This MCQ module is based on: Introduction and Integration as Inverse of Differentiation
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Introduction and Integration as Inverse of Differentiation
🎓 Class 12
Mathematics
CBSE
Theory
Ch 7 — Integrals
⏱ ~25 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Introduction and Integration as Inverse of Differentiation
Targeting Class 12 level in Calculus, with Advanced difficulty.
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7.1 Introduction
Differential Calculus centres on the concept of the derivative?. The original motivation for the derivative was the problem of defining tangent lines to the graphs of functions and calculating the slope of such lines. Integral Calculus? is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions.
If a function \(f\) is differentiable in an interval I, its derivative \(f'\) exists at each point of I. A natural question arises: given \(f'\) at each point of I, can we determine the function? The functions that could possibly have given function as a derivative are called anti derivatives (or primitives?) of the function.
Definition
The formula that gives all these anti derivatives is called the indefinite integral of the function, and the process of finding anti derivatives is called integration.
The development of integral calculus arises from solving two types of problems:
- (a) Finding a function whenever its derivative is given
- (b) Finding the area bounded by the graph of a function under certain conditions
These two problems lead to the two forms of integrals — indefinite and definite integrals — which together constitute the Integral Calculus. There is a connection known as the Fundamental Theorem of Calculus between indefinite and definite integrals, which makes the definite integral a practical tool for science and engineering.
Historical Note
G. W. Leibnitz (1646–1716) and Isaac Newton (1642–1727) independently developed the foundations of calculus. Leibnitz introduced the integral sign \(\int\), derived from an elongated letter "S" (for summa, meaning "sum"). The concept of the definite integral as the limit of a sum was formalised by A. L. Cauchy in the early 19th century.
7.2 Integration as an Inverse Process of Differentiation
Integration is the inverse process of differentiation. Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive?, i.e., the original function. Such a process is called integration or anti differentiation.
Consider the following examples:
We know that:
\[\frac{d}{dx}(\sin x) = \cos x \quad \cdots (1)\] \[\frac{d}{dx}\left(\frac{x^3}{3}\right) = x^2 \quad \cdots (2)\] \[\frac{d}{dx}(e^x) = e^x \quad \cdots (3)\]We observe that in (1), the function \(\cos x\) is the derived function of \(\sin x\). We say that \(\sin x\) is an anti derivative (or an integral) of \(\cos x\). Similarly, from (2) and (3), \(\frac{x^3}{3}\) and \(e^x\) are the anti derivatives of \(x^2\) and \(e^x\), respectively.
Important
Anti derivatives (or integrals) of the above cited functions are not unique. For any real number C, treated as a constant, its derivative is zero. Hence we can write:\(\frac{d}{dx}(\sin x + C) = \cos x\), \(\frac{d}{dx}\left(\frac{x^3}{3} + C\right) = x^2\), \(\frac{d}{dx}(e^x + C) = e^x\)
C is customarily referred to as the arbitrary constant or the parameter.
Notation and Symbols
We introduce a new symbol: \(\int f(x)\,dx\) which will represent the entire class of anti derivatives, read as the indefinite integral of \(f\) with respect to \(x\).
Symbolically, we write \(\int f(x)\,dx = F(x) + C\).
