Definite Integration and Its Properties JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : Definite Integration and Its Properties JEE Notes | EduRev

The document Definite Integration and Its Properties JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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The Area Problem

Use rectangles to estimate the area under the parabola y = x2 from, 0 to 1 Wr first notice that the area of S must be somewhere between 0 and 1 because S is contained in a square with side length 1, but we can certainly do better than that. Suppose we divide S into four strips S1, S2, S3, and S4 by drawing the vertical lines x = 1/4, x = 1/2, x = 3/4,as in Figure (a) 

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

We can approximate each strip by a rectangle whose base is the same as the strip and whose height is the same as the right edge of the strip [see Figure (b)]. In other words, the heights of these rectangle are the values of the function f(x) = x2 at the right end points of the subintervals  Definite Integration and Its Properties JEE Notes | EduRev  Definite Integration and Its Properties JEE Notes | EduRev

Each rectangle has width 1/4th and the heights are  Definite Integration and Its Properties JEE Notes | EduRev  and 12. If we let R4 be the sum of the areas of these approximating rectangles, we get

Definite Integration and Its Properties JEE Notes | EduRev

From the Figure (b) we see that the area A of S is less than R4, so A < 0.46875 Instead of using the rectangles in Figure (b) we could use the smaller rectangles in Figure (c) whose heights are the values of f at the left endpoints of the subintervals. (The leftmost rectangle has collapsed because its heights is 0). The sum of the areas of these approximating rectangles is

Definite Integration and Its Properties JEE Notes | EduRev

We see that the area of S is larger than L4, so we have lower and upper estimates for A , 0.21875 < A < 0.46875 We can repeat this procedure with a larger number of strips. Figure (d), (e) shows what happens when we divide the region S into eight strips of equal width. By computing the sum of the areas of the smaller rectangles (L8) and the sum of the areas of the larger rectangles (R8), we obtain better lower and upper estimates for A :  0.2734375 < A < 0.3984375

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

So one possible answer the question is to say that the true area of S lies somewhere between 0.2734375 and 0.3984375. We coluld obtain better estimates by increasing the number of strips. 

B. Properties Of Definite Integral

P-1 : CHANGE OF VARIABLE : The definite integral Definite Integration and Its Properties JEE Notes | EduRev  is a number, it does not depend on x. In fact, we could use any letter in place of x without changing the value of the integral ;


Definite Integration and Its Properties JEE Notes | EduRev =  Definite Integration and Its Properties JEE Notes | EduRev

P-2 : CHANGE OF LIMIT : When we defined the definite integral Definite Integration and Its Properties JEE Notes | EduRev f(x)dx, we implicity assumed that a < b. But the definition as a limit of sum makes sense even if a > b. Notice that if we reverse a and b, then Δx changes from (b – a)/n to (a – b)/n.

Therefore  Definite Integration and Its Properties JEE Notes | EduRev


P-3 : Additivity With Respect To The Interval Of Integration :

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

This is not easy to prove in general, but for the casewhere f(x) ≥ 0 and a < c < b Property 7 can be seen from the geometric interpretation in Figure : The area under y = f(x) from a to c plus the area from c to b is equal to the total  area from a to b.
Ex.1 If it is known that Definite Integration and Its Properties JEE Notes | EduRev


Sol. We have

Definite Integration and Its Properties JEE Notes | EduRev

= 17 -12 =5
Ex.2 Find  Definite Integration and Its Properties JEE Notes | EduRev


Sol. The function f defined by f(x) = 3 – 2x + x2 is continuous and has antiderivative g defined by
Definite Integration and Its Properties JEE Notes | EduRev  Therefore, by the fundamental theorem of the calculus,
Definite Integration and Its Properties JEE Notes | EduRev

Ex.3 Evaluate  Definite Integration and Its Properties JEE Notes | EduRev

Sol. We can rewirte the integrand as follows  Definite Integration and Its Properties JEE Notes | EduRev

From this, you can rewrite the integral in two parts.
Definite Integration and Its Properties JEE Notes | EduRev

Ex.4 Evaluate Definite Integration and Its Properties JEE Notes | EduRevdx, (where [ * ] denotes greatest integer function)

Sol.Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Ex.5 Find the range of the function f(x) = Definite Integration and Its Properties JEE Notes | EduRev


Sol. Given that

Definite Integration and Its Properties JEE Notes | EduRev 

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev


Definite Integration and Its Properties JEE Notes | EduRev

Proof : Definite Integration and Its Properties JEE Notes | EduRev

In the first integral on the far right side we make the substitution u = –x. Then du = –dx and when x = –a, u = a. 

