IIT JEE Main Maths -Unit 8- Evaluation of Simple and Definite Integrals- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 8- Evaluation of Simple and Definite Integrals – Study Notes – New syllabus
IIT JEE Main Maths -Unit 8- Evaluation of Simple and Definite Integrals – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Evaluation of Simple and Definite Integrals
- Evaluation of Simple Integrals of the Type
- Reduction Formulae and Special Trigonometric Integrals
Evaluation of Simple and Definite Integrals
Integration is the reverse process of differentiation. The value of an integral can be of two types:
- Indefinite Integral: Represents a family of functions and includes a constant of integration \( C \).
- Definite Integral: Represents a numerical value corresponding to the area under a curve between two limits \( a \) and \( b \).
Indefinite Integral (Recap)
$ \int f(x)\,dx = F(x) + C $
Here, \( F'(x) = f(x) \). This represents the general anti-derivative of \( f(x) \).
Definite Integral — Definition
If \( F(x) \) is an anti-derivative of \( f(x) \) on the interval \([a, b]\), then
$ \int_a^b f(x)\,dx = F(b) – F(a) $
This is known as the Fundamental Theorem of Calculus.
Properties of Definite Integrals
| Property | Formula |
|---|---|
| Additivity | \( \displaystyle \int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx \) |
| Reversal of limits | \( \displaystyle \int_a^b f(x)\,dx = -\int_b^a f(x)\,dx \) |
| Zero width | \( \displaystyle \int_a^a f(x)\,dx = 0 \) |
| Constant multiple | \( \displaystyle \int_a^b k f(x)\,dx = k \int_a^b f(x)\,dx \) |
| Linearity | \( \displaystyle \int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx \) |
Geometrical Meaning
The definite integral \( \displaystyle \int_a^b f(x)\,dx \) represents the net area between the curve \( y = f(x) \) and the x-axis from \( x = a \) to \( x = b \):
- Area above x-axis → positive
- Area below x-axis → negative
Evaluation Technique
- Find the indefinite integral \( F(x) \) of \( f(x) \).
- Apply the limits: \( F(b) – F(a) \).
- Simplify to obtain the numerical result.
Example
Evaluate \( \displaystyle \int_0^2 (3x^2 + 2x + 1)\,dx \).
▶️ Answer / Explanation
Step 1: Integrate each term:
\( \displaystyle \int (3x^2 + 2x + 1)\,dx = x^3 + x^2 + x + C. \)
Step 2: Apply limits from 0 to 2:
\( [x^3 + x^2 + x]_0^2 = (8 + 4 + 2) – 0 = 14. \)
Answer: \( 14. \)
Example
Evaluate \( \displaystyle \int_0^{\pi/2} \sin x\,dx \).
▶️ Answer / Explanation
Step 1: \( \displaystyle \int \sin x\,dx = -\cos x + C. \)
Step 2: Apply limits \( 0 \) to \( \pi/2 \):
\( [-\cos x]_0^{\pi/2} = [-(0 – 1)] = 1. \)
Answer: \( 1. \)
Example
Evaluate \( \displaystyle \int_0^1 x e^x\,dx \).
▶️ Answer / Explanation
Step 1: Use integration by parts:
Let \( u = x \Rightarrow du = dx \), \( dv = e^x dx \Rightarrow v = e^x. \)
Then, \( \displaystyle \int x e^x dx = x e^x – \int e^x dx = e^x(x – 1) + C. \)
Step 2: Apply limits \( 0 \) to \( 1 \):
\( [e^x(x – 1)]_0^1 = [e(1 – 1)] – [1(0 – 1)] = 1 – 1/e. \)
Answer: \( 1 – \dfrac{1}{e}. \)
Properties of Definite Integrals (Symmetry and Substitution)
- Property 1: \( \displaystyle \int_0^a f(x)\,dx = \int_0^a f(a – x)\,dx \)
- Property 2: \( \displaystyle \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx \) if \( f(x) \) is even.
- Property 3: \( \displaystyle \int_{-a}^{a} f(x)\,dx = 0 \) if \( f(x) \) is odd.
Example
Evaluate \( \displaystyle \int_{-2}^2 x^3\,dx \).
▶️ Answer / Explanation
\( x^3 \) is an odd function since \( f(-x) = -f(x) \).
Thus, \( \displaystyle \int_{-a}^{a} f(x)\,dx = 0 \) for odd functions.
Answer: \( 0. \)
Evaluation of Simple Integrals of the Type
Below are standard integral forms that are essential for JEE Mains and Advanced. These can be directly used while solving definite or indefinite integrals after applying appropriate substitutions.