| Symbol / Term | Meaning |
|---|---|
| \(\int f(x)\,dx\) | Integral of \(f\) with respect to \(x\) |
| \(f(x)\) in \(\int f(x)\,dx\) | Integrand |
| \(x\) in \(\int f(x)\,dx\) | Variable of integration |
| Integrate | Find the integral |
| An integral of \(f\) | A function F such that \(F'(x) = f(x)\) |
| Integration | The process of finding the integral |
| C | Constant of integration (arbitrary constant) |
Standard Formulae (Derivatives and Their Corresponding Integrals)
From known derivative formulae, we can write down the corresponding integral formulae immediately:
| Derivative | Integral (Anti derivative) |
|---|---|
| \(\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) = x^n\) | \(\int x^n\,dx = \frac{x^{n+1}}{n+1} + C,\; n \neq -1\) |
| \(\frac{d}{dx}(\sin x) = \cos x\) | \(\int \cos x\,dx = \sin x + C\) |
| \(\frac{d}{dx}(-\cos x) = \sin x\) | \(\int \sin x\,dx = -\cos x + C\) |
| \(\frac{d}{dx}(\tan x) = \sec^2 x\) | \(\int \sec^2 x\,dx = \tan x + C\) |
| \(\frac{d}{dx}(-\cot x) = \csc^2 x\) | \(\int \csc^2 x\,dx = -\cot x + C\) |
| \(\frac{d}{dx}(\sec x) = \sec x \tan x\) | \(\int \sec x \tan x\,dx = \sec x + C\) |
| \(\frac{d}{dx}(-\csc x) = \csc x \cot x\) | \(\int \csc x \cot x\,dx = -\csc x + C\) |
| \(\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}\) | \(\int \frac{1}{\sqrt{1-x^2}}\,dx = \sin^{-1} x + C\) |
| \(\frac{d}{dx}(-\cos^{-1} x) = \frac{1}{\sqrt{1-x^2}}\) | (same as above, since \(-\cos^{-1}x = \sin^{-1}x + \text{const}\)) |
| \(\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}\) | \(\int \frac{1}{1+x^2}\,dx = \tan^{-1} x + C\) |
| \(\frac{d}{dx}(-\cot^{-1} x) = \frac{1}{1+x^2}\) | (same as above) |
| \(\frac{d}{dx}(e^x) = e^x\) | \(\int e^x\,dx = e^x + C\) |
| \(\frac{d}{dx}\left(\frac{a^x}{\log a}\right) = a^x\) | \(\int a^x\,dx = \frac{a^x}{\log a} + C\) |
| \(\frac{d}{dx}(\log|x|) = \frac{1}{x}\) | \(\int \frac{1}{x}\,dx = \log|x| + C\) |
Note
In practice, we normally do not mention the interval over which the various functions are defined. However, in any specific problem one has to keep it in mind.
7.2.1 Some Properties of Indefinite Integrals
Property I
The process of differentiation and integration are inverses of each other:\[\frac{d}{dx}\int f(x)\,dx = f(x)\] and \[\int f'(x)\,dx = f(x) + C\] where C is any arbitrary constant.
Property II
Two indefinite integrals with the same derivative lead to the same family of curves and so are equivalent:If \(\frac{d}{dx}\int f(x)\,dx = \frac{d}{dx}\int g(x)\,dx\), then \(\int f(x)\,dx = \int g(x)\,dx + C\).
Property III — Sum Rule
\[\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx\]
Property IV — Scalar Multiple Rule
For any real number \(k\),
\[\int k\,f(x)\,dx = k\int f(x)\,dx\]
Property V — Generalised Sum
Properties (III) and (IV) can be generalised to a finite number of functions \(f_1, f_2, \ldots, f_n\) and real numbers \(k_1, k_2, \ldots, k_n\):\[\int [k_1 f_1(x) + k_2 f_2(x) + \cdots + k_n f_n(x)]\,dx = k_1\int f_1(x)\,dx + k_2\int f_2(x)\,dx + \cdots + k_n\int f_n(x)\,dx\]
Worked Examples
Example 1
Write an anti derivative for each of the following functions using the method of inspection:
(i) \(\cos 2x\) (ii) \(3x^2 + 4x^3\) (iii) \(\frac{1}{x},\; x \neq 0\)
Solution
(i) We look for a function whose derivative is \(\cos 2x\). Recall that \[\frac{d}{dx}(\sin 2x) = 2\cos 2x\] Therefore, an anti derivative of \(\cos 2x\) is \(\frac{1}{2}\sin 2x\).
(ii) We look for a function whose derivative is \(3x^2 + 4x^3\). Note that \[\frac{d}{dx}(x^3 + x^4) = 3x^2 + 4x^3\] Therefore, an anti derivative of \(3x^2 + 4x^3\) is \(x^3 + x^4\).
(iii) We know that \(\frac{d}{dx}(\log|x|) = \frac{1}{x},\; x \neq 0\). Therefore, \(\log|x|\) is an anti derivative of \(\frac{1}{x}\).
