Definite Integral - Calculus

Before going to learn about definite integrals, first, recollect the concept of integral. An integral assigns numbers to functions in mathematics to define displacement, area, volume, and other notions that arise by connecting infinitesimal data. The process of finding integrals is called integration. Definite integrals are used when the limits are defined to generate a unique value. Indefinite integrals are implemented when the boundaries of the integrand are not specified. In case, the lower limit and upper limit of the independent variable of a function are specified, its integration is described using definite integrals. Also, we have several integral formulas to deal with various definite integral problems in maths.

Definite Integral Definition

The definite integral of a real-valued function f(x) with respect to a real variable x on an interval [a, b] is expressed as

Here,

∫ = Integration symbol

a = Lower limit

b = Upper limit

f(x) = Integrand

dx = Integrating agent

Thus, ∫_a^b f(x) dx is read as the definite integral of f(x) with respect to dx from a to b.

Definite Integral as Limit of Sum

The definite integral of any function can be expressed either as the limit of a sum or if there exists an antiderivative F for the interval [a, b], then the definite integral of the function is the difference of the values at points a and b. Let us discuss definite integrals as a limit of a sum. Consider a continuous function f in x defined in the closed interval [a, b]. Assuming that f(x) > 0, the following graph depicts f in x.

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], …… [x_r-1, x_r], [x_n-1, x_n].

Let us consider the width of each subinterval as h such that h → 0, x₀ = a, x₁ = a + h, x₂ = a + 2h,…..,x_r = a + rh, x_n = b = a + nh

and n = (b – a)/h

Also, n→∞ in the above representation.

Now, from the above figure, we write the areas of particular regions and intervals as:

Area of rectangle PQFR < area of the region PQSRP < area of rectangle PQSE ….(1)

Since. h→ 0, therefore x_r–  x_r-1→ 0. The following sums can be established as;

From the first inequality, considering any arbitrary subinterval [x_r-1, x_r] where r = 1, 2, 3….n, it can be said that s_n< area of the region ABCD <S_n

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

From this, it can be said that this area is also the limiting value of an area lying between the rectangles below and above the curve. Therefore,

This is known as the definition of definite integral as the limit of sum.

Definite Integral Properties

Below is the list of some essential properties of definite integrals. These will help evaluate the definite integrals more efficiently.

• ∫_a^b f(x) dx = ∫_a^b f(t) d(t)
• ∫_a^b f(x) dx = – ∫_b^a f(x) dx
• ∫_a^a f(x) dx = 0
• ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx
• ∫_a^b f(x) dx = ∫_a^b f(a + b – x) dx
• ∫₀^a f(x) dx =  f(a – x) dx

Steps for calculating ∫_ab f(x) dx

Step 1: Find the indefinite integral ∫f(x) dx. Let this be F(x). There is no need to keep

integration constant C. This is because if we consider F(x) + C instead of F(x), we get

∫_a^b f(x) dx = [F(x) + C]_a^b = [F(b) + C] – [F(a) + C] = F(b) + C – F(a) – C = F(b) – F(a)

Thus, the arbitrary constant will not appear in evaluating the value of the definite integral.

Step 2: Calculate the value of F(b) – F(a) = [F(x)]_a^b

Hence, the value of ∫_a^b f(x) dx = F(b) – F(a)

Definite Integral by Parts

Below are the formulas to find the definite integral of a function by splitting it into parts.

• ∫₀^2a f (x) dx = ∫₀^a f (x) dx + ∫₀^a f (2a – x) dx
• ∫₀^2a f (x) dx = 2 ∫₀^a f (x) dx … if f(2a – x) = f (x).
• ∫₀^2a f (x) dx = 0 … if f (2a – x) = – f(x)
• ∫-a^a f(x) dx = 2 ∫₀^a f(x) dx … if f(- x) = f(x) or it is an even function
• ∫_-a^a f(x) dx = 0 … if f(- x) = – f(x) or it is an odd function

Definite Integral Examples

Example 1: 

Evaluate the value of ∫₂³ x² dx.

Solution:

Let I = ∫₂³ x² dx

Now, ∫x² dx = (x³)/3

Now, I =  ∫₂³ x² dx = [(x³)/3]₂³

= (3³)/3 – (2³)/3

= (27/3) – (8/3)

= (27 – 8)/3

= 19/3

Therefore, ∫₂³ x² dx = 19/3

Example 2:

Calculate: ∫₀^π/4 sin 2x dx

Solution:

Let I = ∫₀ ^π/4 sin 2x dx

Now, ∫ sin 2x dx = -(½) cos 2x

I = ∫₀ ^π/4 sin 2x dx

= [-(½) cos 2x]₀^π/4

= -(½) cos 2(π/4) – {-(½) cos 2(0)}

= -(½) cos π/2 + (½) cos 0

= -(½) (0) + (½)

= 1/2

Therefore, ∫₀ ^π/4 sin 2x dx = 1/2

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Definite Integration – Short Notes for Quick Revision