Standard Integrals Involving Rational Functions
| Integral | Result |
|---|---|
| \( \displaystyle \int \dfrac{dx}{x^2 + a^2} \) | \( \dfrac{1}{a} \tan^{-1}\!\left(\dfrac{x}{a}\right) + C \) |
| \( \displaystyle \int \dfrac{dx}{x^2 – a^2} \) | \( \dfrac{1}{2a}\ln\!\left|\dfrac{x – a}{x + a}\right| + C \) |
| \( \displaystyle \int \dfrac{dx}{a^2 – x^2} \) | \( \dfrac{1}{2a}\ln\!\left|\dfrac{a + x}{a – x}\right| + C \) |
Standard Integrals Involving Square Roots
| Integral | Result |
|---|---|
| \( \displaystyle \int \dfrac{dx}{\sqrt{x^2 + a^2}} \) | \( \ln\!\left|x + \sqrt{x^2 + a^2}\right| + C \) |
| \( \displaystyle \int \dfrac{dx}{\sqrt{a^2 – x^2}} \) | \( \sin^{-1}\!\left(\dfrac{x}{a}\right) + C \) |
| \( \displaystyle \int \sqrt{x^2 + a^2}\,dx \) | \( \dfrac{x}{2}\sqrt{x^2 + a^2} + \dfrac{a^2}{2}\ln\!\left|x + \sqrt{x^2 + a^2}\right| + C \) |
| \( \displaystyle \int \sqrt{a^2 – x^2}\,dx \) | \( \dfrac{x}{2}\sqrt{a^2 – x^2} + \dfrac{a^2}{2}\sin^{-1}\!\left(\dfrac{x}{a}\right) + C \) |
Integrals Involving Quadratic Expressions
For \( ax^2 + bx + c \), use completing the square before applying standard results:
$ ax^2 + bx + c = a\left[\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac – b^2}{4a^2}\right] $
| Integral | Result |
|---|---|
| \( \displaystyle \int \dfrac{dx}{ax^2 + bx + c} \) | Depends on the discriminant \( D = b^2 – 4ac \): |
| |
Integrals of the Type \( \displaystyle \int \dfrac{px + q}{ax^2 + bx + c}\,dx \)
This can be split as:
$ \int \dfrac{px + q}{ax^2 + bx + c}\,dx = \dfrac{p}{2a}\ln|ax^2 + bx + c| + \int \dfrac{\left(q – \dfrac{bp}{2a}\right)\,dx}{ax^2 + bx + c} $
The remaining integral can be solved using the discriminant-based formula from above.
Integrals Involving \( \sqrt{ax^2 + bx + c} \)
Use substitution or completing the square depending on the form.
- If \( b = 0 \), use standard forms from (2).
- If \( b \neq 0 \), complete the square, then substitute \( x + \dfrac{b}{2a} = t \).
Quick Reference — Common Substitutions
| Expression | Substitute | Resulting Simplification |
|---|---|---|
| \( \sqrt{a^2 – x^2} \) | \( x = a\sin\theta \) | \( \sqrt{a^2 – x^2} = a\cos\theta \) |
| \( \sqrt{x^2 + a^2} \) | \( x = a\tan\theta \) | \( \sqrt{x^2 + a^2} = a\sec\theta \) |
| \( \sqrt{x^2 – a^2} \) | \( x = a\sec\theta \) | \( \sqrt{x^2 – a^2} = a\tan\theta \) |
Example
Evaluate \( \displaystyle \int \dfrac{dx}{\sqrt{x^2 + 9}} \).
▶️ Answer / Explanation
Here \( a = 3 \).
Using standard formula: \( \displaystyle \int \dfrac{dx}{\sqrt{x^2 + a^2}} = \ln|x + \sqrt{x^2 + a^2}| + C. \)
\( \Rightarrow \boxed{\ln|x + \sqrt{x^2 + 9}| + C.} \)
Example
Evaluate \( \displaystyle \int \sqrt{16 – x^2}\,dx \).
▶️ Answer / Explanation
Using \( \displaystyle \int \sqrt{a^2 – x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 – x^2} + \dfrac{a^2}{2}\sin^{-1}\!\left(\dfrac{x}{a}\right) + C. \)
\( a = 4 \Rightarrow \boxed{2x\sqrt{16 – x^2} + 8\sin^{-1}\!\left(\dfrac{x}{4}\right) + C.} \)
Reduction Formulae and Special Trigonometric Integrals
Some integrals contain powers of trigonometric, exponential, or logarithmic functions that are difficult to evaluate directly. In such cases, reduction formulae help express a given integral in terms of another with a lower power, allowing for recursive evaluation.
Reduction Formula — Concept
If \( I_n = \int f_n(x)\,dx \) and \( I_n \) can be expressed in terms of \( I_{n-1} \), then \( I_n \) is said to have a reduction formula.
