Definite Integral

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

Definite Integral as the limit of a sum:

The integral of f(x) is the area of the region bounded by the curve y = f(x). This area is represented by the region ABCD as shown in the above figure. This entire region lying between [a, b] is divided into n equal subintervals given by [x0, x1], [x1, x2], …… [xr-1, xr], [xn-1, xn].

Let us consider the width of each subinterval as h such that h → 0, x= a, x1 = a + h, x2 = a + 2h,…..,xr = a + rh, xn = b = a + nh

and n = (b – a)/h

Also, n→∞ in the above representation.

Since, n→∞, the rectangular strips are very narrow, it can be assumed that the limiting values of sn and Sn are equal and the common limiting value gives us the area under the curve, i.e.,

Question : Find ∫ab x dx as the limit of a sum.

Answer : From equation (6) above, we know that
ab f(x) dx = (b – a) limn → ∞ (1/n) [f(a) + f(a + h) + …. + f(a + {n – 1}h)]
Where, h = (b – a)/n

In this example, we have                      
a = a, b = b, f(x) = x and h = (b – a)/n.

Also, f(a) = a
f(a + h) = a + h
f(a + 2h) = a + 2h
f(a + 3h) = a + 3h ……
f(a + (n – 1)h) = a + (n – 1)h

Therefore,
ab x dx = (b – a) limn → ∞ (1/n) [f(a) + f(a + h) + f(a + 2h) + …. + f(a + {n – 1}h)]
= (b – a) limn → ∞ (1/n) [a + (a + h) + (a + 2h) + …. + (a + (n – 1)h)]
= (b – a) limn → ∞ (1/n) {[a + a + a + … + a(n-times)] + h + 2h + …. + (n – 1)h}
= (b – a) limn → ∞ (1/n) [na + h(1 + 2 + ….. + (n – 1)]

Now, we know that, 1 + 2 + 3 + … + n = n(n + 1)/2. Hence,
1 + 2 + 3 + … + (n – 1) = (n – 1)(n – 1 + 1)/2 = n(n – 1)/2

Therefore,
ab x dx = (b – a) limn → ∞ (1/n) [na + hn(n – 1)/2]
= (b – a) limn → ∞ [na/n + hn(n – 1)/2n]
= (b – a) limn → ∞ [a + (n – 1)h/2]

Replacing h by (b – a)/n, we get
ab x dx = (b – a) limn → ∞ [a + (n – 1) (b – a) / 2n]
= (b – a) limn → ∞ [a + (n/n – 1/n) {(b – a)/2}]
= (b – a) limn → ∞ [a + (1 – 1/n){(b – a)/2}]
= (b – a) [a + (1 – 1/∞) {(b – a)/2}]
= (b – a) [a + (1 – 0){(b – a)/2}]
= (b – a) [a + (b – a)/2]
= (b – a) (2a + b – a)/2
= (b – a) (b + a)/2
= (b2 – a2)/2

Hence, the definite integral ∫ab x dx as the limit of sum is [(b2 – a2)/2].

Example: Find ∫02 (x2 + 1) dx as the limit of a sum.

Solution: From above, we know that
ab f(x) dx = (b – a) limn → ∞ (1/n) [f(a) + f(a + h) + …. + f(a + {n – 1}h)]
Where, h = (b – a)/n

we have a = 0, b = 2, f(x) = (x2 + 1) and h = (2 – 0)/n = 2/n. Therefore,

02 (x2 + 1) dx = 2 limn → ∞ (1/n) [f(0) + f(2/n) + f(4/n) + …. + f(2{n – 1}/n)]
= 2 limn → ∞ (1/n) [1 + {(22/n2) + 1} + {(42/n2) + 1} + …. + {(2n – 2)2/n2 + 1}]
= 2 limn → ∞ (1/n) [1 + 1 + 1 + … + 1(n-times)] + 1/n2) [22 + 42 + … (2n – 2)2]
= 2 limn → ∞ (1/n) [n + 22/n2 (12 + 22 + … (n – 1)2]
= 2 limn → ∞ (1/n) [n + 4/n2 {(n – 1) n (2n – 1) / 6}]
= 2 limn → ∞ (1/n) [n + 2/3 {(n – 1) (2n – 1) / n}]
= 2 limn → ∞ (1/n) [n + 2/3 (1 – 1/n) (2 – 1/n)]
As n → ∞, 1/n → 0. Therefore, we have
02 (x2 + 1) dx = 2 [1 + 4/3] = 14/3.