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Unit - 2


Reduction formula for integrals


Reduction formula for:

 

What is reduction formula?

A formula of the form,

= g(x) +

Where k<n, is called a reduction formula.

 

(a). Reduction formula for

Let,                 I =     =sinxdx , if n>1

Taking  ¹x as first function and sinx as second function, integrate by parts,

I = -¹x cosx – (n -1)cosx(-cosx)dx

= -¹x cosx  + (n - 1)cos²xdx

= -¹x cosx  + (n - 1) ( 1 - sin²x) dx

= -¹x cosx  + (n - 1) - dx

=  -¹x cosx  + (n - 1) [ I ² - I]

By solving above

I  = + Iⁿ‾²

This is the reduction formula for

 

Example: Evaluate by using reduction formula.

Solution:  we know that,

+ Iⁿ‾²,     here   Iⁿ‾² =

Now using this formula,

=

=

=

=

 

(b). Reduction formula for

Similarly we can get the formula for

= + Iⁿ‾²

We can use the following formulas to get the result easily,

= . .    ,       only if n is odd and n ≥3

= . . . ,       only if n is even and n ≥2

We apply same formula for cos x.

 

Example2: evaluate .

Solution: we will apply reduction formula to evaluate,

So, here n = 5, which is odd,   we will get ,

 

(C) Reduction formula for:

Where

 

Example1: Evaluate by using reduction formula.

Solution:  here, m = 4, n = 6,

Now we will apply above reduction formula,

 

Example2: Evaluate by using reduction formula.

Solution:     Here we can see that, m = 3 and n = 5,

Now we will apply reduction formula,

 

(D) Reduction formula for and

Let

Thus, the required reduction formula is-

And for , the reduction formula will be-

 

Example: Evaluate by using reduction formula.

Sol.

Put n = 6, 4, and 2 successively in the reduction formula for , we obtain-

Thus

 

Example: If then prove that

Sol.

The reduction formula for is-

On changing n to n+1, we get-

Or

Hence proved

 

Key takeaways-

  1. I  = + Iⁿ‾²
  2. = + Iⁿ‾²

 


Improper integrals

(1) Let f is function defined on [a, ∞) and it is integrable on [a, t] for all t >a, then

If exists, then we define the improper integral of f over [a, ∞) as follows-

(2) Let f is function defined on (-∞,b] and it is integrable on [t, b] for all t >b, then

If exists, then we define the improper integral of f over (-∞, b] as follows-

(3) Let f is function defined on (-∞, ∞] and it is integrable on [a, b] for every closed and bounded interval [a, b] which is the subset of R., then

If and exist for some c belongs to R, then we define the improper integral of f over (-∞,∞) as follows-

= +

(4) Let f is function defined on (a,∞) and exists for all t>a, then

If exists, then we define the improper integral of f over (a ∞) as follows-

(5)  Let f is function defined on (-∞, b) and exists for all t<b, then

If exists, then we define the improper integral of f over (-∞, b) as follows-

 

Improper integrals over finite intervals-

(1) Let f is function defined on (a, b] and exists for all t (a,b), then

If exists, then we define the improper integral of f over (a, b] as follows-

(2) Let f is function defined on [a, b) and exists for all t (a,b), then

If exists, then we define the improper integral of f over [a, b) as follows-

(3) Let f is function defined on [a, c) and (c, b]. If and exist

Then we define the improper integral of f over [a, b] as follows-

When we apply limits in indefinite integrals are called definite integrals.

If an expression is written as , here ‘b’ is called upper limit and ‘a’ is called lower limit.

If f is an increasing or decreasing function on interval [a, b], then

Where 

 

Properties-

1. The definite integral applies only if a<b, but it would be appropriate to include the case a = b and a>b as well, in that case-

If a = b, then

And if a>b, then

 

2. Integral of a constant function-

 

3. Constant multiple property-

 

4. Interval union property-

If a < c < b, then

 

5. Inequality-

If c and d are constants such that for all x in [a, b], then

c (b – a)

Note- if a function f: [a,b] R is continuous, then the function ‘f’ is always Integrable.

 

Example-1: Evaluate.

