Module 1

Calculus

1.1           Rolle’s theorem

i)      f(x) is continuous in the closed [a, b]

Ii)    f(x) is differentiable in (a, b) &

Iii)  f(a) = f(b)

Then there exist at least one value ‘c’ in (a, b) such that f’(c) = 0.

Exercise 1

Verify Rolle’s theorem for the function f(x) = x2 for

Solution:

Here f(x) = x2;

i)      Since f(x) is algebraic polynomial which is continuous in [-1, 1]

Ii)    Consider f(x) = x2

Diff. w.r.t. x we get

f'(x) = 2x

Clearly f’(x) exist in (-1, 1) and does not becomes infinite.

Iii)  Clearly

f(-1) = (-1)2 = 1

f(1) = (1)2 = 1

f(-1) = f(1).

Hence by Rolle’s theorem, there exist such that

f’(c) = 0

i.e. 2c = 0

c = 0

Thus such that

f'(c) = 0

Hence Rolle’s Theorem is verified.

Exercise 2

Verify Rolle’s Theorem for the function f(x) = ex(sin x – cos x) in

Solution:

Here f(x) = ex(sin x – cos x);

i)      Clearly ex is an exponential function continuous for every also sin x and cos x are Trigonometric functions Hence (sin x – cos x) is continuous in and Hence ex(sin x – cos x) is continuous in .

Ii)    Consider

f(x) = ex(sin x – cos x)

Diff. w.r.t. x we get

f’(x) = ex(cos x + sin x) + ex(sin x + cos x)

= ex[2sin x]

Clearly f’(x) is exist for each & f’(x) is not infinite.

Hence f(x) is differentiable in .

Iii)  Consider

Also,

Thus

Hence all the conditions of Rolle’s theorem are satisfied, so there exist such, that

i.e.

i.e. sin c = 0

But

Hence Rolle’s theorem is verified.

Exercise 3

Verify whether Rolle’s theorem is applicable or not for

Solution:

Here f(x) = x2;

i)      Clearly x2 is an algebraic polynomial hence it is continuous in [2, 3]

Ii)    Consider

Clearly f’(x) is exist for each

Iii)  Consider

Thus .

Thus all conditions of Rolle’s theorem are not satisfied Hence Rolle’s theorem is not applicable for f(x) = x2 in [2, 3]

Exercise 4

1. Show that between any two real root of equation , is at least one real root of .
2. Discuss the applicability of Rolle’s theorem for the function

Lagrange’s Mean value Theorem:-

Statement:- If

i)      f(x) is continuous in [a, b]

Ii)    f(x) is differentiable in (a, b) then there exist at least one value such that

Exercise 5

Verify the Lagrange’s mean value theorem for

Solution:

Here

i)      Clearly f(x) = log x is logarithmic function. Hence it is continuous in [1, e]

Ii)    Consider f(x) = log x.

Diff. w.r.t. x we get,

Clearly f’(x) is exist for each value of & is finite.

Hence all conditions of LMVT are satisfied Hence at least

Such that

i.e.

i.e.

i.e.

i.e.

Since e = 2.7183

Clearly c = 1.7183

Hence LMVT is verified.

Exercise 6

Verify mean value theorem for f(x) = tan-1x in [0, 1]

Solution:

Here ;

i)      Clearly is an inverse trigonometric function and hence it is continuous in [0, 1]

Ii)    Consider

Diff. w.r.t. x we get,

Clearly f’(x) is continuous and differentiable in (0, 1) & is finite

Hence all conditions of LMVT are satisfied, Thus there exist

Such that

i.e.

i.e.

i.e.

i.e.

Clearly

Hence LMVT is verified.

Meaning of sign of Derivative:

Let f(x) satisfied LMVT in [a, b]

Let x1 and x2 be any two points laying (a, b) such that x1 < x2

Hence by LMVT, such that

i.e.       … (1)

Cast I:

If   then

i.e.

is constant function

Case II:

If then from equation (1)

i.e.

means x2 - x1 > 0 and

Thus for x2 > x1

Thus f(x) is increasing function is (a, b)

Case III:

If

Then from equation (1)

i.e.  

Since and then hence f(x) is strictly decreasing function.

Exercise 7

Prove that

And hence show that

Solution:

Let ; 

i)      Clearly is an logarithmic function and hence it is continuous also

Ii)    Consider

Diff. w.r.t. x we get,

Clearly f’(x) exist and finite in (a, b) Hence f(x) is continuous and differentiable in (a, b). Hence by LMVT

Such that

i.e.

i.e.

Since

a < c < b

i.e.

i.e.

i.e.

i.e.

