Integration as an Inverse Process of Differentiation: The integration process is widely used for solving problems in several practical and theoretical scenarios to solve problems. For example, it is used to find an object’s location at an instant from its instantaneous velocity at that time. As per the First Fundamental Theorem of Calculus, the integral function (or antiderivative) ‘F’, of a function ‘f’ may be obtained as the integral of ‘f’ within variable limits (such as x and y).

Thus, it effectively states that integration is the inverse process of differentiation for continuous functions. Read this article to fully understand integration as an inverse process of differentiation along with the meaning of antiderivatives of functions, the definition of indefinite integral, integration, some properties of indefinite integral.

Integration as an Inverse Process of Differentiation

The first law of the Fundamental Theorem of Calculus states that one antiderivative (also known as an indefinite integral), say F, of a function f may be obtained as the integral of f with a variable bound of integration. This implies the existence of antiderivatives for continuous functions in differential calculus. However, before trying to understand integration as an inverse process of differentiation, let us learn some fundamental terms in calculus in detail.

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Primitives or Antiderivatives

A function \(\phi (x)\) is called a primitive \(\left( {\rm{or}}\,{\rm{an}}\,{\rm{antiderivative}}\,{\rm{or}}\,{\rm{an}}\,{\rm{integral}} \right)\) of a function \(f(x)\) if \({\phi ^\prime }(x) = f(x).\)

For example, \(\frac{{{x^4}}}{4}\) is a primitive of \({x^3},\) because \(\frac{d}{{dx}}\left( {\frac{{{x^4}}}{4}} \right) = {x^3}.\)

Let \(\phi(x)\) be a primitive of a function \(f(x)\) and let \(C\) be any constant. Then,

\(\frac{d}{{dx}}[\phi (x) + C] = {\phi ^\prime }(x) = f(x)\)

\(\because \phi (x) + {\text{C}}\) is also a primitive of \(f(x).\)

Thus, if a function \(f(x)\) has a primitive, then it has infinitely many primitives that are contained in the expression

\(\phi (x) + {\text{C}}\)

Example:\(\frac{{{x^4}}}{4},\frac{{{x^4}}}{4} + 2,\)and \(\frac{{{x^4}}}{4} – 1\) are primitives of \({x^3}\)

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Indefinite Integral

Consider the function \(f(x).\) The indefinite integral of \(f(x)\) is thus represented as

\(\int f (x)dx\)

which is the family of all its primitives or antiderivatives.

The expression \(\int f (x)dx\) is read as the indefinite integral of \(f(x)\) with respect to \(x.\)

Thus, 

\(\frac{d}{{dx}}(\phi (x) + C) = f(x) \Leftrightarrow \int f (x)dx = \phi (x) + C\)

 Here,

\(∫\) is the integral sign

\(f(x)\) is the integrand

\(x\) is the integration variable 

\(dx\) is the differential of \(x\)

\(C\) is an arbitrary constant known as the constant of integration and \(\phi (x)\) is primitive of \(f(x)\)

Hence, integration of the function is the process of determining an indefinite integral of a given function.

As a result, we can define integration as the inverse of differentiation.

Thus, we can say that integrating a function \(f(x)\) involves finding a function \(\phi (x)\) such that \(\frac{d}{{dx}}(\phi (x)) = f(x)\)

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Fundamental Integration Formulas

We learnt that 

\(\frac{d}{{dx}}[\phi (x)] = f(x) \Leftrightarrow \int f (x)dx = \phi (x) + C\)

Based on this and the different standard differentiation formulas, we obtain the following integration formulas.

