UNIT 4
INTEGRAL CALCULUS
Beta Function:
The Beta function is defined as
…(1)
Putting
i.e. Beta function is symmetric.
Another form of Beta function
Taking
This is the Euler’s integral of the first kind.
Gamma function:
The gamma function is defined as
This is the Euler’s Integral of the second kind.
Reduction formula for n:
Integration by parts
=
Where .
We have
………………
d. 1=0!=1
Example1: Find the value of 1/2 ?
Let
Then
When
And then t=
Squaring both sides we get
where u and v is any variable in place of t.
Converting it in polar coordinates
Since
Relationship between Beta and Gamma functions
By the definition of gamma function we know that
Putting
Then
When then
then
Similarly
So, ()(
Converting into polar coordinates
Let
Also
Where
using definition
Or
Example1: Compute
We have
.
Example2:Express in terms of gamma function
Given I =
Let
Then
Example3:Prove that
We know that
Putting
….(1)
Again putting
Let
When
Or ….(2)
From (1) and (2) we get
The RMS value of a set of values (or a continuoustime waveform) is the square root of the arithmetic mean of the squares of the values, or the square of the function that defines the continuous waveform. In physics, the RMS current value can also be defined as the "value of the direct current that dissipates the same power in a resistor."
In the case of a set of n values , the RMS is
The corresponding formula for a continuous function (or waveform) f(t) defined over the interval is
and the RMS for a function over all time is
The RMS over all time of a periodic function is equal to the RMS of one period of the function. The RMS value of a continuous function or signal can be approximated by taking the RMS of a sample consisting of equally spaced observations. Additionally, the RMS value of various waveforms can also be determined without calculus, as shown by Cartwright.[4]
In the case of the RMS statistic of a random process, the expected value is used instead of the mean.
Example 1:
Solve the following
Solution:
=
=
= (0+1)
= 1+
=
Example 2:
solve the integral
Solution:
Given,
Leibniz Integral Formula
Example 1:
Solve the following by using Leibniz Rule
Differentiate
Solution:
Using,
Example 2:
Solve the following
Solution:
And Hence,
To get rid of the constant C , notice that I(1)=0, so
Therefore,
Tracing of Cartesian curves
The following rules will help in tracing a Cartesian curve.
Rule 1: Symmetry
(a) Symmetry about Xaxis: If the equation of curve containing all even power terms in ‘y’ then the curve is symmetric about Xaxis.
(b)Symmetry about Yaxis: If the equation of curve containing all even power terms in ‘x’ then the curve is symmetric about Yaxis
(C) Symmetry about both X and Y axes: If the equation of curve containing all even power terms in ‘x’ and ‘y’ then the curve is symmetric about both axes. (d) Symmetry in opposite quadrants: If the equation of curve remains unchanged when x and y are replaced by –x and –y respectively then the curve is symmetric in opposite quadrants. (e) Symmetry about the line :If the equation of curve remains unchanged when x is replaced by y and y is replaced by x then the curve is symmetric about the line y=x. (f) Symmetry about the line :If the equation remains unchanged when x is replaced by y and y is replaced by  x then the curve is symmetric about the line y=x. Rule 2: Points of intersection (a) Origin: If the equation of curve does not contain any absolute constant then the curve passes through the origin. (b) Intersection with the coordinate axes: Intersection with Xaxis: put y=0 in the given equation and find the value of x. Intersection with Yaxis: put x = 0 in the given equation and find the value of y. (c) Points on the line of symmetry: If y=x is the line of symmetry then put y=x to find the points on line of symmetry. Rule 3:Tangents (a) Origin:If the curve passes through origin then the equations of the tangent at origin can be obtained by equating the lowest degree terms taken together to zero. (b) Other points:Tofind nature of tangent at any point find at that point. Case 1: If then tangent is parallel to Xaxis. Case 2: If then tangent is parallel to Yaxis. Case 3: If then tangent makes acute angle with Xaxis. Case 4: If then tangent makes obtuse angle with Xaxis. Rule 4:Asymptotes: (a) Parallel to Xaxis: Asymptotes parallel to Xaxis are obtained by equating the coefficient of highest degree term in x to zero. (b) Parallel to Yaxis: Asymptotes parallel to Yaxis are obtained by equating the coefficient of highest degree term in y to zero. (c) Oblique asymptote: Asymptotes which are not parallel to coordinate axes are called as Oblique asymptotes. Method 1: Let y=mx+c be the asymptote. The point of intersection with the curve f(x, y)=0 are given by f(x, mx+c)=0. Equate to zero the coefficients of two successive highest power of ‘x’, giving equations to determine m & c. Method 2:
Rule 5:Special points on the curve:Find out such points on the curve whose presence can be easily detected. Rule 6:Region of absence of the curve: Findthe values of x(or y) where y(or x) becomes imaginary, then the curve does not exists in that region.

