g(a) = f(a) – Aa

g(b) = f(b) – Ab

so,

g(a) = g(b),

f(a) – Aa = f(b) – Ab ,

which gives,

A =     …………………..(2)

g’(x) = f’(x) – A

here we know that, x = c,

g’(c) = f’(c) – A

as g’(c) = 0, then

f’(c) – A =0

so that,,

f’(c) = A,

from equation (2) , we get

f’(c) =

hence proved.

f(x) = (x-1)(x-2)(x-3)

f(x) =  x-6x²+11x-6

now at x = 0, we get

f(0) = -6  and

at x = 4, we get.

f(4) = 6

diff. the function w.r.t.x , we get

f’(x) = 3x²-6x+11

suppose x = c, we get

f’(c) = 3c²-6c+11

by Lagrange’s mean value theorem,

f’(c) =     =     =   = 3

now we get,

3c²-6c+11 = 3

3c²-6c+8 = 0

On solving the quadratic equation, we get

C = 2

Here we see that the value of c lies between 0 and 4

Therefore the given function is verified.

So that this is the continuous function In [1,e]

Now,

f’(x) = 1/x

which is exists for all x in (1,e)

so that f(x) is differentiable in (1,e).

by Lagrange’s mean value theorem, we get

f’(c) = , let x = c,

then ,

f’(c) =

we get,

c = e-1

e-1 will always lies between 1 and e .

hence the function is verified by Lagrange’s mean value theorem.

h(x) = f(x) – A.g(x) …………………….(1)

so that h(a) = h(b) and A is a constant to be determined.

Now,

h(a) = f(a) – Ag(a)

h(b) = f(b) – A.g(b)

so that,

f(a) – Ag(a) = f(b) – A.g(b) , which gives

A =     …………………………….(2)

Differentiate eq.(1) w.r.t. x , we get

h’(x) = f’(x) – A.g’(x)

At x = c

h’(c) = f’(c) – A.g’(c)

0 = f’(c) – A.g’(c)

A =

So that , we get

where a<c<b

Hence the Cauchy’s mean value theorem is proved.

Sol.  It is given tha,

f(x) = sin x and g(x) = cos x

Now,

  f’(x) = cos x and g’(x) = - sin x

We know that both the functions are continuous in [ 0 , π/2] and differentiable in     ( 0 , π/2 )

Also,g’(x) = -sin x  ≠ 0 for all x ϵ( 0 , π/2 )

By Cauchy’s mean value theorem, we get

for some c: 0< c <

That means

which gives,

Cot c = 1

C =

Now we see that lies between 0 and

Taylor’s Theorem-

If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-

+ …….+ + ……..

Which is called Taylor’s theorem.

If we put x = a, we get-

+ …….+ + ……..     (1)

Maclaurin’s Theorem-

If we put a = 0 and h = x then equation(1) becomes-

+ …….

Which is called Maclaurin’s theorem.

Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)  

We get-

+ ……. 

+ …….     (1)

Sol. Let f(x) = sin x

     Then,

                    =

          By using Taylor’s theorem-

+ …….     (1)

Here f(x) = sin x and a = π/2

f’(x) = cosx  , f’’(x) = - sin x    ,     f’’’(x) = - cos x  and so on.

Putting x = π/2 , we get

f(x) = sin x = = 1

f’(x) = cos x = = 0

f’’(x) = -sin x = = -1

f’’’(x) = -cos x = = 0

from equation (1) put a = and substitute these values, we get-

+ …….    

         = ………………………..

Sol.

Here we have-

Now-

Put these values in Taylor’s theorem-

We get-

Sol.

Let-

Put these values in Maclaurin’s series we get-

Sol.

Let

Put these values in Maclaurin’s series-

Or

1. Taylor’s Theorem-

+ …….+ + ……..

2.     Maclaurin’s Theorem-

+ ……

Then-

Then-

Is an expression of the form  , in that case we can say that f(x)/g(x) is an indeterminate for of the type   at x = a.

Some other indeterminate forms are

Sol. Let f(x) = and g(x) = .