Differentiation Formulae	Integration Formulae
\(\frac{d}{{dx}}\left( {\frac{{{x^{n + 1}}}}{{n + 1}}} \right) = {x^n},n \ne – 1\)	\(\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne – 1\)
\(\frac{d}{{dx}}\left( {{{\log }_e}x} \right) = \frac{1}{x}\)	\(\int {\frac{1}{x}} dx = {\log _e}|x| + C\)
\(\frac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\)	\(\int {{e^x}} dx = {e^x} + C\)
\(\frac{d}{{dx}}\left( {\frac{{{a^x}}}{{{{\log }_e}a}}} \right) = {a^x},a > 0,a \ne 1\)	\(\int {{a^x}} dx = \frac{{{a^x}}}{{{{\log }_e}a}} + C\)
\(\frac{d}{{dx}}( – \cos x) = \sin x\)	\(\int {\sin }\,x\,dx = – \cos x + C\)
\(\frac{d}{{dx}}(\sin x) = \cos x\)	\(\int {\cos }\,x\,dx = \sin x + C\)
\(\frac{d}{{dx}}(\tan x) = {\sec ^2}x\)	\(\int {{{\sec }^2}}\,x\,dx = \tan x + C\)
\(\frac{d}{{dx}}( – \cot x) = {\operatorname{cosec} ^2}x\)	\(\int {{{\operatorname{cosec} }^2}}\,x\,dx = – \cot x + C\)
\(\frac{d}{{dx}}(\sec x) = \sec x\tan x\)	\(\int {\sec } \,x\tan x\,dx = \sec x + C\)
\(\frac{d}{{dx}}( – \operatorname{cosec} x) = \operatorname{cosec} x\cot x\)	\(\int {\operatorname{cosec} } \,x\cot x\,dx = – \operatorname{cosec} x + C\)
\(\frac{d}{{dx}}(\log \sin x) = \cot x\)	\(\int {\cot }\,x\,dx = \log |\sin x| + C\)
\(\frac{d}{{dx}}( – \log \cos x) = \tan x\)	\(\int {\tan } \,x\,dx = – \log |\cos x| + C\)
\(\frac{d}{{dx}}[\log (\sec x + \tan x)] = \sec x\)	\(\int {\sec }\,x\,dx = \log |\sec x + \tan x| + C\)
\(\frac{d}{{dx}}[\log (\operatorname{cosec} x – \cot x)] = \operatorname{cosec} x\)	\(\int {\operatorname{cosec} } \,x\,dx = \log |\operatorname{cosec} \,x – \cot x| + C\)
\(\frac{d}{{dx}}\left( {{{\sin }^{ – 1}}\frac{x}{a}} \right) = \frac{1}{{\sqrt {{a^2} – {x^2}} }}\)	\(\int {\frac{1}{{\sqrt {{a^2} – {x^2}} }}} dx = {\sin ^{ – 1}}\left( {\frac{x}{a}} \right) + C\)
\(\frac{d}{{dx}}\left( {{{\cos }^{ – 1}}\frac{x}{a}} \right) = – \frac{1}{{\sqrt {{a^2} – {x^2}} }}\)	\(\int {\left[ { – \frac{1}{{\sqrt {{a^2} – {x^2}} }}} \right]} dx = {\cos ^{ – 1}}\left( {\frac{x}{a}} \right) + C\)
\(\frac{d}{{dx}}\left( {\frac{1}{a}{{\tan }^{ – 1}}\frac{x}{a}} \right) = \frac{1}{{{a^2} + {x^2}}}\)	\(\int {\frac{1}{{{a^2} + {x^2}}}} dx = \frac{1}{a}{\tan ^{ – 1}}\left( {\frac{x}{a}} \right) + C\)
\(\frac{d}{{dx}}\left( {\frac{1}{a}{{\cot }^{ – 1}}\frac{x}{a}} \right) = – \frac{1}{{{a^2} + {x^2}}}\)	\(\int {\left[ { – \frac{1}{{{a^2} + {x^2}}}} \right]} dx = \frac{1}{a}{\cot ^{ – 1}}\left( {\frac{x}{a}} \right) + C\)
\(\frac{d}{{dx}}\left( {\frac{1}{a}{{\sec }^{ – 1}}\frac{x}{a}} \right) = \frac{1}{{x\sqrt {{x^2} – {a^2}} }}\)	\(\int {\frac{1}{{x\sqrt {{x^2} – {a^2}} }}} dx = \frac{1}{a}{\sec ^{ – 1}}\left( {\frac{x}{a}} \right) + C\)
\(\frac{d}{{dx}}\left( {\frac{1}{a}{{\operatorname{cosec} }^{ – 1}}\frac{x}{a}} \right) = – \frac{1}{{x\sqrt {{x^2} – {a^2}} }}\)	\(\int {\left[ { – \frac{1}{{x\sqrt {{x^2} – {a^2}} }}} \right]} dx = \frac{1}{a}{\operatorname{cosec} ^{ – 1}}\left( {\frac{x}{a}} \right) + C\)

Standard Results on Integration

1. The integrand is the differentiation of an integral, or differentiation and integration are inverse operations. i.e., \(\frac{d}{{dx}}\left( {\int f (x)dx} \right) = f(x).\)
   Let \(\int f (x)dx = \phi (x)\)
   Then, by definition of an integral, we get
   \(\frac{d}{{dx}}(\phi (x)) = f(x) \Rightarrow \frac{d}{{dx}}\left( {\int f (x)dx} \right) = f(x)\)

