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**Definite Integral As Limit Of A Sum**

**Remark :** The symbol was introduced by Leibnitz and is called integral sign. It is an elongated S and was chosen because an integral is a limit of sums. In the notation is called the integrand and a and b are called the limits of integration; a is the lower limit and b is the upper limit. The symbol dx has no official meaning by itself; is all one symbol. The procedure of calculating an integral is called integration.

**Ex.26 Evaluate ** ** using limit of sum.**

**Sol.**

This integral can't be interpreated as an area because f takes on both positive and negative values. But is can be interpreated as the difference of areas A_{1} – A_{2}, where A_{1} and A_{2 }are shown in Figure

**Estimate Of Definite Integration & General Ineθuality**

**STATEMENT : If f is continuous on the interval [a, b], there is atleast one number c between a and b such that **

**Proof : **Suppose M and m are the largest and smallest values of f, respectively, on [a, b]. This means m ≤ f(x) ≤ M when a ≤ x ≤ b

Because f is continuous on the closed interval [a, b] and because the number I = lies between m and M, the intermediate value theorem syas there exists a number c between a and b for which f(c) = I ; that is,

The mean value theorem for integrals does not specify how to determine c. It simply guarantees the existence of atleast one number c in the interval.

Since f(x) = 1 + x^{2} is continuous on the interval [–1, 2], the Mean Value Theorem for Integrals says there is a number c in [–1, 2] such that

In this particular case we can find c explicitly. From previous Example we know that fave = 2, so the value of c satisfies f(c) = = 2

Thus, in this case there happen to be two numbers c = ± 1 in the interval [–1, 2] that work in the mean value theorem for Integrals.

**Walli's Formula & Reduction Formula**

**Ex.27 Prove that, ** **Hence or otherwise evaluate **

**Sol.**

Consider sin nθ + sin (n - 2) θ = 2 sin (n - 1) θ cos θ ⇒ sin nθ sec θ = 2 sin (n-1) θ - sin (n - 2) θ sec θ

Hence

,

**Ex.28 Prove that **

**Sol**.

L.H.S. =

(From Walli's formula)

**Ex.29 If u _{n} = **

**Sol.**

u_{n + 1} – 2u_{n} + u_{n – 1} = (u_{n + 1} – u_{n}) – (u_{n} – u_{n – 1})

∴ u_{n – 1} + u_{n + 1 }= 2u_{n }i.e., u_{n – 1}, u_{n}, u_{n + 1} form an A.P.

⇒ u_{1}, u_{2}, u_{3},.........constitute an A.P.**Ex.30 Evaluate **

**Sol**.

Integrating by parts taking unity as the second function, we have

**Ex.31 Show that ** **Hence or otherwise evaluate **

**Sol**.

adding (1) and (2) then

Put a tan x = t ⇒ a sec^{2}x dx = dt when x = 0 ⇒ t = 0 ; x = p/2 ⇒ t = ∝

Differentiating both side w.r.t. 'a', we get

again differentiating both sides w.r.t. 'a' we get

Put a = √5 on both sides, we get

**Ex.32 Let f be an injective functions such that f(x) f(y) + 2 = f(x) + f(y) + f(xy) for all non negative real x and y with f(0) = 1 and f '(1) = 2 find f(x) and show that ****f(x) dx – x (f(x) + 2) is a constant.**

**Sol.** We have f(x) f(y) + 2 = f(x) + f(y) + f(xy) ......(1)

Putting x = 1 and y = 1 then f(1) f(1) + 2 = 3f(1)

we get f(1) = 1, 2 & f(1) 1 ( f(0) = 1 & function is injective) then f(1) = 2

Replacing y by 1/x in (1) then f(x) f(1/x) + 2 = f(x) + f(1/x) + f(1) ⇒ f(x) f(1/x) = f(x) + f(1/x) [ f(1) = 2)

Hence f(x) is of the type f(x) = 1 ± x^{n} ⇒ f(1) = 1 ± 1 = 2 (given)

∴ f(x) = 1 + x^{n} and f '(x) = nx^{n – 1} ⇒ f '(1) = n = 2 ∴ f(x) = 1 + x^{2}

∴

**Ex.33 Evaluate **

**Sol.**

**Let ** I = Make |sin x| – |cos x| = 0 ∴ |tan x| = 1

∴ tan x = ± 1 and both these values lie in the interval [0, π].

We find for 0 < x < π/4, |sin x| – |cos x| < 0

|sin x| – |cos x| > 0

**Ex.34 Evaluate** **, (where [ * ] is the greatest integer function)**

**Sol**. Let I =

Let f(x) = x^{2} + x + 1 ⇒ f '(x) = 2x - 1 for x > 1/2, f '(x) > 0 and x < 12, f '(x) , 0

Values of f(x) at x = 1/2 and 2 are 3/4 and 3 integers between them an 1, 2 then x^{2} - x + 1 = 1, 2

we get x = 1, and values of f(x) at x = 0 and 1/2 are 1 and 3/4 no integer between them

**Alternative Method :** It is clear from the figure

**Ex.35 If **

**Sol.**

Integrating by parts taking x^{2} as 1 st function, we we get =

[By Prop.]

Adding (1) and (2) we get

**Ex.36 Prove that **

**Sol**.

Intergrating by parts taking x as a first function, we have

**Ex.37 Evaluate**

**Sol.**

Let g(t) = |t – 1| – |t| + |t + 1| =

=

**Ex.38 For all positive integer k, prove that **

**Hence prove that **

**Sol.** We have 2 sinx [cos x + cos 3x + ...+ cos (2k–1)x ]

= 2 sinx cosx + 2 sinx cos 3x + ....+ 2 sinx cos(2k – 1)x

= sin 2x + sin 4x – sin 2x + sin 6x – sin 4x + ....... + sin 2kx – sin (2k – 2) x

= sin 2kx

⇒ 2[cos x + cos 3x + ........ + cos (2k – 1) x ] = sin2kx/sinx

[2 cos^{2}x + 2 cos 3x cos x + ....... + 2 cos (2k – 1 ) x cosx ] dx

[cosx + cos 3x + ....... + cos (2k – 1 ) x ]cosx dx

=

**Ex.39 Let f(x) is periodic function such that ** **Find the function f(x) if (1) = 1 .**

**Sol**.

from (2) and (3)

Differentiating bot sides w.r.t.x, we get

then f(x) = 1/x^{3} But given f(x) is a periodic function Hence f(x) = 1

**Ex.40 Assume** ** then prove that**

**Sol.**

**Ex.41 Use induction to prove that , **

**Sol.**

Now sinkx = sin[(k + 1) x – x ] = sin(k + 1) x cos – cos(k + 1)x sin x

Hence sin (k + 1 ) x cosx = sinkx + cos(k + 1) x sin x

Subistuting P(k + 1) =

Now I. B. P. to get the result

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