Example 2
Find the following integrals:
(i) \(\int \frac{x^3 - 1}{x^2}\,dx\) (ii) \(\int (x^{2/3} + 1)\,dx\) (iii) \(\int \left(x^{3/2} + 2e^x - \frac{1}{x}\right)\,dx\)
Solution
(i) We have: \[\int \frac{x^3 - 1}{x^2}\,dx = \int \left(x - x^{-2}\right)\,dx = \frac{x^2}{2} - \frac{x^{-1}}{-1} + C = \frac{x^2}{2} + \frac{1}{x} + C\]
(ii) We have: \[\int (x^{2/3} + 1)\,dx = \frac{x^{2/3+1}}{2/3+1} + x + C = \frac{x^{5/3}}{5/3} + x + C = \frac{3}{5}x^{5/3} + x + C\]
(iii) We have: \[\int \left(x^{3/2} + 2e^x - \frac{1}{x}\right)\,dx = \frac{x^{5/2}}{5/2} + 2e^x - \log|x| + C = \frac{2}{5}x^{5/2} + 2e^x - \log|x| + C\]
Example 3
Find the following integrals:
(i) \(\int (\sin x + \cos x)\,dx\) (ii) \(\int \text{cosec}\,x(\text{cosec}\,x + \cot x)\,dx\)
Solution
(i) \[\int (\sin x + \cos x)\,dx = -\cos x + \sin x + C\]
(ii) \[\int \text{cosec}\,x(\text{cosec}\,x + \cot x)\,dx = \int (\text{cosec}^2 x + \text{cosec}\,x\cot x)\,dx = -\cot x - \text{cosec}\,x + C\]
Example 4
Find the anti derivative F of \(f\) defined by \(f(x) = 4x^3 - 6\), where \(F(0) = 3\).
Solution
One anti derivative of \(f(x) = 4x^3 - 6\) is \(x^4 - 6x\).
Therefore, \(F(x) = x^4 - 6x + C\), where C is constant.
Given that \(F(0) = 3\), which gives:
\(3 = 0 - 0 + C \Rightarrow C = 3\)
Hence, the required anti derivative is \(F(x) = x^4 - 6x + 3\).
Interactive: Verify Integration by Differentiation
Enter a power \(n\) to see \(\int x^n\,dx\) and verify by differentiating the result
\(\int x^3\,dx = \frac{x^{4}}{4} + C\) Check: \(\frac{d}{dx}\left(\frac{x^{4}}{4}\right) = x^3\) ✓
Exercise 7.1
Find an anti derivative (or integral) of the following functions by the method of inspection:
1. \(\sin 2x\)
We know \(\frac{d}{dx}(-\cos 2x) = 2\sin 2x\). Therefore, \(\int \sin 2x\,dx = -\frac{1}{2}\cos 2x + C\).
An anti derivative is \(-\frac{1}{2}\cos 2x\).
An anti derivative is \(-\frac{1}{2}\cos 2x\).
2. \(\cos 3x\)
\(\frac{d}{dx}(\sin 3x) = 3\cos 3x\). Therefore, an anti derivative of \(\cos 3x\) is \(\frac{1}{3}\sin 3x\).
3. \(e^{2x}\)
\(\frac{d}{dx}(e^{2x}) = 2e^{2x}\). Therefore, an anti derivative of \(e^{2x}\) is \(\frac{1}{2}e^{2x}\).
4. \((ax + b)^2\)
\(\frac{d}{dx}\left(\frac{(ax+b)^3}{3a}\right) = (ax+b)^2\). Therefore, an anti derivative is \(\frac{(ax+b)^3}{3a}\).
5. \(\sin 2x - 4e^{3x}\)
Anti derivative of \(\sin 2x\) is \(-\frac{1}{2}\cos 2x\). Anti derivative of \(4e^{3x}\) is \(\frac{4}{3}e^{3x}\).
Therefore, an anti derivative is \(-\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x}\).
Therefore, an anti derivative is \(-\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x}\).