2. (The constant) (integral of the function) is equal to the integral of the product of a constant and a function, i.e., \(\int k f(x)dx = k\int f (x)dx,\), where \(k\) is a constant.
   Let \(\int f (x)dx = \phi (x)\)
   Then, by definition of an integral, we get
   \(\frac{d}{{dx}}(\phi (x)) = f(x) \ldots \ldots (i)\)
   \(\therefore \frac{d}{{dx}}[k\phi (x)] = k \cdot \frac{d}{{dx}}(\phi (x)) = kf(x)\,\,\,\left[ {{\text{Using}}\,\left( i \right)} \right]\)
   \( \Rightarrow \int k f(x)dx = k\phi (x)\)            [By definition of an integral].
   \(\therefore \int k f(x)dx = k\int f (x)dx\left[ {\because \int f (x)dx = \phi (x)} \right]\)

3. The integral of a finite number of functions sum or difference is equal to the sum or difference of the integrals of each functions. i.e., \(\int {} \left[ {f(x) \pm g(x)} \right]dx = \int f (x)dx \pm \int g (x)dx\)
   Let \(\int f (x)dx = \phi (x)\) and \(\int g (x)dx = \psi (x)\)
   Then,
   \(\frac{d}{{dx}}(\phi (x)) = f(x)\) and \(\frac{d}{{dx}}(\psi (x)) = g(x)\)
   \( \Rightarrow \frac{d}{{dx}}(\phi (x)) \pm \frac{d}{{dx}}(\psi (x)) = f(x) \pm g(x)\)
   \( \Rightarrow \frac{d}{{dx}}[\phi (x) \pm \psi (x)] = f(x) \pm g(x)\)
   \( \Rightarrow \int {} \left[ {f(x) \pm g(x)} \right]dx = \phi (x) \pm \psi (x)\)        [By definition of integral]
   \( \therefore \int {} \left[ {f(x) \pm g(x)} \right]dx = \int f (x)dx \pm \int g (x)dx\)

Geometrical Interpretation of Indefinite Integral

In order to understand the geometrical meaning of an indefinite integral, let us consider a function \(f\) given by \(f(x)=-2x.\)

Clearly, \(∫f(x)dx=-x^2+C,\) where \(C\) is the constant of integration.

Let us now consider the family of curves given by \(y=∫f(x)dx=-x^2+C.\)

Clearly, \(y=-x^2+C\) represents a family of parabolas having their common axis of symmetry along \(y-\)axis as shown in the figure given below:

In other words, each integral of \(f(x)=-2x\) represents a parabola with its axis of symmetry along \(y-\)axis.

Let \(x=a\) be a line parallel to \(y-\)axis which cuts the parabolas \(y=-x^2,y=-x^2+1,y=-x^2+2,y=-x^2+3\) etc.  respectively at points \(P_1,P_2,P_3,P_4\) etc.  At each of these points, we have \(\frac{{dy}}{{dx}} = – 2a.\)

That is the slopes of the tangents to the parabolas at \(P_1,P_2,P_3,P_4\) etc are same. Consequently, tangents at \(P_1,P_2,P_3,P_4\) etc are parallel.

Thus, the indefinite integral of a function may be interpreted geometrically as follows: 

A family of curves with parallel tangents at their points of intersection with lines orthogonal to the axis representing the variable of integration represents the indefinite integral of a function.

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Differentiation vs. Integration

The following points will help us understand the difference between differentiation and integration.

1. The operations of differentiation and integration are defined for functions.
2. When it exists, the derivative of a function is a unique function, whereas the integral of a function is not uncommon. There are infinitely many integrals of a position such that any two integrals differ by a constant.
3. The derivative of a function at a point (if it exists) is meaningful, but the integral of a function at a point does not make any sense.
4. Every function is not differentiable. Similarly, every function is not integrable.
5. The derivative of a function at a point determines the slope of the tangent to the corresponding curve at that point. The integral of a function represents a family of curves with parallel tangents at the ends of the intersection. These tangents are orthogonal to the axis representing the variable of integration.
6. The processes of differentiation and integration are inverse of each other.
7. Differentiation and integration are operations on functions.
8. Both the operations are linear,
   i.e. \(\frac{d}{{dx}}(f(x) + g(x)) = \frac{d}{{dx}}(f(x)) + \frac{d}{{dx}}(g(x))\) and, \(\int {(f(} x) + g(x))dx = \int f (x)dx + \int g (x)dx\)
9. The constant can be taken outside the differential as well as integral sign, i.e.\(\frac{d}{{dx}}(kf(x)) = k\frac{d}{{dx}}(f(x))\) and \(\int k f(x)dx = k\int f (x)dx\)
10. Differentiation and integration both involve limits.

