Integral Calculus

In Differential Calculus, you find the derivative of a given function using various rules for differentiation. In Integral Calculus, you find the function whose derivative is given. The process of finding the function given its derivative is called integration. As you can see, integration is the reverse process of differentiation.

Let $f(x)=x^2$, then the derivative of $f\left(x\right)$ with respect to x is 2x. So, if we integrate $2x$ with respect to x, we get $x^2$.

And, we say that the integral of 2x with respect to x is $x^2$. And, we write it as follows:

$${\displaystyle \int 2xdx=x^2}$$.

(NOTE: You can read $\int$ as "integral of " and dx as with "respect to x".)

One can argue that the integral of 2x with respect to x is $x^2+1$, or $x^2-4$ or, $x^2+\frac{4}{5}$ or $x^2-7$ because their derivatives with respect to x is 2x. And, that is a valid arguement. In other words, there is no single answer for the integral of a function of x with respect to x, there are as many answers as you like. But, as you can already see, all those answers are only different from each other by a constant number. The reason for it is that the derivative of a constant number (constant function) with respect to x is always zero.

So, we say that the integral of 2x with respect to x is $x^2+\mathrm{C}$, where C is any constant number called the constant of integration and write it as follows: $${\displaystyle \int 2xdx=x^2+C}$$.

It is absolutely necessary that you put that constant of integration C when you write the integral of a function.

Since there is no definite answer to the integral problems we have been talking about, we call this type of integral as indefinite integral.

I am sure most of you must be wondering if there is another kind of integral which is called definite itegral. And, yes, that is the case. Definite Integrals are defined in a different way, but calculations of definite integral involve finding the indefinite integral or simply the integral function. The definite integrals are always over an real interval (both finite or infinite) or over the union of two or more real intervals.

The definite integral of 2x with respect to x over the interval (3,5) is basically calculated by considering any function F(x) whose derivative with respect to x is 2x (the usual integral without the constant of integration C), in this case $F(x)=x^2$ and finding the difference F(5) - F(3), and written as follows: $${\displaystyle \underset{3}{\overset{5}{\int }}2xdx=F(5)-F(3)}=5^2-3^2=25-9=16$$.