Find the following integrals (Exercises 6 to 20):
6. \(\int (4e^{3x} + 1)\,dx\)
\[\int (4e^{3x} + 1)\,dx = \frac{4}{3}e^{3x} + x + C\]
7. \(\int x^2\left(1 - \frac{1}{x^2}\right)\,dx\)
\[\int x^2\left(1 - \frac{1}{x^2}\right)\,dx = \int (x^2 - 1)\,dx = \frac{x^3}{3} - x + C\]
8. \(\int (ax^2 + bx + c)\,dx\)
\[\int (ax^2 + bx + c)\,dx = \frac{ax^3}{3} + \frac{bx^2}{2} + cx + C\]
9. \(\int (2x^2 + e^x)\,dx\)
\[\int (2x^2 + e^x)\,dx = \frac{2x^3}{3} + e^x + C\]
10. \(\int \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2\,dx\)
\[\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 = x - 2 + \frac{1}{x}\]
\[\int \left(x - 2 + \frac{1}{x}\right)\,dx = \frac{x^2}{2} - 2x + \log|x| + C\]
11. \(\int \frac{x^3 + 5x^2 - 4}{x^2}\,dx\)
\[\int \frac{x^3 + 5x^2 - 4}{x^2}\,dx = \int \left(x + 5 - 4x^{-2}\right)\,dx = \frac{x^2}{2} + 5x + \frac{4}{x} + C\]
12. \(\int \frac{x^3 + 3x + 4}{\sqrt{x}}\,dx\)
\[\int \frac{x^3 + 3x + 4}{\sqrt{x}}\,dx = \int (x^{5/2} + 3x^{1/2} + 4x^{-1/2})\,dx = \frac{2}{7}x^{7/2} + 2x^{3/2} + 8x^{1/2} + C\]
13. \(\int \frac{x^3 - x^2 + x - 1}{x - 1}\,dx\)
Factoring: \(x^3 - x^2 + x - 1 = x^2(x-1) + 1(x-1) = (x^2+1)(x-1)\).
\[\int \frac{(x^2+1)(x-1)}{x-1}\,dx = \int (x^2 + 1)\,dx = \frac{x^3}{3} + x + C\]
\[\int \frac{(x^2+1)(x-1)}{x-1}\,dx = \int (x^2 + 1)\,dx = \frac{x^3}{3} + x + C\]
14. \(\int (1 - x)\sqrt{x}\,dx\)
\[\int (1-x)\sqrt{x}\,dx = \int (x^{1/2} - x^{3/2})\,dx = \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C\]
15. \(\int \sqrt{x}\left(3x^2 + 2x + 3\right)\,dx\)
\[\int (3x^{5/2} + 2x^{3/2} + 3x^{1/2})\,dx = \frac{6}{7}x^{7/2} + \frac{4}{5}x^{5/2} + 2x^{3/2} + C\]
16. \(\int \frac{2x - 3\cos x + e^x}{1}\,dx\)
\[\int (2x - 3\cos x + e^x)\,dx = x^2 - 3\sin x + e^x + C\]
17. \(\int (2x^2 - 3\sin x + 5\sqrt{x})\,dx\)
\[\int (2x^2 - 3\sin x + 5\sqrt{x})\,dx = \frac{2x^3}{3} + 3\cos x + \frac{10}{3}x^{3/2} + C\]
18. \(\int \sec x(\sec x + \tan x)\,dx\)
\[\int (\sec^2 x + \sec x\tan x)\,dx = \tan x + \sec x + C\]
19. \(\int \frac{\sec^2 x}{\text{cosec}^2 x}\,dx\)
\[\frac{\sec^2 x}{\text{cosec}^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x = \sec^2 x - 1\]
\[\int (\sec^2 x - 1)\,dx = \tan x - x + C\]
20. \(\int \frac{3 - 2\sin x}{\cos^2 x}\,dx\)
\[\int \left(\frac{3}{\cos^2 x} - \frac{2\sin x}{\cos^2 x}\right)\,dx = \int (3\sec^2 x - 2\sec x \tan x)\,dx = 3\tan x - 2\sec x + C\]
Choose the correct answer in Exercises 21 and 22:
21. The anti derivative of \(\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)\) equals:
(A) \(\frac{1}{3}x^{1/3} + 2x^{1/2} + C\) (B) \(\frac{2}{3}x^{3/2} + 2x^{1/2} + C\) (C) \(\frac{2}{3}x^{3/2} + \frac{1}{2}x^{1/2} + C\) (D) \(\frac{3}{2}x^{3/2} + \frac{1}{2}x^{1/2} + C\)
(A) \(\frac{1}{3}x^{1/3} + 2x^{1/2} + C\) (B) \(\frac{2}{3}x^{3/2} + 2x^{1/2} + C\) (C) \(\frac{2}{3}x^{3/2} + \frac{1}{2}x^{1/2} + C\) (D) \(\frac{3}{2}x^{3/2} + \frac{1}{2}x^{1/2} + C\)
\(\int (x^{1/2} + x^{-1/2})\,dx = \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C = \frac{2}{3}x^{3/2} + 2x^{1/2} + C\).