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Solved Examples – Integration as an Inverse Process of Differentiation

Below, we have provided a few solved examples to elaborate Integration as an Inverse Process of Differentiation.

Q.1. Write an antiderivative of \(\cos 2x\) using the method of inspection.
Ans: We know that if \({\phi ^\prime }(x) = f(x)\) then \(\phi (x)\) is said to be an antiderivative of \(f(x).\)
Now, we know that
\(\frac{d}{{dx}}(\sin 2x) = 2\cos 2x\)
\( \Rightarrow \cos 2x = \frac{1}{2}\frac{d}{{dx}}(\sin 2x)\)
\( = \frac{d}{{dx}}\left( {\frac{1}{2}\sin 2x} \right)\)
Hence, the antiderivative of \(\cos 2x\) is \(\frac{1}{2}\sin 2x.\)

Q.2. Write an antiderivative of \(3x^2+4x^3.\)
Ans: Let \(f(x) = 3{x^2} + 4{x^3}.\)
We know that \({\phi ^\prime }(x) = f(x)\) if and only if \(\int f (x)dx = \phi (x) + C\)
Now,
\(\int {\left[ {3{x^2} + 4{x^3}} \right]} dx = 3\int {{x^2}} dx + 4\int {{x^3}} dx\)
\( = 3 \times \frac{{{x^3}}}{3} + 4 \times \frac{{{x^4}}}{4} + C\quad \left[ {\because \int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne – 1} \right]\)
\( = {x^3} + {x^4} + C\)
Hence, an antiderivative of \(3x^2+4x^3\) is \(x^3+x^4+C.\)

Q.3.Write an antiderivative of \(\frac{{{x^3} – 1}}{{{x^2}}}.\)
Ans: Let \(f(x)\frac{{{x^3} – 1}}{{{x^2}}}\)
We know that \({\phi ^\prime }(x) = f(x)\) if and only if \(\int f (x)dx = \phi (x) + C\)
Now, \(\int {\left( {\frac{{{x^3} – 1}}{{{x^2}}}} \right)} dx = \int x dx – \int {{x^{ – 2}}} dx\)
\( = \frac{{{x^2}}}{2} – \frac{{{x^{ – 1}}}}{{ – 2}} + C\)
\( = \frac{{{x^2}}}{2} + \frac{1}{{2x}} + C\)
Hence, an antiderivative of \(\frac{{{x^3} – 1}}{{{x^2}}}\) is \(\frac{{{x^2}}}{2} + \frac{1}{{2x}} + C.\)

Q.4. Evaluate:
\(\int {\frac{{{e^{5{{\log }_e}x}} – {e^{4{{\log }_e}x}}}}{{{e^{3{{\log }_e}x}} – {e^{2{{\log }_e}x}}}}} dx\)
Ans: Since
\({e^{a{{\log }_e}x}} = {x^a}\)
\(\therefore \int {\frac{{{e^{5{{\log }_e}x}} – {e^{4{{\log }_e}x}}}}{{{e^{3{{\log }_e}x}} – {e^{2{{\log }_e}x}}}}} dx = \int {\frac{{{x^5} – {x^4}}}{{{x^3} – {x^2}}}} dx\)
\( = \int {\frac{{{x^4}(x – 1)}}{{{x^2}(x – 1)}}} dx\)
\( = \int {{x^2}} dx\)
\(\therefore \int {\frac{{{e^{5{{\log }_e}x}} – {e^{4{{\log }_e}x}}}}{{{e^{3{{\log }_e}x}} – {e^{2{{\log }_e}x}}}}} dx = \frac{{{x^3}}}{3} + C\)

Q.5. Evaluate: \(\int {\frac{2}{{1 + \cos 2x}}} dx\)
Ans: Given: \(\int {\frac{2}{{1 + \cos 2x}}} dx\)
We know that \(1 + \cos 2x = 2{\cos ^2}x\)
\(\therefore \int {\frac{2}{{1 + \cos 2x}}} dx = \int {\frac{2}{{2{{\cos }^2}x}}} dx\)
\( = \int {{{\sec }^2}} x\,dx\)
\( = \tan x + C\)

Q.6. Evaluate: \(\int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}} dx\)
Ans: Given: \(\int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}} dx\)
We know that \(\cos 2x = 1 – 2{\sin ^2}x\)
\(\therefore \int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}} = \int {\frac{{1 – 2{{\sin }^2}x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}} dx\)
\( = \int {\frac{1}{{{{\cos }^2}x}}} dx\)
\( = \int {{{\sec }^2}} x\,dx\)
\(\therefore \int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}} = \tan x + C\)