Hence the result

Now put a = 5, b = 6 we get

Hence the result

Exercise 8

Prove that , use mean value theorem to prove that,

Hence show that

Solution:

Let f(x) = sin-1x; 

i)      Clearly f(x) is inverse trigonometric function and hence it is continuous in [a, b]

Ii)    Consider f(x) = sin-1x

Diff. w.r.t. x we get,

Clearly f’(x) is finite and exist for . Hence by LMVT, such that

i.e.

Since a < c < b

i.e.

i.e.

i.e.

i.e.

Hence the result

Put we get

i.e.

i.e.

i.e.

i.e.

Hence the result

1.2           Mean Value theorems

Statement:-

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

a)     f(x) and g(x) are continuous in (a, b)

b)    both f(x) and g(x) are derivable in (a, b)

c)    

Then for any value of , at least such that

Exercise

Verify Cauchy mean value theorems for & in

Solution:

Let & ;

i)      Clearly f(x) and g(x) both are trigonometric functions. Hence continuous in

Ii)    Since &

Diff. w.r.t. x we get,

&

Clearly both f’(x) and g’(x) exist & finite in . Hence f(x) and g(x) is derivable in and

Iii)   

Hence by Cauchy mean value theorem, there exist at least such that

i.e.

i.e. 1 = cot c

i.e.

Clearly

Hence Cauchy mean value theorem is verified.

Exercise

Considering the functions ex and e-x, show that c is arithmetic mean of a & b.

Solution:

i)      Clearly f(x) and g(x) are exponential functions Hence they are continuous in [a, b].

Ii)    Consider &

Diff. w.r.t. x we get

and

Clearly f(x) and g(x) are derivable in (a, b)

By Cauchy’s mean value theorem such that

i.e.

i.e.

i.e.

i.e.

i.e.

i.e.

Thus

i.e. c is arithmetic mean of a & b.

Hence the result

Exercise

Show that

Prove that if

and Hence show that

Verify Cauchy’s mean value theorem for the function x2 and x4 in [a, b] where a, b > 0

If for then prove that,

[Hint:, ]

Expansions of functions

In this topic we learn two important series expansions namely

i)      Maclaurin’s series

Ii)    Taylors Series

1.3           Expansion of functions by Mc

Maclaurin’s Series Expansions

Statement:-

Maclaurin’s series of f(x) at x = 0 is given by,

Expansion of some standard functions

i)      f(x) = ex then

Proof:-

Here

By Maclaurin’s series we get,

i.e.

Note that

1. Replace x by –x we get

2.     f(x) = sin x then

Proof:

Let (x) = sin x

Then by Maclaurin’s series,

    … (1)

Since

By equation (i) we get,

3.     Then

Proof:

Let f(x) = cos x

Then by Maclaurin’s series,

   … (1)

Since

From Equation (1)

4.       then

Proof:

Here f(x) = tan x

By Maclaurin’s expansion,

  … (1)

Since

…..

By equation (1)

5.     Then

Proof:-

Here f(x) = sin hx.

By Maclaurin’s expansion,

  (1)

By equation (1) we get,

6.     . Then

Proof:-

Here f(x) = cos hx

By Maclaurin’s expansion

 (1)

By equation (1)

7.     f(x) = tan hx

Proof:

Here f(x) = tan hx

By Maclaurin’s series expansion,

    … (1)

By equation (1)

8.        then

Proof:-

Here f(x) = log (1 + x)

By Maclaurin’s series expansion,

  … (1)

By equation (1)

9.    

In above result we replace x by -x

Then

10. Expansion of tan h-1x

We know that

Thus

11. Expansion of (1 + x)m

Proof:-

Let f(x) = (1 + x)m

By Maclaurin’s series.

  … (1)

By equation (1) we get,

Note that in above expansion if we replace m = -1 then we get,

Now replace x by -x in above we get,

Expand by, Maclaurin’s theorem

Solution:

Here f(x) = log (1 + sin x)

By Maclaurin’s Theorem,

   … (1)

……..

equation (1) becomes,

Expand by Maclaurin’s theorem

Log sec x

Solution:

Let f(x) = log sec x

By Maclaurin’s Expansion’s,

   (1)

By equation (1)

Prove that

Solution:

Here f(x)  = x cosec x

=

Now we know that

Expand upto x6

Solution:

Here

Now we know that

    … (1)

    … (2)

Adding (1) and (2) we get

Show that

Solution:

Here

Thus

1.4           Taylor’s theorem for function of two variables

a)     The expansion of f(x+h) in ascending power of x is

b)    The expansion of f(x+h) in ascending power of h is

c)     The expansion of f(x) in ascending powers of (x-a) is,

Using the above series expansion we get series expansion of f(x+h) or f(x).

