Integration as an Inverse Process of Differentiation

Integration as the Reverse Process of Differentiation

Integration is one of the most fundamental concepts in calculus. It represents the reverse process of differentiation, helping us determine the original function from its derivative. In other words, if differentiation measures how quantities change, integration helps us accumulate or reconstruct those quantities. Geometrically, it also represents the area under a curve or the total accumulation of change over an interval — a principle that finds use in physics, economics, and engineering.

Understanding Integration

In simple terms, integration is the process of finding a function whose differential coefficient is given. The rate of change of a quantity $y$ with respect to another quantity $x$ is called the derivative or differential coefficient of $y$ concerning $x$. Geometrically, differentiation gives the slope of the tangent to the graph of a function at any point.

However, integration does the opposite. It finds the function from which that slope was derived. The integral of a function is based on a limiting process that approximates the area under a curve by dividing it into infinitely small vertical strips.

Example of Integration and Differentiation Relationship

$\begin{aligned}
& \frac{d}{d x}(\sin x)=\cos x \\
& \frac{d}{d x}\left(x^2\right)=2 x \\
& \frac{d}{d x}\left(e^x\right)=e^x
\end{aligned}$

Here, $ \cos(x) $ is the derivative of $ \sin(x) $, which means $ \sin(x) $ is the antiderivative (or integral) of $ \cos(x) $. Similarly, $ x^2 $ and $ e^x $ are the antiderivatives of $ 2x $ and $ e^x $, respectively.

Integration with Constant Term

The derivative of a constant ($C$) is always zero. Hence,

$\begin{aligned}
& \frac{d}{d x}(\sin x+c)=\cos x \\
& \frac{d}{d x}\left(x^2+c\right)=2 x \\
& \frac{d}{d x}\left(e^x+c\right)=e^x
\end{aligned}$

This means the antiderivative is not unique — there are infinitely many integrals for a given function, differing only by a constant $C$. The constant of integration represents this family of functions.

Definition of Indefinite Integral

If $F(x)$ is an antiderivative of $f(x)$, then $\int f(x),dx = F(x) + C$ where $F'(x) = f(x)$ and $C$ is an arbitrary constant.

Basic Terms in Integration

Rules of Integration

Let $f(x)$ and $g(x)$ be functions having antiderivatives $\int f(x),dx$ and $\int g(x),dx$. Then:

1. $\int kf(x),dx = k\int f(x),dx$, for any constant $k$

2. $\int (f(x) + g(x)),dx = \int f(x),dx + \int g(x),dx$

3. $\int (f(x) - g(x)),dx = \int f(x),dx - \int g(x),dx$

These rules form the foundation for solving more complex integration problems.

Standard Integration Formulae

Since $\frac{d}{dx}(F(x)) = f(x) \Leftrightarrow \int f(x),dx = F(x) + C$,
we can derive the following standard integration formulas:

1. $\int k,dx = kx + C$, where $k$ is a constant

2. $\int x^n,dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$

3. $\int \frac{1}{x},dx = \log|x| + C$, where $x \neq 0$

4. $\int e^x,dx = e^x + C$

5. $\int a^x,dx = \frac{a^x}{\log_e a} + C$, where $a > 0$ and $a \neq 1$

Why Integration is Important in Calculus

Integration helps in calculating:

• Area under curves and between curves

• Displacement and velocity from acceleration (in physics)

• Total growth, accumulation, and cost in economics

• Work done by variable forces and probability distributions

It’s a crucial tool that connects mathematics with the real world — from measuring distance covered to predicting change over time.

Definite vs Indefinite Integrals

This section compares definite and indefinite integrals, explaining their fundamental differences, uses, and interpretations. It highlights how indefinite integrals represent a family of functions with an arbitrary constant, while definite integrals calculate the exact area under a curve between two limits.

Solved Examples Based On Integration

Example 1: If a curve passes through the point $(1,-2)$ and has a slope of the tangent at any point $(x, y)$ on it as $\frac{x^2-2 y}{x}$, then the curve also passes through the point:
1) $(\sqrt{3}, 0)$
2) $(3,0)$
3) $(-1,2)$
4) $(-\sqrt{2}, 1)$

Solution
Equation of the tangent -
To find the equation of the tangent we need either one slope + one point or two points.

$
\therefore\left(y-y_0\right)=m\left(x_0-y_0\right)
$

or $\left(y-y_2\right)=\frac{y_2-y_1}{x_2-x_1}\left(x-x_2\right)$
where $\left(x_0, y_0\right)$ is the point on the curve

Geometrical interpretation of dy / dx -
$
\begin{aligned}
& \therefore \frac{d y}{d x}=\tan \theta \\
& \quad \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 y}{x}=x \\
& \text { If } e^{\int \frac{2}{x} d x}=e^{2 \ln x}=x^2 \\
& y\left(x^2\right)=\int x \cdot x^2 d x \\
& x^2 y=\frac{x^4}{4}+c \\
& \therefore y(1)=-2 \\
& \Rightarrow c=-\frac{9}{4}
\end{aligned}
$

$y=\frac{x^4-9}{4 x^2}$ which passes through $(\sqrt{3}, 0)$
Hence, the answer is the option (1).