Q1. Trace the following curves (i) Solution:                (1) We check the following points for tracing of the above curve. 1. Symmetry:Since the power of y is even. The curve is symmetry about xaxis. No any other symmetry. 2. Origin: Since there is no constant term in the given equation. The curve passes through origin. (i) Tangent at origin is given by
(ii) Nature of origin: Since there is only one common tangent at origin. Origin is cusp. (iii) Intersection with coordinate axes: Put in (1), we have. Put in (1), we have. Hence curve meets coordinate axes only at origin. 3. Asymptotes: (i) Asymptotes parallel to xaxis: Equating the coefficient of highest power of x, we obtain the asymptotes parallel to xaxis. Here highest power of x is and whose coefficient is 1 cannot equate to zero. Hence no asymptotes parallel to xaxis. (ii) Asymptotes parallel to yaxis: Equating the coefficient of highest power of y, we obtain the asymptotes parallel to yaxis. Here highest power of y is and whose coefficient is. Hence the asymptotes parallel to yaxis is
(iii) Oblique asymptotes: Asymptotes not parallel to x and y axes are called oblique asymptotes. Here since parallel asymptotes are present, so no oblique asymptotes. 4. Region: From (1), we have
From the above expression, it is clear that (i) If x is negative, then y will be imaginary. So, there is no part of the curve for which x is negative i.e. left hand side of yaxis. (ii) If, then y will be imaginary. So, there is no part of the curve right hand side of the line. Hence the approximate shape of the curve is as follows: 
TRACING OF POLAR CURVES
The following rules will help in tracing a Polar curve.
Rule 1:Symmetry
(a) Symmetry about pole: If the equation of the curve remains unchanged by replacing then curve is symmetric to the pole.
(b) Symmetry about initial line: If the equation of the curve remains unchanged by replacing, then curve is symmetric about the initial line.
(c)Symmetry about :
1. If the equation of the curve remains unchanged by replacing
and, then curve is symmetric about the line .
2. If the equation of the curve remains unchanged by replacing
then curve is symmetric about the line .
Rule 2:Pole: If for some value of, r becomes zero then the pole will lie on the curve.
Rule 3:Tangents: To find tangents at the pole, put r = 0 in the equation, the values of gives the tangent at the pole.
Rule 4: Angle between radius vector and tangent : Use the formula and find and also the points where Rule 5: Form the table showing values of r for some values of Rule 6: Find the region of absence of the curve.
 
Q1. Trace the following curve: Solution: We check the following points for tracing of the above curve
(i) About the Pole: If we replace by, then the equation of the curve is remains unchanged. The curve is symmetry about pole. (ii) About initial line : If we replace by, then the equation of the curve is remains unchanged. The curve is symmetry about the initial line. (iii) About the line perpendicular to the initial line at pole or about the line: If we replace byand by, then the equation of the curve is remains unchanged. The curve is symmetry about the line perpendicular to the initial line at pole or about the line. 3. Pole: (i) For. Hence the curve passes through the pole. (ii) Tangent at Pole: If we put , then we get the tangent at pole. Putting in (1), we have
4. Tangent:
5. Asymptotes:No asymptotes. 6. Table values:
It is clear that at, and the tangent is perpendicular to the initial line at and. Again at. Hence there is no part of the curve between. Also, the curve is symmetry about pole, initial line and the line perpendicular to initial line. Hence the approximate shape of the curve is as follows: 
TRACING OF ROSE CURVES .
Rule 1: No. of loops:
Rule 2: Symmetry (a) Symmetry about initial line: If the equation of the curve remains unchanged by replacing, then the curve is symmetric about the initial line (b) Symmetry about the line : 1. If the equation of the curve remains unchanged by replacing and, then curve is symmetric about the line . 3. If the equation of the curve remains unchanged by replacing then curve is symmetric about the line . Rule 3:Pole:Find in particular values of, which give r = 0. Rule 4:Tangents: To find tangents at the pole, put r =0 in the equation, the values of gives the tangent at the pole. Rule 5:Angle between radius vector and tangent : Use the formula and find and also the points where Rule 6:Form the table showing values of r for some values of
 
Q1. Trace the following curve:
Solution: We check the following points for tracing of the above curve
(i) About the line perpendicular to initial line i.e. the line: If we replace by and bythen the equation of the curve is remains unchanged. The curve is symmetry about the line. Pole: (i) For . Hence the curve passes through the pole. (ii) Tangent at pole: If we put , then we get the tangent at pole. Putting in (1), we have
4. Asymptotes:No asymptotes. 5. Table values:
It is clear that for, the value of r is zero therefore these are tangents at pole and for the value of r is maximum i.e. ‘a’. Hence, we get four loops at those points. Hence the approximate shape of the curve is as follows.
 
Q. 2 Trace the following curve: Solution: We check the following points for tracing of the above curve
(i) About initial line : If we replace by, then the equation of the curve is remains unchanged. The curve is symmetry about the initial line. 4. Pole: (i) For Hence the curve passes through the pole. (ii) Tangent at pole: If we put , then we get the tangent at pole. Putting in (1), we have
5. Asymptotes:No asymptotes. 6. Table values:
It is clear that for, the value of r is zero therefore these are tangents at pole and for the value of r is maximum i.e. ‘a’. Hence we get three loops at those points. Hence the approximate shape of the curve is as follows.  

Reference Books:
1. A text book of Applied Mathematics Volume I and II by J.N. Wartikar and P.N. Wartikar
2. Higher Engineering Mathematics by Dr. B. S. Grewal
3. Advanced Engineering Mathematics by H. K. Dass
4. Advanced Engineering Mathematics by Erwins Kreyszig