Answer: (B)
Answer: (B)
22. If \(\frac{d}{dx}f(x) = 4x^3 - \frac{3}{x^4}\) such that \(f(2) = 0\). Then \(f(x)\) is:
(A) \(x^4 + \frac{1}{x^3} - \frac{129}{8}\) (B) \(x^3 + \frac{1}{x^4} + \frac{129}{8}\) (C) \(x^4 + \frac{1}{x^3} + \frac{129}{8}\) (D) \(x^3 + \frac{1}{x^4} - \frac{129}{8}\)
(A) \(x^4 + \frac{1}{x^3} - \frac{129}{8}\) (B) \(x^3 + \frac{1}{x^4} + \frac{129}{8}\) (C) \(x^4 + \frac{1}{x^3} + \frac{129}{8}\) (D) \(x^3 + \frac{1}{x^4} - \frac{129}{8}\)
\(f(x) = \int (4x^3 - 3x^{-4})\,dx = x^4 + x^{-3} + C = x^4 + \frac{1}{x^3} + C\).
Given \(f(2) = 0\): \(16 + \frac{1}{8} + C = 0 \Rightarrow C = -\frac{129}{8}\).
\(f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}\). Answer: (A)
Given \(f(2) = 0\): \(16 + \frac{1}{8} + C = 0 \Rightarrow C = -\frac{129}{8}\).
\(f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}\). Answer: (A)
Activity: Matching Derivatives and Integrals
L3 Apply
Materials: Paper, pencil
Predict: Can you match each derivative to its original function (anti derivative) without looking at the formula table?
- Write down these derivatives on separate slips: \(\cos x\), \(2x\), \(e^x\), \(\sec^2 x\), \(\frac{1}{x}\), \(3x^2\)
- Write their anti derivatives on different slips: \(\sin x\), \(x^2\), \(e^x\), \(\tan x\), \(\log|x|\), \(x^3\)
- Shuffle both sets and try to match them as quickly as possible.
- Time yourself. Repeat until you can match all 6 pairs in under 30 seconds.
Observation: Integration reverses differentiation. The key insight is that the power rule in reverse adds 1 to the exponent and divides by the new exponent. Memorising the standard formulae table is essential for speed in board exams.
Competency-Based Questions
Scenario: A physics student models the velocity of a particle at time \(t\) seconds as \(v(t) = 6t^2 - 4t + 1\) m/s. The displacement \(s(t)\) is the anti derivative of velocity, and \(s(0) = 2\) m.
Q1. The displacement function \(s(t)\) is:
L3 ApplyAnswer: (a). \(s(t) = \int (6t^2 - 4t + 1)\,dt = 2t^3 - 2t^2 + t + C\). Since \(s(0) = 2\), we get \(C = 2\). So \(s(t) = 2t^3 - 2t^2 + t + 2\).
Q2. At what time \(t\) is the velocity zero? Interpret what this means physically.
L4 AnalyseAnswer: Setting \(v(t) = 0\): \(6t^2 - 4t + 1 = 0\). Discriminant = \(16 - 24 = -8 < 0\). Since the discriminant is negative, the velocity is never zero. The particle never stops; it always moves in the positive direction since \(v(t) > 0\) for all \(t\) (as the leading coefficient is positive and there are no real roots).
Q3. The acceleration \(a(t) = v'(t)\). Find when the acceleration is zero and explain whether the particle speeds up or slows down at that instant.
L5 EvaluateAnswer: \(a(t) = \frac{d}{dt}(6t^2 - 4t + 1) = 12t - 4\). Setting \(a = 0\): \(t = \frac{1}{3}\) s. At \(t = \frac{1}{3}\), \(v = 6 \cdot \frac{1}{9} - 4 \cdot \frac{1}{3} + 1 = \frac{2}{3} - \frac{4}{3} + 1 = \frac{1}{3} > 0\). Since velocity is positive and acceleration changes from negative to positive at this point, the particle is momentarily neither speeding up nor slowing down; it is at the minimum velocity.
