e^{x}+C]
[int a^{x}dx=frac{a^{x}}{lna}+C;a> 0,aneq1]
[int frac{1}{sqrt{1-x^{2}}}dx=sin^{-1}x+C]
[int frac{1}{sqrt{1+x^{2}}}dx=tan^{-1}x+C]
[int frac{1}{mid xmid sqrt{1-x^{2}}}dx=sec^{-1}x+C]
[int sin^n(x)dx=frac{-1}{n}sin^{n-1}(x)cos(x)+frac{n-1}{n}int sin^{n-2}(x)dx]
[int cos^n(x)dx=frac{1}{n}cos^{n-1}(x)sin(x)+frac{n-1}{n}int cos^{n-2}(x)dx]
[int tan^n(x)dx=frac{1}{n}tan^{n-1}(x)+int tan^{n-2}(x)dx]
[int sec^n(x)dx=frac{1}{n}sec^{n-2}(x)tan(x)+frac{n-2}{n-1}int sec^{n-2}(x
)dx]
[int csc^n(x)dx=frac{-1}{n}csc^{n-2}(x)cot(x)+frac{n-2}{n-1}int csc^{n-2}(x)dx]

Definite Integral as a Limit of Sum:

Let us consider a continuous function ‘f’ defined by the variable ‘x’ in a closed duration of m,n. If f (x) is greater than zero, then f (x) can be graphically represented as (graphical representation will )

The area enclosed by the curve y = f (x) within x = a and x = b gives the definite integral as the limit of sum. The total area enclosed by the curve is represented in the graph as ABCD. The entire region of ABCD is divided into equal intervals of ‘n’ divisions. The limiting values of smaller rectangles are obtained to give the area of these rectangles. So, definite integral as a limit of sum can be written as

[int_{a}^{b}f(x)=displaystylelimlimits_{nrightarrow infty}h[f(a)+f(a+h)+….+f(a+(n-1)h]]

[Rightarrowint_{a}^{b}f(x)=(b-a)limlimits_{nrightarrowinfty}frac{1}{n}[f(a)+f(a+h)+….+f(a+(n-1)h]]

The area enclosed by a curve is also the limiting value of the area lying between the rectangles below or above the curve.

In the above equation, when

[n{text{ }} to {text{ }}infty ,{text{ }}left( {b{text{ }} – {text{ }}a} right){text{ }}/{text{ }}n{text{ }} to {text{ }}0]

This fact is named as the definite integral as the limit of a sum.

Applications of Definite Integrals:

Definite integrals are used in various Mathematical computations including

Determination of area between two linear, quadratic or cubic curves
Find the volumes
Estimating the length of a plane curve
Finding the surface area of revolution

Apart from these computations, definite integrals also find their applications in various other fields such as Physics, Engineering and Statistics.

Definite Integral Example Problems:

Evaluate [_1{smallint ^6}left( {5{z^2} – {text{ }}7z{text{ }} + {text{ }}2} right){text{ }}dx;]

Solution:

The integration of [5{z^2} – 7z + 2 = frac{{5{z^3}}}{3} – frac{{7{z^2}}}{2} + 2z]

Applying upper limit to the above integral value i.e. [z = 6frac{{5{{left( 6 right)}^3}}}{3} – frac{{7{{left( 6 right)}^2}}}{2} + 2left( 6 right)]

= 360 – 126 + 12

= 246

Applying the value of lower limit i.e. z = 1, [frac{{5{{left( 1 right)}^3}}}{3} – frac{{7{{left( 1 right)}^2}}}{2} + 2left( 1 right)]

= 1.67 – 3.5 + 2

= 7.17

Therefore, [_1{smallint ^6}left( {5{z^2} – 7z + 2} right)dx; = 246 – 7.17 = 238.83]

Find the value of [_0{smallint ^pi }left( {Cos{text{ }}A{text{ }} + {text{ }}Sin{text{ }}A} right){text{ }}dA]

Solution:

The definite integral of Cos A + Sin A from 0 to π is calculated as:

∫ (Cos A + Sin A). dA [ = smallint CosA.{text{ }}dA + smallint SinA.{text{ }}dA]

= [Sin A]0π + [- Cos A]0π

= Sin π – Sin 0 – Cos π + Cos 0

= 0 – 0 – (-1) + 1

= 2

Fun Facts:

Definite integrals are used to compute the physical quantities which change over some time and are a sum of infinitesimal data like speed, displacement, area etc.
Definite integrals always give the signed area.

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