Module 1
Calculus
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
 Show that between any two real root of equation , is at least one real root of .
 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) = tan1x 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) = sin1x;
i) Clearly f(x) is inverse trigonometric function and hence it is continuous in [a, b]
Ii) Consider f(x) = sin1x
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
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 ex, 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
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
 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 h1x
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
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 (xa) 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.
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 5/21 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 = 02 x2 / (2 – x ) . Dx
Solution:
Letting x = 2y, we get
I = (8/2) 01 y 2 (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 x4 (a2 – x2 ) . Dx Solution: Letting x2 = a2 y , we get I = (a6 / 2) 01 y 3/2 (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 y1/3 (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 ,;,
Also
Example(5): Show that Solution : = (0<p<1)(by above result)Exercise :  Q. Show that 1. 2. 
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 .
 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 McGrawHill, 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.