Definite Integral As Limit Of A Sum And Estimate Of Definite Integrals | Mathematics (Maths) for JEE Main & Advanced PDF Download

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.

Estimate Of Definite Integration & General Inequality

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 + x² 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

Solved Examples

Ex.1 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 A₁ – A₂, where A₁ and A₂ are shown in Figure

Ex.2 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.3 Prove that  

Sol.

L.H.S. =  

 (From Walli's formula)

Ex.4 If u_n =     then show that u₁, u₂, u₃ ,...... constitute an arithmetic progression. Hence or otherwise find the value of 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.

⇒ u₁, u₂, u₃,.........constitute an A.P.
Ex.5 Evaluate  

Sol.

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

Ex.6 Show that    Hence or otherwise evaluate 

Sol.

adding (1) and (2) then 

Put a tan x = t ⇒ a sec²x 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.7 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 ± xⁿ ⇒ f(1) = 1 ± 1 = 2 (given)

∴ f(x) = 1 + xⁿ  and f '(x) = nx^{n – 1} ⇒ f '(1) = n = 2 ∴ f(x) = 1 + x²

∴  

Ex.8 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.9 Evaluate  , (where [ * ] is the greatest integer function)

Sol. Let I =

Let f(x) = x² + 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 x² - 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.10 If  

Sol.

Integrating by parts taking x² as 1 st function, we we get  =  

 [By Prop.] 

Adding (1) and (2) we get  

Ex.11 Prove that  

Sol.

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

Ex.12 Evaluate 

Sol.

Let g(t) = |t – 1| – |t| + |t + 1| =  

=  

Ex.13 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 cos²x + 2 cos 3x cos x + ....... + 2 cos (2k – 1 ) x cosx ] dx

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

=  
 

Ex.14 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/x³ But given f(x) is a periodic function Hence f(x) = 1

Ex.15 Assume   then prove that

Sol.

Ex.16 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