Here we see that this is the indeterminate form of 0/0 at x = 0.

Now by using L’Hospital rule, we get-

                                                        =

=

                                                       = = 1

Let

   …

By L – Hospital rule

   …

   …

    … (1)

    …

But

From equation (1)

Note- Suppose we get an indeterminate form even after finding first derivative, then in that case , we use the other form of L’Hospital’s rule.

If we have f(x) and g(x) are two functions such that 

.

If exist or (∞ , -∞), then

Example-3: Evaluate

Sol. Let f(x) = , then

And

= 0

= 0

But if we use L’Hospital rule again, then we get-

Example-4: Evaluate

Sol. We can see that this is an indeterminate form of type 0/0.

Apply L’Hospital’s rule, we get

But this is again an indeterminate form, so that we will again apply L’Hospital’s rule-

We get

=

Let f and g are two differentiable functions on an open interval containing x = a, except possibly at x = a and that

If has a finite limit, or if it is , then

Theorem- If we have f(x) and g(x) are two functions such that  .

If exist or (∞ , -∞), then

Example-5: Find   , n>0.

Sol. Let f(x) = log x and g(x) =

These two functions satisfied the theorem that we have discussed above-

So that,

Example-6: Evaluate

Sol. Apply L’Hospital rule as we can see that this is the form of

=

Note- In some cases like above example, we can not apply L’Hospital’s rule.

Example-7: Evaluate

Sol. Here we find that-

So that this limit is the form of 0.

Now,

Change to obtain the limit-

Now this is the form of 0/0,

Apply L’Hospital’s rule-

Example: Evaluate

Solution:

Let

   … 0o form

Taking log on both sides we get,

  …

   …

By L – Hospital Rule

i.e.

Example: Evaluate

Solution:

Let

   …

Taking log on both sides,

  …

By L – Hospital rule,

i.e.

Example: Evaluate

Solution:

Let

    …

Taking log on both sides, we get

    …

By L – Hospital Rule,

Note- If we have f(x) and g(x) are two functions such that 

.

If exist or (∞ , -∞), then

2.If we have f(x) and g(x) are two functions such that  .

If exist or (∞ , -∞), then

Let

Then first derivative-

So that-

Now-

So that-

So that-

Therefore f(x) has a maximum at .

Sol.

Suppose r is the radius OA of the base and be the semi-vertical angle of the given cone.

Now inscribe a cylinder in it with base radius OL = x

Then the height of the cylinder LP = LA cot

The CSA of cylinder-

Now

Therefore S is maximum when x = r/2.

Where 

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:x→cos x  is a decreasing function on [a , b],

Therefore by the definition of the definite integrals-

Then

Now,

Here

Thus

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] asfollows-

(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-

Key takeaways-

1. 
2. 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

3.      Integral of a constant function-

4.      Constant multiple property-

5.      Interval union property-

6.      If a < c < b, then

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 .

dt

dt

Ex. 5: Evaluate   dx.

Solution:   Let   dx.

                                        Put or  ;             4x dx = dt

dx

2.      

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 :

2.    

3.    

4.    

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

Relation between Beta and Gamma functions :

Let the area between the two curve is A, then

                                                     A =

Let’s do some examples,

Example-1: find the area under the curves where f(x) =  x+4 and g(x) =  3 – x/2 over the interval [1,4]

Sol. Here limits are given a = 1 , b = 4,

We know that, area under the curve,

A =

= 

=

= 

= (16 – 7/4) =  57/4

So that the area of the region is 57/4 unit square.

Example-2: find the area under the curves where y =  x and y =  x + 1,         x = 2 and y-axis.

Sol. When we draw the graph, curve does not meet, but depend on two vertical lines,

Here boundary is [0 , 2]

Area under the curve,

A =

Example-2: Determine the area under the curves y = 2x² + 10 and y = 4x + 16.

Sol. Here we will find out the intersection, which are the boundaries of the curve,

2x² + 10 =  4x + 16

2x² - 4x – 6 = 0

2 (x + 1) ( x – 3 ) = 0

We get, x = -1  and x = 3,

We know that, Area under the curve,

A =

(2) Average function value-

We can use definite integrals to find out the average value of a function.