Therefore   Definite Integration and Its Properties JEE Notes | EduRev   Definite Integration and Its Properties JEE Notes | EduRev

(a) If f is even, then  Definite Integration and Its Properties JEE Notes | EduRev

(b) If f is odd, then Definite Integration and Its Properties JEE Notes | EduRev

Theorem is illustrated by Figure(a, b) For the case where f is positive and even, part(a) says that the area under y= f(x) from –a to a is twice the area from 0 to a because of symmetry. Thus, part (b) says the integral is 0 because the areas cancel.

Definite Integration and Its Properties JEE Notes | EduRev

Since f(x) = x6 + 1 satisfies f(–x) = f(x), it is even and so

Definite Integration and Its Properties JEE Notes | EduRev  Definite Integration and Its Properties JEE Notes | EduRev

Since f(x) = (tan x)/(1 + x2 + x4) satisfies f(–x) = –f(x), it is odd and so  Definite Integration and Its Properties JEE Notes | EduRev


Definite Integration and Its Properties JEE Notes | EduRev


Ex.6 Evaluate  Definite Integration and Its Properties JEE Notes | EduRev


Sol. The integrand, f(x) = x3 – 3x, is an an odd function; i.e., the equation f(–x) is satisifed for every x. Its graph, drwan in Figure, is therefore symmetric under reflection first about the x-axis and then about the y-axis. If follows that the region above the x-axis has the same areas as the region below it.

Definite Integration and Its Properties JEE Notes | EduRev

We conclude that Definite Integration and Its Properties JEE Notes | EduRev = 0.


Ex.7 Evaluate  Definite Integration and Its Properties JEE Notes | EduRev where [x] denotes the greatest integer function less than or equal to x.

Sol.

Let Definite Integration and Its Properties JEE Notes | EduRev  Suppose f(x) = (–1)[x]

∴ f(–x) = (–1)[–x] = (–1)–1–[x], x ∉ l = - (-1)-[x]

Definite Integration and Its Properties JEE Notes | EduRev = – f(x), f(x) is odd function.

Definite Integration and Its Properties JEE Notes | EduRev


Ex.8 Show that  Definite Integration and Its Properties JEE Notes | EduRev

Sol.

Definite Integration and Its Properties JEE Notes | EduRev  (By prop.) ........(1)

Definite Integration and Its Properties JEE Notes | EduRev

adding (1) and (2) we get 2l  Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev (By prop.)  Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev ...(3)

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev ...(4)

adding (3) and (4) we get Definite Integration and Its Properties JEE Notes | EduRev


Ex.9 Evalute Definite Integration and Its Properties JEE Notes | EduRev

Sol.

Definite Integration and Its Properties JEE Notes | EduRev ....(1)

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Adding (1) and (2) we get 2l  Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev
 
Ex.10 Evaluate Definite Integration and Its Properties JEE Notes | EduRev (where { * } denotes the fractional part function)

Sol.

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev = 1(0 + 1) + 15(1 – 0) = 16 (∴ {x} is a periodic function)


Ex.11 Let the function f be defined by f(x) = Definite Integration and Its Properties JEE Notes | EduRev
 Evaluate  

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev


Sol.

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev


Ex.12 Evaluate Definite Integration and Its Properties JEE Notes | EduRev  where [x] and {x} denotes the integral part, and fractional part function of x and n ε N.

Sol.

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

 

Ex.13 Show that  Definite Integration and Its Properties JEE Notes | EduRev

Sol.

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

= q{(1 – 0) – (0 – 1)} + sin p – sin 0 = 2q + sin p

 

Ex.14 Evaluate  Definite Integration and Its Properties JEE Notes | EduRev   (where [ * ] is the greatest integer function)


Sol.

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Since sin x + cos x is periodic function with period 2π, so
Definite Integration and Its Properties JEE Notes | EduRev

Alternative Method : (Graphical Method)

It is clear from the figure.

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

Definite Integration and Its Properties JEE Notes | EduRev

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