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Integrals

Definite Integral as a Limit of a Sum

Imagine a curve above the x-axis. The function of this graph is a continuous function defined on a closed interval [a, b], where all the values of the function are non-negative. The area bound between the curve, the points ‘x = a’ and ‘x = b’ and the x-axis is the definite integral ∫ab f(x) dx of any such continuous function ‘f’.

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Definite Integral as a Limit of a Sum

Look at the following graph:

definite integral

To understand this, let’s evaluate the area PRSQP between the curve y = f(x), x-axis and the coordinates ‘x = a’ and ‘x = b’. Now, divide the interval [a, b] into ‘n’ equal sub-intervals denoted as:

[x0, x1], [x1, x2], [x2, x3] …. [xn – 1, xn], where,
x0 = a, x1 = a + h, x2 = a + 2h, x3 = a + 3h ….. xr = a + rh and xn = b = a + nh
Or, n = (b – a)/h. Note that as n → ∞, h → 0.

Now, the region PRSQP under consideration is the sum of all the ‘n’ sub-regions, where each sub-region is defined on subintervals [xr – 1, xr], r = 1, 2, 3 … n. Now, look at the region ABDM in the figure above. We can make the following observation:

Area of the rectangle (ABLC) < Area of the region (ABDCA) < Area of the rectangle (ABDM)        (1)

Also, note that as, h → 0 or xr – xr – 1 → 0, all these three areas become nearly equal to each other. Hence, we have

sn = h [f(x0) + f(x1) + f(x2) + …. f(xn – 1)] = h r=0n–1 f(xr) … (2)
and, Sn = h [f(x1) + f(x2) + f(x3) + …. f(xn)] = h r=1n f(xr) … (3)

Here, sn and Sn denote the sum of areas of all lower rectangles and upper rectangles raised over subintervals [xr–1, xr] for r = 1, 2, 3, …, n, respectively. To bring it into perspective, equation (1) can be re-written as:

sn < area of the region (PRSQP) < Sn … (4)

Browse more Topics under Integrals

As n → ∞, these strips become narrower

Further, 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 have

limn → ∞ Sn = limn → ∞ sn = area of the region (PRSQP) = ∫ab f(x) dx … (5)

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 convenience, we shall take the rectangles having height equal to that of the curve at the left-hand-edge of each sub-interval. Hence, equation (%) is re-written as:

ab f(x) dx = limn → ∞ h [f(a) + f(a + h) + …. + f(a + {n – 1}h)]
Or, ∫ab f(x) dx = (b – a) limn → ∞ (1/n) [f(a) + f(a + h) + …. + f(a + {n – 1}h)] … (6)

Where, h = (b – a)/n → 0 as n → ∞. This equation is the definition of Definite Integral as the limit of a sum.

Note: 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. Hence, the variable of integration is called a dummy variable.

Example 1

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

Solution: 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 = 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.

More Solved Examples for You

Question 1: 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].

Question 2: Define a definite integral?

Answer:  We can define it as an exact limit and summation that we looked at in the last section to find the net area between a function and the x-axis. In addition, note that the notation for the definite integral is very similar to the notation for indefinite integrals.

Question 3: Why we use integrals?

Answer: We use integrals to find volumes, areas, central points, and many useful things. However, often we use it to find the area under the graph of a function like this: We can find this area by adding slices that approach zero in width. Moreover, there are rules of integration that helps to get the answer.

Question 4: What is definite and indefinite integral?

Answer: Definite integral refers to an integral that has limits of integration and the answer is a specific area. On the other hand, indefinite integral returns a function of the independent variable/s.

Question 5: Can a definite integral be negative?

Answer: A definite integral can be negative because integrals measure the area between the x-axis and the curve over a specified interval. Furthermore, if all of the areas within the interval exist above the x-axis yet below the curve then the result is positive.

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