Q.7. Evaluate: \(\int {\frac{{2{{\cos }^2}x – \cos 2x}}{{{{\sin }^2}x}}} dx\)
Ans: Given: \(\int {\frac{{2{{\cos }^2}x – \cos 2x}}{{{{\sin }^2}x}}} dx\)
We know that \(\cos 2x = 2{\cos ^2}x – 1\)
\(\therefore \int {\frac{{2{{\cos }^2}x – \cos 2x}}{{{{\sin }^2}x}}} dx = \int {\frac{{\cos 2x + 1 – \cos 2x}}{{{{\sin }^2}x}}} dx\)
\( = \int {\frac{1}{{{{\sin }^2}x}}} dx\)
\( = \int {{{\operatorname{cosec} }^2}} x\,dx\)
\(\therefore \int {\frac{{2{{\cos }^2}x – \cos 2x}}{{{{\sin }^2}x}}} dx = – \cot x + C\)

Q.8. If \(a>0\) and \(a \ne 1\) evaluate the following integral,
\(\int {{e^x}} {a^x}dx\)
Ans: Given: \(\int {{e^x}} {a^x}dx\)
\( = \int {{{(ae)}^x}} dx\)
\( = \frac{{{{(ae)}^x}}}{{\log (ae)}} + C\)
\(\therefore \int {{e^x}} {a^x}dx = \frac{{{{(ae)}^x}}}{{\log (ae)}} + C\)

Q.9. Evaluate the following integral.
\(\int {{a^{3{{\log }_a}x}}} dx\)
Ans: Given: \(\int {{a^{3{{\log }_a}x}}} dx\)
\( = \int {{a^{{{\log }_{\text{a}}}{x^3}}}} dx\)
\( = \int {{x^3}} dx\)
\(\therefore \int {{a^{3{{\log }_a}x}}} dx = \frac{{{x^4}}}{4} + C\)

Q.10. Evaluate: \(\int {\sqrt {1 + \cos 2x} }\,dx\)
Ans: Given: \(\int {\sqrt {1 + \cos 2x} }\,dx\)
We know that \(\cos 2x = 2{\cos ^2}x – 1\)
\(\therefore \int {\sqrt {1 + \cos 2x} \,} dx = \int {\sqrt {1 + 2{{\cos }^2}x – 1} } \,dx\)
\( = \int {\sqrt {2{{\cos }^2}x} \,} dx\)
\( = \sqrt 2 \int {\cos }\,x\,dx\)
\(\therefore \int {\sqrt {1 + \cos 2x\,} } dx = \sqrt 2 \sin x + C\)

Summary

A function \(\phi (x)\) is called a primitive (or an antiderivative or an integral) of a function \(f(x)\) if \({\phi ^\prime }(x) = f(x)\) Finding an indefinite integral of a given function is called the integration of the function. Thus, we can say that integration is an inverse process of differentiation. The indefinite integral of a function geometrically represents a family of curves with parallel tangents at their points of intersection with lines orthogonal to the axis representing the variable of integration. We have provided some fundamental integral formulas and some standard results of integration.

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Frequently Asked Questions (FAQs)

The most frequently raised doubts on Integration as an Inverse Process of Differentiation are answered here:

Q.1. Why is integration the inverse of differentiation?
Ans: Integration is a reversal of the differentiating process. We are given the derivative of a function and asked to discover the primitive, i.e. the original function, rather than differentiating it. Such a process is called antidifferentiation or integration.

Q.2. Is integral and integration the same?
Ans: The symbol \(\int f (x)dx\) is read as the indefinite integral of \(f(x)\) with respect to \(x.\) Here \(\int\) is the integral sign. Finding the indefinite integral of a given function is called the integration of the function.

Q.3. Where are integration and differentiation used in real life?
Ans: Differentiation and integration can assist us in resolving a variety of real-world issues. We utilise the derivative to determine a function’s maximum and minimum values (e.g. cost, strength, amount of material used in a building, profit, loss, etc.). Integrals are utilised in various sectors in real life, including engineering, where engineers use integrals to determine the geometry of a building. In physics, it’s used to describe the centre of gravity, among other things.

Q.4. Why are we doing integration?
Ans: Integration is mostly used to compute the volumes of three-dimensional objects and calculate the areas of two-dimensional regions. Finding the area of the curve with respect to the \(x-\)axis is the same as finding the integral of a function with respect to the \(x-\)axis.

Q.5. What’s the difference between differentiation and integration?
Ans: Differentiation is a technique for analysing tiny changes in one quantity in relation to a unit change in another, whereas the formula and process for finding the area under the curve are referred to as integration in calculus.

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