Integration = Area under the curve?

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What is the concept of integration? What is the purpose of integration? Why does integrating give you area under the curve?

Integration,also known as anti-derivative of a function, gives the area under the curve of the function. Most of you already know all this thing but did you ever wonder why integration gives you area under the curve? How does it work actually?

Well, the answer is quite simple and basic.

We will take some polynomials as examples and the integration rule for a polynomial term of an arbitrary degree n "$ax^n $ " is given by \[\int ax^n dx=\frac{ax^{n+1}}{n+1}+C\]

1.   $y= \int_{0}^{a} x\,\, dx$ 

Soln: 
Integration
What is the area of y=x??

As we know,
\[\int_{0}^{a} x \, dx = \frac{x^{2}}{2}|_{0}^{a}\Rightarrow\, \, \frac{a^{2}-0}{2}\]
Now, let's find the red-colored area in the given graph
The required area is half of the area of the square of side a 
and the area of a square is given as,
 \[Area\, of\, square = side^2\]
\[\therefore \, Area\, of\, red\, region\,  = \frac{a^2}{2}\]
Which is what we got from the integration.




2.   $y=\int_{a}^{b} Constant \,\,dx$
Soln: 
Integration
What is the Area of y=constant??

We will find the area under the curve from a to b

integration

As we know,
\[y= \int_{a}^{b}C\,\, dx=C \times x\mid_{a}^{b} \,  \, \, \Rightarrow C \times(b-a)\]
Now, let's find the area under the curve of the given function. 
The area that we want to find is the area of a rectangle of length b-a and breadth C.
The Area of a rectangle is given as,
\[Area\,of\,rectangle=length\times breadth\]
\[\Rightarrow Area \, of\, rectangle= C\times ( b-a)\]
which is the same as we got by Integrating the given function.
 
The above two examples made It clear that integration gives the area under the curve of a function.


Integration of Trigonometric functions:

The Trigonometric functions can be represented in terms of polynomials and then their Integration can be done using the polynomial integration formula, 
The polynomial expansion of sin(x) is,
\[sin(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-.....\] 
Now, the integration will be given as,
\[\int sin(x) \,\,dx = C+\frac{x^{2}}{2}-\frac{x^{4}}{4!}+\frac{x^{6}}{6!}-....\]
 \[ \,\,= C'-\left \{ 1-\frac{x^{2}}{2}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+..... \right \}\]
 \[     = -cos(x)+C'\]
where $C'=C+1$
other trigonometric functions can also be Integrated similarly.


Integration of Exponential functions:

The Exponential  functions can be represented as the following polynomials 
\[exp(ax)=1+ax+\frac{(ax)^2}{2!}+\frac{(ax)^3}{3!}+...\]
and its integration will be,
\[\int exp(ax) \,\,dx=C+x+\frac{ax^2}{2}+\frac{a^2x^3}{3!}+\frac{a^3x^4}{4!}+....\]
\[=C'+\frac{1}{a}\left\{ 1+ax+\frac{(ax)^2}{2!}+\frac{(ax)^3}{3!}+\frac{(ax)^4}{4!}+...    \right\}\]
\[where\,\, C'=C-\frac{1}{a}\]
\[=\frac{exp(ax)}{a}+C'\]
and for a=1, we get,
\[\int exp(x)\,\, dx=exp(x)+C'\]

Constant of Integration:

Here, you would have noticed +C in the integration of sin(x) and exp(ax) but not in the polynomial cases so why is it there in the last two integrations? And why is it not in the starting two? Do we have a rule when to use it?

+C is known as the Constant of integration.

We know that the derivative of a constant is zero i.e., \[C'=0\].
Now consider a function G(x) given as, 
\[G(x)=F(x)+C\]
then its derivative is
\[G'(x)={\left (F(x)+C\right )}'=F'(x)+C'=F'(x)\]
This implies that the derivative of G(x) is always equal to F'(x) and is independent of C.
I'll be taking an example to explain this concept.
Let us take the derivative of $\frac{x^2}{2}$ 
 \[\frac {d\,}{dx\,}[\frac {x^{2}}{2}]=x\] 
However, there are many functions whose derivative is x like  derivative of $\frac{x^2}{2}+1000$ is\[\frac {d\,}{dx\,}[\frac {x^{2}}{2}+1000]=x\] 
Hence,the constant implies that every function with at least one antiderivative has an infinite number of antiderivatives.
 
You must be wondering why we ignored +C in the first two examples.

The idea is that we can ignore +C for definite integral (the integral in which the limits are given).
I'll explain you this with an another example
As integral of $x^2$ is given as  \[ \int x^{2} dx= \frac{x^{3}}{3}+C\]
Now,let's evaluate the integral by taking limits
\[\int_1^2x^2dx=[\frac{x^3}{3}+C]_1^2 \]
\[=\frac{8}{3}+C-\left\{\frac{1}{3}+C\right\}\]
\[=\frac{7}{3}+C-C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\]
\[=\frac{7}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\]
that's why we only use +C for indefinite integral (the integral who's limit is not specified).


Finding the area under the curve for straight lines is quite straight forward but in reality, we have to integrate curves and to estimate their area. We can divide the curve in some non-overlapping rectangles and then add their areas as shown in fig (i). It can be observed that, the smaller the width of the rectangles on the x-axis the better the approximated solution will be. It is shown in fig(ii). 

integration
fig(i)
integration
fig(ii)

Hence, as the width of the rectangles approaches to zero the area approaches to the correct solution. 



Hey there, This is Hifza Siddiqui, an aspiring Mathematician pursuing masters from the University of Mumbai. I'm writing this blog to share the knowledge and the way I understand mathematics. I hope it helps you to get a better insight. My blog's frequency will be twice per week.

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