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Difference and relation between Differentiation and Integration

Relation between Differentiation and Integration

    Look at the information given below.

    \[\mathbf{ y=f(x)}\]

    \[ \mathbf{ f'(x)\rightarrow \text{Derivatives of f(x)}}\]

    \[ \mathbf{\displaystyle \int_a^b f’(x) = ?}\]

    Can you tell me the value of above integral?
    Yes, it will be equal to f(b)-f(a).

    We have already known this result. It tells us that integration is just the reverse of differentiation, integral of the derivative of the function f(x) is just equal to the difference in the function f(x) evaluated at the limits of integration.

    Part 1

    Now with this topic, we will understand how to apply this result to find the integral of a function. 

    Consider this function \(\mathbf{g(x)=x^2}\)
    Let's find the integral of this function from a to b i.e \(\mathbf{\displaystyle \int_a^b g(x) \, dx}\). 

    Can you think how we can apply this result to find the integral. Let's see. Suppose we find the function G(x), such that its derivative is this function g(x), then we will get the integral of g(x) is equal to the integral of the derivative of G(x).

    \[\mathbf{G(x) \rightarrow G'(x)=g(x)=x^2}\tag1\]

    \[\mathbf{\displaystyle \int_a^b g(x) \, dx=\displaystyle \int_a^b G'(x) \, dx }\tag2\]

    Now, according to this result, We can see that this integral will be equal to G(b)-G(a). So we see that the integral of the function g(x) is equal to the difference between the values of the function, G(x).

    Do you know any such function. Let's try to find the function G(x). So we have only equation 1 and 2. We need to find the function G(x). Now we have seen earlier that if we have a function f1(x) equal to x, then its derivative is equal to one. If we have a function f2(x) equal to x², then its derivative is equal to two times x. 

    \[\mathbf{f_1(x)=x \qquad f'_1(x)=1}\]\[\mathbf{f_2(x)=x^2 \qquad f'_2(x)=2x}\]

    Notice that each time we take the derivative, the power of x is reduced by one. Now we wanted the function whose derivative = x². So maybe the derivative of the function \(\mathbf{f_3(x)=x^3}\) cubed will give us this function.

    By now, we are familiar with how to find the derivative of a function. Can you find the derivative of \(\mathbf{f_3(x)}\). The derivative of \(\mathbf{f_3(x)}\) will be given by this limit. 

    \[\mathbf{f_3(x) = \displaystyle \lim_{\Delta x \to 0} \dfrac {(x+\Delta x)^3-x^3}{\Delta x}}\]

    This is Derivatives of Function using first principle. We know the formula for the cube of sum of two numbers, and the cube of difference of two numbers.\[\mathbf{  (x+y)^3=x^3+y^3+3xy(x+y)}\]

    \[\mathbf{(x-y)^3=x^3-y^3-3xy(x-y)}\]

     Using this here, we can easily find that the derivative will be equal to three times x squared. So we get the derivative of x cubed, to be three times x squared.

    \[\mathbf{f_3(x) = \displaystyle \lim_{\Delta x \to 0} \dfrac {(x+\Delta x)^3-x^3}{\Delta x}=3x^2}\]

    But we want the derivative of this function to be only x squared. Okay, so what if we consider the function G(x) equal to x cubed over three i.e \(\mathbf{G(x)=\dfrac{x^3}{3}}\). What will be its derivative. Now for this function, this average rate will get multiplied by one over three. So we will get the derivative to be equal to x squared.

    \[\mathbf{G(x)= \dfrac {x^3}{3} = \dfrac{1}{3}\displaystyle \lim_{\Delta x \to 0} \dfrac {(x+\Delta x)^3-x^3}{\Delta x}=x^2}\]

    So we get the function G(x) to be equal to x cubed over three. Now for finding the value of this integral. We just have to find G(b)-G(a) difference. But wait, What if we add a constant. Let's say one to this function G(x), what will be the derivative of the function, G1(x)\[\mathbf{G_1(x)= G(x)+1}\] 

    Part 2

    Let's continue this in the next part. So what will be the derivative of the function, \(G_1(x)\). Yes. Notice that the function  \(G_1(x)\) is just the function G(x) plus one. We know that the derivative of a function tells us its instantaneous rate of change. So adding a constant to a function will not affect its rate of change. We can find its derivative by taking the limit of its average rate of change. We will surely get it to be equal to the derivative of G(x).

