# Geometrical Interpretation of Definite Integral

## What is Geometrical Interpretation of Definite Integral?

Let us consider a curve that is above the x-axis, that is a continuous function on an interval [a,b] where all the values are positive. The definite integral of any continuous function is the area that is bounded by the curve and the x-axis.

$\int_{a}^{b} f(x) dx = F(b) - F(a)$

In the above formula, a and b are the limits, d/dx(F(x)) = f(x).

There are some distinctions between the definite integral and the indefinite integral, let’s discuss those:

The base for the definite integral is laid by indefinite integral.

Indefinite integral defines the calculation of indefinite area whereas an area of specified limit is calculated by the definite integral.

### Important Properties

The following are the properties of definite integrals:

1. $\int_{a}^{b}f(x) dx = \int_{a}^{b}f(t) dt$

2. $\int_{a}^{b}f(x) dx = -\int_{b}^{a}f(x) dx$ …[ Also, $\int_{a}^{a}f(x) dx = 0$]

3. $\int_{a}^{b}f(x) dx = \int_{a}^{c}f(x) dx + \int_{c}^{b}f(x) dx$

4. $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$

5. $\int_{0}^{a}f(x) dx = \int_{0}^{a}f(a - x) dx …..$

6. $\int_{0}^{2a}f(x) dx = \int_{0}^{a}f(x) dx + \int_{0}^{a}f(2a - x) dx$

7. Two parts:

$\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a}f(x) dx ….$ If f(2a - x) = f(x)

$\int_{0}^{2a} f(x) dx = 0 …$ if (2a - x) = -f(x)

1. Two parts:

$\int_{-a}^{a}f(x) dx = 2 \int_{0}^{a}f(x) dx$ ... if f(-x) = f(x) or it is an even function.

$\int_{a}^{-a}f(x) dx = 0$... if f(-x) = - f(x) or it is an odd function.

1. Walli’s Formula

$\int_{0}^{\pi/2} sin^{n} x cos^{m} x dx$ = [(n - 1)(n - 3)(n - 5)..... 1 or 2][(m - 1)(m - 3)....1 or 2]. K/(m + n)(m + n - 2) ...1 or 2),

Given, if both m and n are even, then K = π/2

1. Leibnitz’s Rule

If g is continuous on [a,b] and f1(x) and f2(x) are the two differentiable functions whose values lie in [a,b], then d/dx ∫f2(x) f1(x) g(t) dt = g(f2(x)) f2’(x) - g(f1(x)) f1’(x).

### Measuring Definite Integral

The best way to find out definite integral through geometric interpretation is by finding out the area that is bounded by the x-axis and a curve. The definite integral is the sum of the height of the function and the product of the length of the intervals, the height being integrated with the interval that includes the formula of the area of the rectangle.

### Solved Problems

Example 1:

$\int_{0}^{\pi/2} ([\sqrt{cotx}] / [\sqrt{cotx} + \sqrt{tanx}]) dx$

Solution:

$[I = \int_{0}^{\pi/2} ([\sqrt{cotx}] / [\sqrt{cotx} + \sqrt{tanx}]) dx$ ….(i)

= $[\int_{0}^{\pi/2}(\sqrt{[cot(\pi /2 - x)}] / \sqrt{cot (\pi /2 - x)} + \sqrt{tan(\pi /2 - x)} dx$

= $\int_{0}^{\pi/2}([\sqrt{tanx]}/ [\sqrt{tanx} + \sqrt{cotx}]) dx$ …… (ii)

$2l = \int_{0}^{\pi/2} ([\sqrt{cotx} + \sqrt{tanx}] / [\sqrt{tanx} + \sqrt{cotx}]) dx$

$= [x]_{0}^{\pi/2} \Rightarrow I = \pi / 4$

Example 2 :

$\int_{0}^{\pi/2}$ dθ / [1 + tanθ]

Solution:

$I = \int_{0}^{\pi/2}$ dθ / [1 + tanθ]

= $\int_{0}^{\pi/2}$ dθ / [1 + tan(π / 2 - θ)]

= $\int_{0}^{\pi/2}$ dθ / [1 + cotθ]

$2I = \int_{0}^{\pi/2}$ (1 / [1 + tanθ]) + (1 / [1 + cotθ]) dθ

=  $\int_{0}^{\pi/2}$ dθ = $[\theta]_{0}^{\pi/2}$

= π/2 ⇒ I = π/4

Example 3:

$\int_{0}^{\pi/2}$ ([cosx - sinx] / [1 + sinx.cosx]) dx

Solution:

$I = \int_{0}^{\pi/2}$ ([cosx - sinx] / [1 + sinx.cosx]) dx … (i)

Now $I = \int_{0}^{\pi/2}$ cos( π / 2 - x) - sin( π / 2 - x) 1 + sin( π / 2 - x).cos( π/2 - x) dx

= $\int_{0}^{\pi/2}$ ([sinx - cosx] / [1 + sinx.cosx]) dx….(ii)

On adding, 2I = 0 ⇒ I = 0

Example 4:

$\int_{-1}^{1}$(log[2 - x] / [2 + x]) dx

Solution:

Let f(x) = (log [2 - x] / [2 + x])

⇒ f(-x) = (log [2 - x] / [2 + x]-1

= - (log [2 - x] / [2 + x]) = -f(x)

Therefore,

$\int_{1}^{-1}$ (log[2 - x] / [2 + x]) dx = 0

Example 5:

What is the smallest interval [a,b] such that $\int_{0}^{1}$ dx / √[1 + x4] ∈ [a,b]?

Solution:

Let $I = \int_{0}^{1}$ dx / √[1 + x4]

Here,

0 ≤ x ≤ 1 1 ≤ (1 + x4) ≤ 2

1 ≤ √[1 + x4] ≤ √2 1 / √2 ≤ 1 / √[1 + x4] ≤ 1

1 / √2 ≤  $\int_{0}^{1}$ dx / √[1 + x4] ≤ 1

Hence, [ 1 / √2, 1] is the smallest interval, such that I ∈  [ 1 / √2, 1].