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.


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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)


Now adding (i) and (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]


On adding,


\[2I = \int_{0}^{\pi/2}\] (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].

FAQs (Frequently Asked Questions)

1. What is a Geometric Interpretation?

Answer: Geometric interpretation literally means to interpret the solution of a math problem from the perspective of geometry. Geometric interpretation in simple words means to interpret something geometrically that is not originally a part of geometry or within the realm of geometry and represent it with something different but not with numbers and tables. The representation can be created in the form of a xy(n) graph, but it is also possible that it will come in the form of a shape or any other geometric entity in n-dimensional space. Geometric interpretation is about giving an object of math an analogy with geometry so that the properties of the Math object are seen in geometry.

2. What is an Integral Value?

Answer: In mathematics, the integral value can be defined as a numerical value that is equal to the area under the graph of a function (the area between the x-axis and the curve) for some interval. This is also is known as a definite integral. If the interval is not known it is known as indefinite integral. An integral value is an area or volume under or above a given mathematical function that is derived from an equation. The integral value can be either a two-dimensional or three-dimensional.