Mathematics Definite Integral For CBSE-NCERT
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star Definite Integral

### Definite Integral

● A definite integral is denoted by ∫_a^b f(x)dx , where a is called the lower limit of the integral and b is called the upper limit of the integral.

● The definite integral is introduced either as the limit of a sum or if it has an anti derivative F in the interval [a, b], then its value is the difference between the values of F at the end points, i.e., F(b) – F(a)

color{red}{"Definite integral as the limit of a sum :"}

● Let f be a continuous function defined on close interval [a,b]. Assume all the values taken by the function are non negative, so the graph of the function is a curve above the x-axis.

● So The definite integral int_a^b f(x)dx is the area bounded by the curve y = f(x), the ordinates x=a  and x=b and the x-axis.

### evaluation of Area

=> To evaluate this area, consider the region PRSQP between this curve, x-axis and the ordinates x = a and x = b.

=>1. Divide the interval [a, b] into n equal subintervals denoted by [x_0, x_1], [x_1,x_2] ,...., [x_(r-1), x_r],..., [x_(n-1), x_n], where x_0 = a, x_1 = a + h, x_2 = a +2h, ..., x_r = a + rh and x_n = b = a + nh or n = (b-a)/h. We note that as n → ∞, h → 0.

=>2. The region PRSQP under consideration is the sum of n subregions, where each subregion is defined on subintervals [x_(r-1), x_r] , r =1,2,3 ...n

From Fig =>

=>color{green}{"Area of the rectangle (ABLC) < area of the region (ABDCA) < area of the rectangle (ABDM)"}

=>3. Evidently as x_r – x_(r–1 )→ 0, i.e., h → 0 all the three areas shown in (1) become nearly equal to each other. Now we form the following sums.

color{brown}{s_n = h [f(x_0) + ..... + f(x_(n-1) )] = h sum_(r=0)^(n-1) f(x_r)}........(2)

and color{brown} {S_n = h[ f(x_1) +f(x_2) +......+ f(x_n)] = h sum_(r=1)^n f(x_r)}........(3)

=>4. Here, s_n and S_n denote the sum of areas of all lower rectangles and upper rectangles raised over sub intervals [x_(r–1), x_r] for r = 1, 2, 3, …, n, respectively.

In view of the inequality (1) for an arbitrary subinterval [x_(r-1), x_r] ,we have

color{blue}{=>} s_n < area of the region PRSQP < S_n

=> As n -> oo strips become narrower and narrower, it is assumed that the limiting values of (2) and (3) are the same in both cases and the common limiting value is the required area under the curve.

Symbolically, we write

color{orange}{lim_(n->oo) S_n = lim_(n->oo) s_n = "area of the region PRSQP "= int_a^b f(x) dx} ... (5)

=>5.  It follows that this area is also the limiting value of any area which is between that of the rectangles below the curve and that of the rectangles above the curve.

=> For the sake of convenience, we shall take rectangles with height equal to that of the curve at the left hand edge of each subinterval. Thus, we rewrite (5) as

color {brown} {int_a^b f(x) dx = lim_(h ->0) h [ f(a) + f(a +h) +....+ f(a + (n -1) h ]}

or  int_a^b f(x) dx = (b -a) lim_(n -> oo) 1/n [ f(a) + f(a +h) +.... + f(a + ( n-1) h ] ......(6)

where h = (b -a)/n -> 0 as n-> oo

The above expression (6) is known as the definition of definite integral as the limit of sum.

color {blue} text{Remarks}

=>The value of the definite integral of a function over any particular interval depends on the function and the interval, but not on the variable of integration that we choose to represent the independent variable. If the independent variable is denoted by t or u instead of x, we simply write the integral as  int_a^b f(t) dt or int_a^b f(u) du instead of int_a^b f(x) dx. Hence, the the variable of integration is called a dummy variable.
Q 3105380268

Evaluate  int_(0)^2 e^x dx as the limit of a sum.
Class 12 Chapter 7 Example 26
Solution:

By definition

 int_(0)^2 e^x dx = (2-0 ) lim_(n->oo) 1/n [ e^0 + e^(2/n) + e^(4/n) + ..... + e^( (2n -2)/n ) ]

Using the sum to n terms of a G.P., where a = 1, r = e^2/n , we have

 int_(0)^2 e^x dx = 2 lim_(n->oo) 1/n [ ( e^( (2n)/n) -1)/( e^(2/n) -1 ) ] =2 lim_(n-> oo ) 1/n [ (e^2 -1)/( e^(2/n) -1) ]

= (2 (e^2 -1 ) )/( lim_(n->oo) [ ( e^(2/n) -1)/(2/n) ]*2 ) = e^(2) -1 [ using  lim_(h->0) (e^h -1)/h =1  ]
Q 3105280168

Find  ∫_(0)^2 (x^2 +1) dx as the limit of a sum.
Class 12 Chapter 7 Example 25
Solution:

By definition

 ∫_(a)^b f(x ) dx = (b-a) lim_(n->oo) 1/n [f(a) + f(a+ h ) + ..... + f (a + (n-1) h ] ,

where,  h = ( b-a )/n

In this example, a = 0, b = 2, f (x) = x^2 + 1, h = (2-0)/n = 2/n

Therefore,

 int_(0)^2 (x^2 +1) dx = 2 lim_(n->oo) 1/n [ f(0) + f (2/n) + f ( 4/n) + ..... + f (2 (n-1) )/(n) ]

= 2 lim_(n->oo) 1/n [1+ ( (2^2)/(n^2) +1 ) + ( (4^2)/(n^2) +1) + ..... + ( ( (2n -2)^2 )/(n^2 ) +1 ) ]

 = 2 lim_(n->oo) 1/n [ underset ( n - terms) underbrace ( (1 + 1 + .... +1 ) ) +1/n^2 (2^2 + 4^2 + ... + (2n-2)^2 ]

 =2 lim_(n->oo) 1/n [ n + 2^2/n^2 ( 1^2 + 2^2 + .... + (n-1)^2 ) ]

= 2 lim_(n->oo) 1/n [ n +4/n^2 ( (n-1) n (2n -1) )/6 ]

 = 2 lim_(n->oo )1/n [ n + 2/3 ( (n-1) (2n-1) )/n ]

= 2 lim_(n->oo) [1 + 2/3 (1- 1/n ) (2- 1/n ) ] = 2 [1+4/3] = 14/3