The average value of a function over the interval [a , b]  is given by the following expression ,

f( avg.) =

Example-1: find the average value of the function f(x) = x³ over the interval[0,1].

Sol.  We know that

f( avg.) =

                                    =      =  = 1 / 4

Example-2: find the average value of f(x) = cosx  over the interval [0 , 𝛑/2]

Sol. We know that,

f( avg.) =

Put the values ,  we get

(3) Work-

This one is the easiest use of definite integral

we know that work done, when a constant force F is applied to move the object over a distance d.

W = Fd

Now suppose that the force at any point x is F(x) , the the work done by to move the object from x = a to x = b is,

W =

Example-1: Find the work done on a spring when we compress it from its natural length of 1m to the 0.75m , if the spring constant k = 16 N / m.

Sol. The force is given by,

F =16x

Here we start to compress it at x = 0 and finish at x= 0.25 from its natural length.

So we take lower limit and upper limit 0 and 0.25 respectively.

We know that,

W =

=

= 0.5 N.m

Example-2: A force of 1200 N compresses a spring from its natural length of 18 cm to a length of 16 cm. How much work is done in compressing it from 16 cm to 14 cm?

Sol.  Here,

F = kx

So that,

1200 = 2k

K = 600 N/cm

In that case,

F = 600x

 We know that,

W =

W = , which gives

W  = 3600 N.cm

Area under and between the curves-

Total shaded area will be as follows of the given figure( by using definite integrals)-

Total shaded area =

Example-1: Determine the area enclosed by the curves-

Sol. We know that the curves are equal at the points of interaction, thus equating the values of y of each curve-

Which gives-

By factorization,

Which means,

x = 2 and x = -3

By determining the intersection points the range the values of x has been found-

And

We get the following figure by using above two tables-

Area of shaded region =

=

= ( 12 – 2 –  8/3 ) – (-18 – 9/2 + 9)

=

= 125/6 square unit

Example-2: Determine the area bounded by three straight lines y = 4 – x, y = 3x and 3y = x

Sol. We get the following figure by using the equations of three straight lines-

                                         y = 4 – x, y = 3x and 3y = x

Area of shaded region-

Example-3: Find the area enclosed by the two functions-

and g(x) = 6 – x

Sol. We get the following figure by using these two equations

To find the intersection points of two functions f(x) and g(x)-

f(x) = g(x)

On factorizing, we get-

x = 6, -2

Now

Then, area under the curve-

                                        A =

Therefore the area under the curve is  64/3  square unit.

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-

                                                  =

1. Area between the curves-

A =

2.     f( avg.) =

3.     Work- W =

Sequence –

A function f : N , where S is a non-empty set , is called sequence , for each nϵN.

The sequence is written as f(1) , f(2) , f(3) , f(4)……….f(n).

Any sequence f(n) can be denoted as <f(n)> or {f(n)} or (f(n)).

Suppose

f(n) =

Then it can be written as -    and can be denoted as <>or  {} or ()

is the n’th term of the sequence.

Example:  suppose we have a sequence – 1 , 4 , 9 , 16 ,……….. and its n’th term is

This sequence can be written as  -<>

1          

Example: its n’th term will be .and can be written as <>

Types of sequences –

1. Finite sequence- A sequence which has finite number of terms is called finite sequence.

2. Infinite sequence-  A sequence which is not finite , called infinite sequence.

Limit of a Sequence- A sequence <> is said to tend to limit “l” , when given any positive number ‘ϵ’ , however small , we can always find a integer ‘m’ such that | – l| <ϵ , for every for all, n≥m , and we can define this as follows,

Example: If    , then the limit of will be,

  = =  = ½

Hence the limit of the sequence is 1/2  .

Some important limits to remember-

Convergent sequence- A sequence Sn is said to be convergent when it tends to a finite limit. That means the limit of a sequence Sn will be always finite in case of convergent sequence.

Divergent sequence- when a sequence tends to ±∞ then it is called divergent sequence.