Sol. Here we notice that f:xcos x is a decreasing function on [a, b],

Therefore, by the definition of the definite integrals-

Then

Now,

Here

Thus

 

Example-2: Evaluate

Sol. Here is an increasing function on [1, 2]

So that,

  …. (1)

We know that-

And
 

Then equation (1) becomes-

Note- we can find the definite integral directly as-

 

Example-3:  Evaluate-

Sol.         

 

Improper integrals

(1) Let f is function defined on [a,∞) and it is integrable on [a, t] for all t >a, then

If exists, then we define the improper integral of f over [a,∞) as follows-

 

(2) Let f is function defined on (-∞,b] and it is integrable on [t, b] for all t >b, then

If exists, then we define the improper integral of f over (-∞, b] as follows-

 

(3)Let f is function defined on (-∞, ∞] and it is integrable on [a, b] for every closed and bounded interval [a, b] which is the subset of R., then

If and exist for some c belongs to R, then we define the improper integral of f over (-∞,∞) as follows-

= +

 

(4) Let f is function defined on (a,∞) and exists for all t>a, then

If exists, then we define the improper integral of f over (a, ∞) as follows-

 

(5) Let f is function defined on (-∞, b) and exists for all t<b, then

If exists, then we define the improper integral of f over (-∞, b) as follows-

 

Improper integrals over finite intervals-

(1) Let f is function defined on (a, b] and exists for all t (a,b), then

If exists, then we define the improper integral of f over (a, b] as follows-

 

(2) Let f is function defined on [a, b) and exists for all t (a,b), then

If exists, then we define the improper integral of f over [a, b) as follows-

 

(3) Let f is function defined on [a, c) and (c, b]. If and exist

Then we define the improper integral of f over [a, b] as follows-

 


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 

X

0

t

0

Put or ;dx =2t dt

dt

dt

 

Ex. 5: Evaluate   dx.

Solution:   Let   dx.

x

0

t

0

Put or  ;             4x dx = dt

dx

Evaluation of beta function 𝛃 (m, n)-

Here we have-

Or

Again, integrate by parts, we get-

Repeating the process above, integrating by parts we get-

Or

 

Evaluation of gamma function-

Integrating by parts, we take as first function-

We get-

Replace n by n+1,

 

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 = 02 x / (2 – x). Dx

Solution:

Letting   x = 2y, we get

I   = (8/2) 01 y2 (1 – y)-1/2dy

= (8/2). B (3, 1/2)= 642 /15 

 

Beta Function More Problems

Relation between Beta and Gamma functions:

 

Example(1):   Evaluate:  I = 0a x (a2 – x2). Dx

Solution:  Letting   x2 = a2y, we get

I   = (a6 / 2) 01 y3/2 (1 – y )1/2dy

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

= a6 /3 2

 

Example (2):   Evaluate:  I = 02 x (8 – x3). Dx

Solution: Let   x3 = 8y

I = (8/3) 01 y-1/3 (1 – y) 1/3. Dy

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

= 16 π / (93)  

 

Example(3):   Prove that

Solution: Let

Put  or 

 

Example(4):  Evaluate

Solution: Let

Put   or  ,,

When,;,

Also

 

Example(5):   Show that

Solution:

=

(0<p<1)

(By above result)

 


RULE I: If

where
is parameter and
are constants then

=

 

Q1) Evaluate:

(i)               

.

 

Sol)

Let 

  - - - - - - - - - - - - - - - -  (1)

Differentiating by DUIS rule

 

 

 

 

 

 

 

 

     

Separating vectors and integrating, we have

 

 

 

 

 
   - - - - - - - - - - - - - - - -  (2)

Putting

in equation (1) and (2), we get

and

 

 

(ii) 

Solution:

 Let 

   - - - - - - - - - - - - - - - -  (1)

Differentiating by DUIS rule

 

 

Put

 

 

Integrating, we get  

  

 

 
   

 

 

 

 
   - - - - - - - - - - - - - - - -  (2)

Putting

in equation (1) and (2), we get

    

 

 

DIFFERENTIATION UNDER INTEGRAL SIGN RULE – II

(Leibnitz’s Rule)

RULE II: If

where
are function of parameter
then

=

 

Q)Prove that:

(i)                If

, then show that
.