Expansion of functions using standard expansions

Expand in power of (x – 3)

Solution:

Let

Here a = 3

Now by Taylor’s series expansion,

 … (1)

equation (1) becomes.

Using Taylors series method expand

in powers of (x + 2)

Solution:

Here

a = -2

By Taylors series,

   … (1)

Since

,  , …..

Thus equation (1) becomes

Expand in ascending powers of x.

Solution:

Here

i.e.

Here h = -2

By Taylors series,

    … (1)

equation (1) becomes,

Thus

Expand in powers of x using Taylor’s theorem,

Solution:

Here

i.e.

Here

h = 2

By Taylors series

  … (1)

By equation (1)

Exercise

a)     Expand in powers of (x – 2)

b)    Expand in powers of (x + 2)

c)     Expand in powers of (x – 1)

d)    Using Taylors series, express in ascending powers of x.

e)     Expand in powers of x, using Taylor’s theorem.

1.5           Partial Differentiation, Maxima & Minima (two and three variables)

Partial Differentiation of composite function

a)    Let and , then z becomes a function of , In this case z is called composite function of .

i.e.

b)   Let possess continuous partial derivatives and let possess continuous partial derivatives, then z is called composite function of x and y.

i.e.

&

Continuing in this way, …..

Ex. If Then prove that

Ex. If then prove that

Where is function of x, y, z.

Ex. If where ,

then show that,

i)     

Ii)   

Notations of partial derivatives of variable to be treated as a constant

Let

and

i.e.

Then means partial derivative of u w.r.t. x treating y const.

To find from given reactions we first express x in terms of u & v.

i.e. & then diff. x w.r.t. u treating v constant.

To find express v as a function of y and u i.e. then diff. v w.r.t. y treating u as a const.

Ex. If , then find the value of

.

Ex. If , then prove that

Ex.1: Evaluate   0∞  x3/2  e -x dx

Solution:               0∞  x3/2  e -x dx          =   0∞  x 5/2-1  e -x 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

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  y 2  (1 – y ) -1/2dy

=  (8/2) . B(3 , 1/2 )

= 642 /15 

BETA FUNCTION MORE PROBLEMS

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 , ; , o 1 0 Also   Example(5):   Show that Solution : = ( 0<p<1) (by above result)   Exercise : - Q. Show that 1.   2.

1.6           Method of Lagranges Multipliers

Let be a function of x, y, z which to be discussed for stationary value.

Let be a relation in x, y, z

for stationary values we have,

i.e.    … (1)

Also from we have

    … (2)

Let ‘’ be undetermined multiplier then multiplying equation (2) by and adding in equation (1) we get,

     … (3)

     … (4)

      … (5)

Solving equation (3), (4) (5) & we get values of x, y, z and .

1. Decampere a positive number ‘a’ in to three parts, so their product is maximum

Solution:

Let x, y, z be the three parts of ‘a’ then we get.

    … (1)

Here we have to maximize the product

i.e.

By Lagrange’s undetermined multiplier, we get,

       … (2)

       … (3)

        … (4)

i.e.

        … (2)’

        … (3)’

        … (4)

And

From (1)

Thus .

Hence their maximum product is  .

2.     Find the point on plane nearest to the point (1, 1, 1) using Lagrange’s method of multipliers.

Solution:

Let be the point on sphere which is nearest to the point . Then shortest distance.

Let

Under the condition    … (1)

By method of Lagrange’s undetermined multipliers we have

       … (2)

       … (3)

i.e. &

       … (4)

From (2) we get

From (3) we get

From (4) we get

Equation (1) becomes

i.e.

y = 2

If where x + y + z = 1.

Prove that the stationary value of u is given by,

References

1. G.B. Thomas and R.L. Finney, Calculus and Analytic geometry, 9th Edition,Pearson, Reprint, 2002.

2. Erwin kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley & Sons, 2006.

3. Veerarajan T., Engineering Mathematics for first year, Tata McGraw-Hill, New Delhi, 2008.

4. Ramana B.V., Higher Engineering Mathematics, Tata McGraw Hill New Delhi, 11thReprint, 2010.

5. D. Poole, Linear Algebra: A Modern Introduction, 2nd Edition, Brooks/Cole, 2005.

6. N.P. Bali and Manish Goyal, A text book of Engineering Mathematics, Laxmi Publications, Reprint, 2008.

7. B.S. Grewal, Higher Engineering Mathematics, Khanna Publishers, 36th Edition, 2010.