    \[\mathbf{G^{'}_1 (x)= \displaystyle \lim_{\Delta x \to 0}\dfrac {G_1(x+\Delta x)-G_1(x)}{\Delta x}=G^{'}(x)}\]

    So now we have two functions whose derivative is equal to the function g(x) that is x squared. So what have to find \(\mathbf{\displaystyle \int_a^b g(x) \, dx=G(b)-G(a)}\)  integral, we use the function \(\text{G$_1$(x)}\) instead of  G(x). Notice that since this function is just G(x) plus a constant. We will again get that same value for the integral. That means we can use any of these two functions, to find the value of the integral. We found these functions by going in the reverse direction. That is, we started with the derivative g(x), and found the original functions whose derivative will be g(x). So these functions are called anti derivatives of the function g(x).

    Now let me ask you a question. Are there any other anti derivatives of g(x). Correct. Any function which is equal to G(x), plus some constant will be the antiderivative of g(x)=x²

    \[\mathbf{G_1(x)= G(x)+1=\dfrac{x^3}{3}+1}\] \[\mathbf{G_2(x)= G(x)+2=\dfrac{x^3}{3}+2}\]\[\mathbf{G_3(x)= G(x)+3=\dfrac{x^3}{3}+3}\]\[\cdot\]\[\cdot\]\[\cdot\]

    Now we can use any of these anti derivatives to find the integral of the function. We will always get the value of this integral to be equal to G(b)-G(a). So for this reason, we denote all these anti derivatives by one general antiderivative  G(x) plus a constant C, called the indefinite integral of g(x), we denote it by the integral symbol, without the limits of integration.

    \[\mathbf{\displaystyle \int g(x) \, dx=G(x)+c=\dfrac{x^3}{3}+c}\]\[c\rightarrow\text{Arbitrary Constant}\]

    We can see that once we know the indefinite integral, We can easily find this integral, the indefinite integral of x squared is equal to x cubed over three plus a constant C, and evaluating it at the limits of integration, we will get this \[\mathbf{\displaystyle \int_a^b x^2 \, dx=\left[ \dfrac{x^3}{3}+c\right]_a^b}\]\[\mathbf{=\left(\dfrac{b^3}{3}+c\right)-\left(\dfrac{a^3}{3}+c\right)}\]\[\mathbf{=\dfrac{b^3-a^3}{3}+c}\]

    Notice that for this integral, we will get a number as our answer, whereas for the indefinite integral, we get a function. So this integral with the limits of integration is called as the definite integral.

    Let's quickly review the whole process.

    Indefinite integration and definite Integration

    To find the definite integral of the function, we first find its indefinite integral. Now to find the indefinite integral of the function g(x), we find a function G(x), such that its derivative is equal to g(x). And now to find the definite integral, we just have to evaluate it at the limits of integration. We can see that once we know the indefinite integral, we can find any definite integral of the function. So usually the indefinite integral is itself, called the integral of the function. So we see that related to a function, we have two types of integrals, indefinite integral, and definite integral.

    Now let's see if you can find the indefinite integral of this function h(x)= x²+6. How can we do that, the indefinite integral of h(x) will be given by a function, H(x)+C. The function H(x) should be such that its derivative is equal to h(x). Can you find the function H(x).

    Tell us your answers in the comment section below and to keep learning such interesting things bookmark this website.


    FAQ Question

    What is the physical meaning of differntiation and integration ?

    It tells us that integration is just the reverse of differentiation, integral of the derivative of the function f(x) is just equal to the difference in the function f(x) evaluated at the limits of integration.

    What is the logical relation between Integration and Differntiation?

    Differentiation is basically instantaneous rate of change of slope of a Function. Where Integration represent area under that curve. Also integration is just the reverse of differentiation.

    Where is Integration used ?

    Some easy application is finding the centre of mass, moment of inertia, area under the curve, Impulse or many more.

    How can I understand differntiation and Integration ?

    To understand, you must have some knowledge of Function and Algebra. Also terminology like Instantaneous change, variable, slope etc. And solving easy problem make you understand.

    What is the meaning of integration ?

    It is basically the Algebraic sum of area or reverse of differntiation.


    Mathematics-Important Notes, concept & Formula

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