 

Sol)  

We have given that

 

  - - - - - - - - - - - - - - - -  (1)

Then we have to prove that

.

Differentiating by DUIS rule (Leibnitz’s Rule), we have

 

 

 

 

Again differentiating by DUIS rule (Leibnitz’s Rule), we have

 

 

 

 

       

 

This is the desired result.

 

(ii)If

, then prove that
.

Solution:

We have given that

 

  - - - - - - - - - - - - - - - -  (1)

Then we have to prove that

.

Differentiating by DUIS rule (Leibnitz’s Rule), we have

 

 

 

 

Again differentiating by DUIS rule (Leibnitz’s Rule), we have

 

 

 

 

 

 

Again differentiating by DUIS rule (Leibnitz’s Rule), we have

 

 

 

 

 

 

 

 

ERROR FUNCTION

Introduction to error function and its properties.

Error Function:

  1. Error function is denoted by and defined as

 

2.     Complementary Error function is denoted by and defined as

 

3.     Properties of Error function

(i)               

(ii)            

(iii)         

(iv)           i.e. error function is odd function.

(v)             

 

Q) Prove that

(i)                is an odd function and hence deduce .

Sol)

We have to prove thatis an odd function i.e.

We know that

  - - - - - - - - - - - - - - - -  (1)

Replace by, we have

Put , then and

 

 

 

Hence the error function is odd function.

Now we have to prove that

Consider

   

   

 

This is the desired result.

 

(ii)              

.

 

Solution:

We have to prove that

Consider

   

   

 

This is the desired result.

 

(iii) 

Solution:

We have to prove that

Consider

   

   

Put , then and

 

    

 

 

This is the desired result.

 

(iv)  

Solution:

We have to prove that

We know that

 

                         

 

   

 

    

 

 


Area in Cartesian coordinates-

Example-1: Find the area enclosed by two curves using double integration.

y = 2 – x and y² = 2 (2 – x)

Sol.  Let,

y =  2 – x  ………………..(1)

And                y² = 2 (2 – x) ………………..(2)

On solving eq. (1) and (2)

We get the intersection points  (2,0) and (0,2) ,

We know that,

Area =

 

Here we will find the area as below,

Area =

Which gives,

= ( - 4 + 4 /2 ) + 8 / 3  = 2 / 3.

 

Example-2: Find the area between the parabola y ² = 4ax and another parabola x² = 4ay.

Sol. Let,

y² = 4ax ………………..(1)

And

x² = 4ay…………………..(2)

Then if we solve these equations , we get the values of points where these two curves intersect

x varies from y²/4a to and y varies from o to 4a,

Now using the conceot of double integral,

Area = 

 

Area in polar coordinates-

Example-3:  Find the area lying inside the cardioid r = a(1+cosθ) and outside the circle r = a, by using double integration.

Sol. We have,

r = a(1+cosθ)   …………………….(1)

And

r = a ……………………………….(2)

On solving these equations by eliminating r , we get

a(1+cosθ)    =   a

(1+cosθ) = 1

Cosθ = 0

Here a θ varies from – π/2  to    π/2 

Limit of r will be a and 1+cosθ)   

Which is the required area.

 

Example-4: Find the are lying inside a cardioid r = 1 + cosθ and ouside the parabola r(1 + cosθ) = 1.

Sol. Let,

r = 1 + cosθ   ……………………..(1)

r(1 + cosθ) = 1……………………..(2)

Solving these equestions , we get

(1 + cosθ )(1 + cosθ ) = 1

(1 + cosθ )² = 1

1 + cosθ    = 1

Cosθ = 0  

θ = ±π / 2  

So that, limits of r are,

1 + cosθ    and 1 / 1 + cosθ  

The area can be founded as below,

 

Double integrals as volumes

Suppose we have a curve  y = f(x) is revolved about an axis , then a solid is generated , now we need to find out the volume of the solid generated ,

The formula for volume of the solid generated about x-axis,

 

Example-1: Calculate the volume generated by the revolution of a cardioid,

r = a ( 1 – cosθ)  about its axis

Sol. Here,              r = a ( 1 – cosθ

Volume  =

=

 

Which is the volume of generated by cardioid.

 

Example-2: Find the volume generated by revolving a circle x ² + y² = 4 about the line x= 3.