Example 2: If the area (in sq. units) of the region $\left\{(x, y): y^2 \leq 4 x, x+y \leq 1, x \geq 0, y \geq 0\right\}$ is $a \sqrt{2}+b$, then $a-b$ is equal to :
1) $-\frac{2}{3}$
2) $\frac{10}{3}$
3) 6
4) $\frac{8}{3}$

Solution

Area between two curves -

If we have two functions intersection each other.First find the point of intersection. Then integrate to find area

$\int_o^a[f(x)-9(x)] d x$

- wherein

Indefinite integrals for Algebraic functions -
$\frac{\mathrm{d}}{\mathrm{d} x} \frac{\left(x^{\mathrm{n}+1}\right)}{n+1}=x^n \mathbb{m}^{\mathrm{j} 0} d x=\frac{x^{n+1}}{n+1}+c$

- wherein

Where $n \neq-1$

$\left\{(x, y): y^2 \leqslant 4 x ; x+y \leqslant 1, x \geqslant 0, y \geqslant 0\right\}$

$
\begin{aligned}
A & =\int_0^{3-2 \sqrt{2}} 2 \sqrt{x} d x+\frac{1}{2}(1-(3-2 \sqrt{2}))(1-(3-2 \sqrt{2})) \\
& =\frac{2\left[x^{\frac{3}{2}}\right]_0^{3-2 \sqrt{2}}}{\frac{3}{2}}+\frac{1}{2}(2 \sqrt{2}-2)(2 \sqrt{2}-2) \\
& =\frac{8 \sqrt{2}}{3}+\left(-\frac{10}{3}\right) \\
a & =\frac{8}{3}, b=\frac{-10}{3} \\
a & -b=\frac{8}{3}-\left(-\frac{10}{3}\right)=\frac{18}{3}=6
\end{aligned}
$

Example 3: The area (in sq. units) of the region $A=\left\{(x, y): x^2 \leq y \leq x+2\right\}$ is:
1) $\frac{10}{3}$
2) $\frac{9}{2}$
3) $\frac{31}{6}$
4) $\frac{13}{6}$

Solution

Area between two curves -

If we have two functions intersection each other.First find the point of intersection. Then integrate to find area
$
\int_a^b[f(x)-g(x)] d x
$
Here,
Point of intersection of $x^2=y$ and $y=x+2$ are $(-1,1)$ and $(2,4)$

$y=x+2$

Required Area,

$
\begin{aligned}
& =\int_{-1}^2\left(x+2-x^2\right) d x \\
& =\left[\frac{x^2}{2}+2 x-\frac{x^3}{3}\right]_{-1}^2 \\
& =\frac{9}{2}
\end{aligned}
$

Example 4: If the area ( in sq. units) bounded by the parabola $y^2=4 \lambda x$ and the line $y=\lambda x, \lambda>0$, is $\frac{1}{9}$, then $\lambda$ is equal to:
1) $2 \sqrt{6}$
2) 48
3) 24
4) $4 \sqrt{3}$

Solution

Area between two curves -

If we have two functions intersection each other. First find the point of intersection. Then integrate to find area

$\int_0^a[f(x)-9(x)] d x$

- wherein

Indefinite integrals for Algebraic functions -

$\frac{\mathrm{d}}{\mathrm{d} x} \frac{\left(x^{\mathrm{n}+1}\right)}{n+1}=x^n \int_{\mathrm{s}} x^n d x=\frac{x^{n+1}}{n+1}+c$

-wherein
Where $n \neq-1$

the parabola $y^2=4 \lambda x$ and the line $y=\lambda x$
If $\lambda>0$, then

$\begin{aligned}
\text { Area } & =\int_0^{4 / \lambda}(2 \sqrt{\lambda} \sqrt{x}-\lambda x) d x=\frac{1}{9} \\
& \Rightarrow\left[\frac{2 \sqrt{\lambda} x^{\frac{3}{2}}}{3 / 2}-\frac{\lambda x^2}{2}\right]_0^{4 / \lambda}=\frac{1}{9} \\
& =\frac{4}{3} \sqrt{\lambda} \frac{8}{\lambda^{\frac{1}{2}}}-\lambda \frac{8}{\lambda^2}=\frac{1}{9} \\
& \Rightarrow \frac{32}{3 \lambda}-\frac{8}{\lambda}=\frac{1}{9} \\
& \Rightarrow \frac{8}{3 \lambda}=\frac{1}{9} \\
& \Rightarrow \lambda=24
\end{aligned}$

$\text { If } f\left(\frac{x-4}{x+2}\right)=2 x+1, \quad(x \in \mathbb{R}-[1,-2])$
Example 5:
Then $\int f(x) d x$ equals
(where C is a constant of integration)
1) $12 \log _e|1-x|+3 x+C$
2) $-12 \log _e|1-x|-3 x+C$
3) $12 \log _c|1-x|-3 x+C$
4) $-12 \log _e|1-x|+3 x+C$

Solution

$\begin{aligned}
& \frac{x-4}{x+2}=y \Rightarrow x=\frac{2 y+4}{1-y} \\
& f(y)=2\left(\frac{2 y+4}{1-y}\right)+1=\frac{3 y+9}{1-y} \\
& \text { Now }_{\uparrow} f(x)=\frac{3 x+9}{1-x}
\end{aligned}$

Thus

$\begin{aligned}
& \int f(x) d x=\int \frac{3 x+9}{1-x} d x=\int\left(-3+\frac{12}{-x+1}\right) d x \\
& =-3 x-12 \ln |-x+1|+C
\end{aligned}$

Hence, the answer is the option 2.

List of Topics related to Integration as Reverse Process of Differentiation

This section provides a structured overview of all major subtopics connected to integration as the reverse process of differentiation, helping you understand the fundamental links between integral calculus and its practical applications.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration by Parts

Application of Inequality in Definite Integration

NCERT Resources

Here you’ll find curated NCERT-based study materials, including solved examples, exemplar problems, detailed notes, and exercises for mastering the Integrals chapter effectively.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on Integration as the Reverse Process of Differentiation

This section offers topic-wise practice questions and MCQs designed to strengthen conceptual understanding and improve problem-solving accuracy in integration.

Integration As An Inverse Process Of Differentiation- Practice Question MCQ

We have provided below the practice questions related to different concepts of integration to improve your understanding: