Module 1

Calculus

- 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.

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.

3. 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.

4. 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.

5. then

Proof:

Here f(x) = tan x

By Maclaurin’s expansion,

… (1)

Since

…..

By equation (1)

6. Then

Proof:-

Here f(x) = sin hx.

By Maclaurin’s expansion,

(1)

By equation (1) we get,

7. 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

8. Show that

Solution : =

=

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

=

=

9. Evaluate dx.

Solution : Let dx

X | 0 | |

t | 0 |

Put or ;dx =2t dt .

dt

dt

10. 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