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.
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
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 + 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
Ex.27 Prove that, Hence or otherwise evaluate
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
Ex.28 Prove that
(From Walli's formula)
Ex.29 If un = then show that u1, u2, u3 ,...... constitute an arithmetic progression. Hence or otherwise find the value of un.
un + 1 – 2un + un – 1 = (un + 1 – un) – (un – un – 1)
∴ un – 1 + un + 1 = 2un i.e., un – 1, un, un + 1 form an A.P.
⇒ u1, u2, u3,.........constitute an A.P.
Integrating by parts taking unity as the second function, we have
Ex.31 Show that Hence or otherwise evaluate
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.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 ± xn ⇒ f(1) = 1 ± 1 = 2 (given)
∴ f(x) = 1 + xn and f '(x) = nxn – 1 ⇒ f '(1) = n = 2 ∴ f(x) = 1 + x2
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) = 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
Integrating by parts taking x2 as 1 st function, we we get =
Adding (1) and (2) we get
Ex.36 Prove that
Intergrating by parts taking x as a first function, we have
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 cos2x + 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 .
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.40 Assume then prove that
Ex.41 Use induction to prove that ,
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