Oscillatory sequence-  when a sequence neither converges nor diverges then it is called oscillatory sequence.

Note- a sequence which neither converges nor diverges , is called oscillatory sequence.

A sequence converges to zero Is called null.

Example-1:  consider a sequence 2, 3/2, 4/3 , 5/4, …….. here Sn =  1 + 1/n

Sol. As we can see that the sequence Sn is convergent and has limit 1.

According to def.

Example-2: consider a sequence Sn=  n² + (-1)ⁿ.

Sol. Here we can see that, the sequence Sn is divergent  as it has infinite limit.

Example: suppose , here the sequence is said to be oscillate. Because it is between -2 and 2.

Series

Infinite series- If  is a sequence , then is called the infinite series.

It is denoted by .

Examples of infinite series-

Covergent series -  suppose n→∞ , Sn→ a finite limit  ‘s’  , then the seiresSn is said to be convergent .

We can denote it as,

Divergent series-  when Sn tends to infinity then the series is said to be divergent.

Oscillatory series-  whenSn does not tends to a unique limit (finite or infinite) , then it is called Oscillatory series.

Properties of infinite series –

1. the convergence and divergence of an infinite series is unchanged addition od deletion of a finite number of term from it.

2. if positive terms of convergent series change their sign , then the series will be convergent.

3. let   converges to s , let k be a non-zero fixed number  then converges to ks.

4. let converges to ‘l’ and converges to ‘m’.

Example-1: check whether the series is convergent or divergent. Find its value in case of convergent.

Sol.  As we know that,

Sn = 

Therefore,

Sn = 

Now find out the limit of the sequence,

  = ∞

Here the value of the limit is infinity, so that the series is divergent as sequence diverges.

Example-2: check whether the series  is convergent or divergent. Find its value in case of convergent.

Sol. The general formula for this series is given by,

Sn =  = )

           We get,

) = 3/2

Hence the series is convergent and its values is 3/2.

Example-3: check whether the series  is convergent or divergent.         

Sol. The general formula can be written as,

We get on applying limits,

= 3/4

This is the convergent series and its value is 3 / 4

Example-4: check whether the following series is convergent or divergent. If convergent , find its value.

Sol. n’th term of the series will be,

]  = ½

Key takeaways-

1. Finite sequence- A sequence which has finite number of terms is called finite sequence.
2. Limit of a Sequence- A sequence <> is said to tend to limit “l” , when given any positive number ‘ϵ’ , however small , we can always find a integer ‘m’ such that | – l| <ϵ , for every for all, n≥m , and we can define this as follows,

3.    

4.    

5.    

6.     a sequence which neither converges nor diverges , is called oscillatory sequence.

7.     A sequence converges to zero Is called null.

8.     If  is a sequence , then is called the infinite series.

It is denoted by

The series will be

1. Convergent if the value of p is :   p>1

2. Divergent if the value of p is:    p≤1

Proof;

Let p>1

Similarly,

And so on.

Adding we get,

The right hand side of the above inequality is an infinite geometric series , the sum of the geometric series is finite hence the given series is convergent

Example-1: Test the convergence of the following series:

Sol.

Here we take,

Which not zero and finite ,

So by comparison test , and both converges or diverges, but by p-series test

Is convergent. so that  is convergent.

Example-2: Test the convergence of the given series:

Sol.  Here,

Take,

is divergent by p-series test. (p = 0 <1)

Example-3: Test the convergence of the series:

Sol.

By p-series test,

Is convergent ( p = 3>1).

Comparison test-

Statement-

Suppose we have two series of positive terms and then,

, where k is a finite number , then both series converges or diverges together.

Proof-

We know that by the definition of limits, there exist a positive number epsilon(ε)

Which is very small. Such that

According to definition( comparison test)

||<ε for n>m  , that means

k-ε<    for n>m

ignoring the first m terms of the series,

we get

k-ε<    for n>m  for all n  ………………..(1)

there will be two cases-

case-1: is convergent , then

() = r (say)  , where r is finite number

From (1),

()<() =

Therefore  is also convergent.