Sol. We know that,

Volume =

Here , PQ = 3 – x,

=

 

The volume is 24π².

 

Example-3: Calculate the volume under the surface z = 3 + x² - 2y over the region R defined as 0≤ x ≤ 1 and  -x ≤ y ≤ x

Sol. The is a double integral of  z = 3 + x² - 2y over the region R .

Volume will be,

 

Areas and volumes of revolutions

The volume of revolution (V) is obtained by rotating area A through one revolution about the x-axis is given by-

Suppose the curve x = f(y) is rotated about y-axis between the limits y = c and y = d, then the volume generated V, is given by-

 

Example-1:  Find the area enclosed by the curves and if the area is rotated about the x-axis then determine the volume of the solid of revolution.

Sol. We know that, at the point of intersection the coordinates of the curve are equal. So that first we will find the point of intersection-

We get,

x = 0 and x = 2

The curve of the given equations will look like as follows-

Then,

The area of the shaded region will be-

A =

So that the area will be 8/3 square unit.

The volume will be

= (volume produced by revolving – (volume produced by revolving

=

 

Method of cylindrical shells-

Let f(x) be a continuous and positive function. Define R as the region bounded above by the graph f(x), below by the x-axis, on the left by the line x = a and on the right x = b, then the volume of the solid of revolution formed by revolving R around the y-axis is given by

 

Example-2: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 1/x over the interval [1 , 3].

Sol. The graph of the function f(x) = 1/x will look like-

The volume of the solid of revolution generated by revolving R(violet region) about the y-axis over the interval [1 , 3]

Then the volume of the solid will be-

 

Example-3: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 2x - x² over the interval [0 , 2].

Sol. The graph of the function f(x) = 2x - x² will be-

The volume of the solid is given by-

=

 

Area of surfaces of revolution

Suppose we have a curve y = f(x) and if we revolve this curve around x – axis then the area formed is known as surface area of revolution.

Let f(x) be a non-negative smooth function over the interval[ [ a , b] , then the surface area of the surface of revolution formed by revolving the graph of f(x) around x-axis is given by,

S =

 

Example-1: Find the surface of the surface generated by revolving the graph f(x) = over the interval [1,4] around x-axis.

Sol.  The graph of the function will look like

f(x) = then f’(x) = 1/2 (

(f’(x))² = 1/4x

Then,

S =

S =

S = ,

Let u =   , then du =  dx, when  x = 1 , u = 5/4 and when x = 4 , u = 17/4,

This will give,

=

On solving the integral we get,

= 30.84 (approx.)

 

Example-1: find the surface of the surface generated by revolving the graph g(y) = over the interval [0,2] around y-axis.

Sol.  

Hint -  use formula

S =

Ans. = 24.118

 

CENTER OF MASS AND GRAVITY-

Centre of mass-

We have-

Mass = volume

 

Example: Find the mass of a plate formed by the coordinate planes and the plane-

The variable density

 

Sol.

Here we the mass will be-

The limits of y are from y = 0 to y = b(1 – x/a) and limits of x are from 0 to a.

 

Therefore the required mass will be-

Which is the required answer.

 

Centre of gravity-

The centre of gravity of a lamina is the point where it balances perfectly which means the lamina’s centre of mass.

If and are the coordinates of the centroid C of area A, then-

And

 

Example: Determine the coordinates of the centroid of the area lying between the curve y = 5x - x² and the x-axis.

Sol.

Here y = 5x - x²when y = 0, x = 0 or x = 5

Therefore the curve cuts the x-axis at 0 and 5 as in the figure-

 

Now

= 2.5

Therefore the centroid of the area lies at (2.5, 2.5)

 

Key takeaways-

  1. The formula for volume of the solid generated about x-axis,

2.     Centre of mass-

We have-

Mass = volume

 

References:

1. G.B. Thomas and R.L. Finney, “Calculus and Analytic Geometry”, Pearson, 2002.

2. T. Veerarajan, “Engineering Mathematics”, McGraw-Hill, New Delhi, 2008.

3. B. V. Ramana, “Higher Engineering Mathematics”, McGraw Hill, New Delhi, 2010.

4. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

5. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.

6. HK Dass

 


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