Q4. If another particle has velocity \(w(t) = k \cdot v(t)\) where \(k\) is a constant, write its displacement using Property IV of integrals. What does the constant \(k\) physically represent?
L6 CreateAnswer: By Property IV: \(\int k \cdot v(t)\,dt = k \int v(t)\,dt = k \cdot s(t) + C'\). So displacement is \(k(2t^3 - 2t^2 + t) + C'\). The constant \(k\) represents a scaling factor for speed — e.g., if \(k = 2\), the second particle moves twice as fast and covers twice the displacement in the same time interval.
Assertion–Reason Questions
Assertion (A): \(\int x^{-1}\,dx = \frac{x^0}{0} + C\) is undefined, so we use \(\log|x| + C\) instead.
Reason (R): The power rule \(\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\) is not valid when \(n = -1\).
Reason (R): The power rule \(\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\) is not valid when \(n = -1\).
Answer: (a) — Both statements are true. When \(n = -1\), the denominator \(n+1 = 0\) makes the power rule undefined. This is precisely why we need the separate formula \(\int \frac{1}{x}\,dx = \log|x| + C\). R correctly explains A.
Assertion (A): \(\int (\sin x + \cos x)\,dx = -\cos x + \sin x + C\).
Reason (R): \(\int [f(x) + g(x)]\,dx = \int f(x)\,dx \cdot \int g(x)\,dx\).
Reason (R): \(\int [f(x) + g(x)]\,dx = \int f(x)\,dx \cdot \int g(x)\,dx\).
Answer: (c) — A is true: \(\int \sin x\,dx + \int \cos x\,dx = -\cos x + \sin x + C\). R is false: the integral of a sum equals the sum of integrals (not the product). The correct property is \(\int [f + g]\,dx = \int f\,dx + \int g\,dx\).
Assertion (A): If \(F(x) = x^4 - 6x + 3\), then \(F'(x) = 4x^3 - 6\).
Reason (R): The derivative of a constant is zero.
Reason (R): The derivative of a constant is zero.
Answer: (a) — Both true. \(F'(x) = 4x^3 - 6 + 0 = 4x^3 - 6\). The derivative of the constant 3 is zero (as R states), which is why the "+3" disappears. R explains A.
Frequently Asked Questions
What is integration in mathematics?
Integration is the inverse process of differentiation. Given a derivative, integration finds the original function. The result includes a constant of integration C because many functions share the same derivative.
What are the basic integration formulas?
Key formulas: integral of x^n = x^(n+1)/(n+1) + C, integral of sin x = -cos x + C, integral of cos x = sin x + C, integral of e^x = e^x + C, integral of 1/x = log|x| + C.
What is the constant of integration?
The constant C accounts for the fact that many functions have the same derivative. Since the derivative of any constant is zero, all antiderivatives differ by a constant.
What is the difference between indefinite and definite integrals?
An indefinite integral gives a family of functions with constant C. A definite integral has limits and gives a numerical value representing area under the curve.
What are the properties of indefinite integrals?
Linearity: integral of sum equals sum of integrals. Scalar multiplication: constants factor out. The derivative of the integral of f(x) equals f(x).
Who invented integral calculus?
Newton and Leibniz independently developed calculus. Leibniz introduced the integral sign from elongated S for summa. Cauchy later formalized the definite integral as a limit of sums.
Frequently Asked Questions — Integrals
What is Introduction and Integration as Inverse of Differentiation in NCERT Class 12 Mathematics?
Introduction and Integration as Inverse of Differentiation is a key concept covered in NCERT Class 12 Mathematics, Chapter 7: Integrals. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Introduction and Integration as Inverse of Differentiation step by step?
To solve problems on Introduction and Integration as Inverse of Differentiation, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 7: Integrals?
The essential formulas of Chapter 7 (Integrals) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Introduction and Integration as Inverse of Differentiation important for the Class 12 board exam?
Introduction and Integration as Inverse of Differentiation is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Introduction and Integration as Inverse of Differentiation?
Common mistakes in Introduction and Integration as Inverse of Differentiation include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Introduction and Integration as Inverse of Differentiation?
End-of-chapter NCERT exercises for Introduction and Integration as Inverse of Differentiation cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 7, and solve at least one previous-year board paper to consolidate your understanding.
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