Case-2: : is divergent, then

()→∞   …………………………..(2)

From eq. (1)

Then

()<() 

From(2)

()→∞  

Hence, is also divergent.

Example: Test the convergence of the following series.

Sol. We have  

First we will find  and the

And

Here, we can see that, the limit is finite and not zero,

Therefore, and converges or diverges together.

Since is of the form  where p = 2>1

So that , we can say that,

is convergent , so that will also be convergent

Example: Test the convergence of the following series-

Sol. Here we have the series,

Now,

Now comapare

We can see that the limit is finite and not zero.

Here  and converges or diverges together since ,

is the form of here p = 1,

So that,

is divergent then is also divergent.

Example: Show that the following series is convergent.

Sol. 

Suppose,

Which is finite and not zero.

By comparison test and converge or diverge together.

But,

Is convergent. So that is also convergent.

Example: Test the series:

Sol. The series is,

Now,

Take,

Which is finite and not zero.

Which is finite and not zero.

By comparison test and converge or diverge together.

But,

Is divergent. ( p = ½)

 So that is divergent.

D-Alembert’s ratio test

Statement- suppose is a series of positive terms  such that then,

1. if k<1 , the series will be convergent.

2. if k>1 , then the series will be divergent.

Proof: Proof:

Case-1:

We know that from the definition of limits ,it follows,

But,

Is the finite quantity. So is convergent.

Case-2:

There could be some finite terms in starting which will never satisfy the condition,

In this case we can find a number ‘m’,

Ignoring the first ‘m’ terms, if we write the series as

We have , in this case

So that is divergent.

Example-1: Test the convergence of the series whose n’th term is given below-

n’th term =

sol. We have    and

byD’Alembert ratio test,

So that by D’Alembert ratio test , the series will be convergent.

Example-2: Test for the convergence of the n’th term of the series given below-

Sol. We have ,

Now , by D’Almbert ratio test converges if and diverges if 

At x = 1 , this test fails.

Now , when x = 1

The limit is finite and not zero.

Then by comparison test, converges or diverges together.

Since is the form of , in which

Hence diverges then will also diverge.

Therefore in the given series converges if x<1 and diverges if x≥1

Power series

A series of the form

Where all the ‘a’ are independent of x, is known as power series in x

Interval for convergence-

In this power series-

D’Almbert’s ratio test-

If , then by ratio test, the power series is convergent, when is less than 1.

Or in other way if |x|<1/ then the series converges and diverges for other values.

Thus the interval of convergence of power series is-

Example-1: If the series converges, then find the value of x.

Sol. Here

Then,

By D’Almbert’s ratio test the series is convergent for |x|<1 and divergent if |x|>1.

So at x = 1

The series becomes-

At x = -1

This is an alternately convergent series.

This is also convergent series, p = 2

Here, the interval of convergence is

Example-2: If the series converges, then find the value of x.

Sol. Here

Then,

By D’Almbert’s ratio test the series is convergent for ||<1 or |1-x|>1

Or

At x = 0, the series becomes-  which is divergent harmonic series.

At x = 2, the series becomes- 

It is an alternate series which is convergent by Leibnitz rule.

So that the series  .

Key takeaways-

1. The series will be

1. Convergent if the value of p is :   p>1

2. Divergent if the value of p is:    p≤1

2. Comparison test-

Suppose we have two series of positive terms and then,

, where k is a finite number , then both series converges or diverges together.

3. D-Alembert’s ratio test

Statement- suppose is a series of positive terms  such that then,

1. if k<1 , the series will be convergent.

2. if k>1 , then the series will be divergent.

The power series for

The value of can be defined in terms of the following power series:

If we obtain an actual value of by adding all the terms of the series, then the series is said to be convergent.

The more terms we take in the series, the closer will be the value of to its actual value.

Example: Find the value of 5, correct to five decimal places by using the power series for .

Sol. As we know that the exponential series is-

Here we get-

Now

Example: Expand as far as the term in .

Sol. We know that the power series for is-

Here we have to find- 

So that-

On solving, we get-