Unit 1 | unit 1 integral calculus i

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M 2

Unit - 1

Integral calculus-I

1.1 Beta and Gamma functions

The beta and gamma functions are defined as-

And

These integrals are also known as first and second Eulerian integrals.

Note- Beta function is symmetrical with respect to m and n.

Some important results-

Ex.1: Evaluate dx

Solution dx = dx

= γ(5/2)

= γ(3/2+ 1)

= 3/2 γ(3/2 )

= 3/2. ½ γ(½ )

= 3/2. ½ π

= ¾ π

Ex. 2: Find γ(-½)

Solution: (-½) + 1= ½

γ(-1/2) = γ(-½ + 1) / (-½)

= - 2 γ(1/2 )

= - 2 π

Ex. 3. Show that

Solution:

= ) .......................

Ex. 4: Evaluate dx.

Solution: Let dx

Put or ;dx =2t dt

Ex. 5: Evaluate dx.

Solution: Let dx.

Put or ; 4x dx = dt

Evaluation of beta function 𝛃 (m, n)-

Here we have-

Again integrate by parts, we get-

Repeating the process above, integrating by parts we get-

Evaluation of gamma function-

Integrating by parts, we take as first function-

We get-

Replace n by n+1,

Relation between beta and gamma functions

Relation between beta and gamma function-

We know that-

………… (1)

…………………..(2)

Multiply equation (1) by , we get-

Integrate both sides with respect to x within limits x = 0 to x = , we get-

But

By putting λ = 1 + y and n = m + n

We get by using this result in (2)-

So that-

Definition: Beta function

Properties of Beta function:

Example(1): Evaluate I =

Solution:

= 2 π/3

Example(2): Evaluate: I = 02 x2 / (2 – x ) . Dx

Solution:

Letting x = 2y, we get

I = (8/2) 01 y2 (1 – y ) -1/2dy

= (8/2). B (3, 1/2 )

= 642 /15

Evaluation of integrals

by using beta and gamma functions-

Evaluation of the first interal-

Evaluation of the second integral-

Example: Prove that-

Where a>0.

Sol.

We know that-

Put m – 1 = 1 or m = 2

We get-

Hence proved

Relation between Beta and Gamma functions:

Example(1): Evaluate: I = 0a x4  (a2 – x2 ) . Dx

Solution: Letting x2 = a2 y , we get

I = (a6 / 2) 01 y 3/2 (1 – y )1/2dy

= (a6 / 2) . B(5/2 , 3/2 )

= a6 /3 2

Example(2): Evaluate: I = 02 x (8 – x3 ) . Dx

Solution: Let x3 = 8y

I = (8/3) 01 y-1/3 (1 – y ) 1/3 . Dy

= (8/3) B(2/3 , 4/3 )

= 16 π / ( 9 3 )

Example(3): Prove that

Solution: Let

Put or ,

Example(4): Evaluate

Solution: Let

Put or ,,

When,;,

Example(5): Show that

Solution:

(0<p<1)

(by above result)

Elliptic integral-

The integral is called elliptic function.

This is derived from the determination of the perimeter of the ellipse.

To evaluate this integral, we first expand the integrand as power series and then integrate term by term.

Elliptic integral of first kind

Elliptic integral of second kind

Here k is modulus and is amplitude.

Jacobi’s form of elliptic integrals

Example: Express in terms of elliptic integral.

Sol.

Here we have-

Put

Error function-

Error function of x is written as erf(x) and defined as below-

The complementary function of error function of x is written as and defined as below-

Note-

Example: Find erf(0).

Sol.

We know that-

Then

Hence-

Example: Prove that erf (-x) = - erf (x)

Sol.

We know that-

Put t = -u

Key takeaways-

The integral is called elliptic function.
Elliptic integral of first kind

3. Elliptic integral of second kind

4. Jacobi’s form of elliptic integrals

10.

11.

1.2 Differentiation of definite integral

Differentiation under integral sign: Leibnitz’s rule:

We know from the fundamental theorem on integral calculus that if f (x) is a continuous function and

Then

i.e., the derivative of a definite integral w.r.t. The upper

Limit is equal to the integrand in which the variable of integration t is replaced by the upper limit ‘x’

Differentiating integrals depend on a parameter:

Consider the definite integral:

In which the integrand f (x, α) is dependent on a parameter α. The value of the definite integral (2) changes as the parameter α varies. It is difficult to integrate (2) in many cases.

Now to differentiate the integral (2) w.r.t. The parameter α, without having to first carry out integration and then differentiation, we use the Leibnitz’s rule.

Leibnitz’s formula (rule):

If are continuous functions when c and a

General Leibnitz’s rule:

If and are continuous and if the limits of integration a and b are functions of then

Example: Apply Leibnitz’s rule

Sol: here

Now applying Leibnitz’s rule

1.3 Mean Value and Root Mean Square Values

We know that the voltages and current vary with time and if we are interested to find the mean value of current or voltage over some particular time interval. We define the mean value of a function in terms of an integral.

An associated quantity is the root-mean-square.

Average value of a function:

Suppose a time-varying function f(t) is defined on the interval . The area, A, under the graph of f(t) is given by the integral A

Illustration of this is given in the figure below:

The area under the curve from t = a to t = b

2. The area under the curve and the area of the rectangle are equal

The value of m is the mean value of the function across the interval

Note-

The mean value of a function f(t) in the interval

The mean value depends upon the interval chosen. If the values of a or b are changed, then the

Mean value of the function across the interval from a to b will in general change as well.

Example: Find the mean value of f(t) = over the interval

Sol:

As we know that

Here

a = 1 and b = 3 and f(t) =

Root-mean-square-value of a function

If f(t) is defined on the interval , the mean-square value is given by the expression:

This is simply the mean value of over the given interval.

The related quantity: the root-mean-square (r.m.s.) value is given by the following formula.

The r.m.s. Value depends upon the interval chosen. If the values of a or b are changed, then the r.m.s. Value of the function across the interval from a to b will in general change as well. Note that when finding an r.m.s. Value the function must be squared before it is integrated.

Example: Find the r.m.s. Value of f(t) = across the interval from t = 1 to t = 3.