That is,
$ I_n = g(x, n) + h(n)\,I_{n-1} $
Reduction Formulae for Trigonometric Integrals
| Integral Type | Reduction Formula |
|---|---|
| \( I_n = \displaystyle \int \sin^n x \, dx \) | \( I_n = -\dfrac{\sin^{n-1}x\cos x}{n} + \dfrac{n-1}{n}I_{n-2} \) |
| \( I_n = \displaystyle \int \cos^n x \, dx \) | \( I_n = \dfrac{\sin x\cos^{n-1}x}{n} + \dfrac{n-1}{n}I_{n-2} \) |
| \( I_n = \displaystyle \int \tan^n x \, dx \) | \( I_n = \dfrac{1}{n-1}\tan^{n-1}x – I_{n-2} \) |
| \( I_n = \displaystyle \int \sec^n x \, dx \) | \( I_n = \dfrac{\sec^{n-2}x\tan x}{n-1} + \dfrac{n-2}{n-1}I_{n-2} \) |
Reduction Formula for Exponential and Logarithmic Forms
| Integral Type | Reduction Formula |
|---|---|
| \( I_n = \displaystyle \int x^n e^{ax}\,dx \) | \( I_n = \dfrac{x^n e^{ax}}{a} – \dfrac{n}{a}I_{n-1} \) |
| \( I_n = \displaystyle \int x^n \sin(ax)\,dx \) | \( I_n = -\dfrac{x^n \cos(ax)}{a} + \dfrac{n}{a}I_{n-1}^\cos \) |
| \( I_n = \displaystyle \int x^n \cos(ax)\,dx \) | \( I_n = \dfrac{x^n \sin(ax)}{a} – \dfrac{n}{a}I_{n-1}^\sin \) |
| \( I_n = \displaystyle \int (\ln x)^n \, dx \) | \( I_n = x(\ln x)^n – nI_{n-1} \) |
Reduction Formula for Powers of \( (a^2 + x^2) \)
For \( I_n = \displaystyle \int \dfrac{dx}{(a^2 + x^2)^n} \):
$ I_n = \dfrac{x}{2a^2(n – 1)(a^2 + x^2)^{n – 1}} + \dfrac{2n – 3}{2a^2(n – 1)}I_{n-1} $
Reduction Formula for \( \displaystyle \int \sin^m x \cos^n x \, dx \)
- If \( n \) is odd, use \( \cos^2x = 1 – \sin^2x \) and substitute \( \sin x = t \).
- If \( m \) is odd, use \( \sin^2x = 1 – \cos^2x \) and substitute \( \cos x = t \).
- If both are even, use the power-reduction identities:
$ \sin^2x = \dfrac{1 – \cos(2x)}{2}, \quad \cos^2x = \dfrac{1 + \cos(2x)}{2} $
Important Standard Results
- \( \displaystyle \int \sec^2x\,dx = \tan x + C \)
- \( \displaystyle \int \csc^2x\,dx = -\cot x + C \)
- \( \displaystyle \int \sec x \tan x\,dx = \sec x + C \)
- \( \displaystyle \int \csc x \cot x\,dx = -\csc x + C \)
Example
Evaluate \( \displaystyle \int \sin^4x\,dx \).
▶️ Answer / Explanation
Use \( I_n = -\dfrac{\sin^{n-1}x\cos x}{n} + \dfrac{n-1}{n}I_{n-2} \).
Here, \( n = 4 \): \( I_4 = -\dfrac{\sin^3x\cos x}{4} + \dfrac{3}{4}I_2. \)
Now, \( I_2 = \int \sin^2x\,dx = \dfrac{x}{2} – \dfrac{\sin 2x}{4}. \)
\( \Rightarrow I_4 = -\dfrac{\sin^3x\cos x}{4} + \dfrac{3}{4}\left(\dfrac{x}{2} – \dfrac{\sin 2x}{4}\right) + C. \)
Example
Evaluate \( \displaystyle \int x^2 e^{3x}\,dx \).
▶️ Answer / Explanation
Use reduction formula \( I_n = \dfrac{x^n e^{3x}}{3} – \dfrac{n}{3}I_{n-1}. \)
Here, \( I_2 = \dfrac{x^2 e^{3x}}{3} – \dfrac{2}{3}I_1. \)
Now, \( I_1 = \dfrac{x e^{3x}}{3} – \dfrac{1}{3}I_0 \), and \( I_0 = \dfrac{e^{3x}}{3}. \)
Substitute back: \( I_1 = \dfrac{e^{3x}}{9}(3x – 1) \). Hence, \( I_2 = \dfrac{e^{3x}}{27}(9x^2 – 6x + 2) + C. \)
Example
Find \( \displaystyle \int \sec^5x\,dx \).
▶️ Answer / Explanation
Use \( I_n = \dfrac{\sec^{n-2}x\tan x}{n-1} + \dfrac{n-2}{n-1}I_{n-2}. \)
Here, \( n = 5 \): \( I_5 = \dfrac{\sec^3x\tan x}{4} + \dfrac{3}{4}I_3. \)
Now, \( I_3 = \dfrac{\sec x\tan x}{2} + \dfrac{1}{2}I_1 = \dfrac{\sec x\tan x}{2} + \dfrac{1}{2}\ln|\sec x + \tan x|. \)
Substitute back: \( I_5 = \dfrac{\sec^3x\tan x}{4} + \dfrac{3}{8}\sec x\tan x + \dfrac{3}{8}\ln|\sec x + \tan x| + C. \)
Notes and Study Materials
- Concepts of definite integrals
- definite integrals Master File
- definite integrals Revision Notes
- definite integrals Formulae
- definite integrals Reference Book
- definite integrals Past Many Years Questions and Answer
Examples and Exercise
IIT JEE (Main) Mathematics ,”definite integrals” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian
About this unit
Integral as an anti – derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by parts, and by partial fractions. Integration using trigonometric identities Integral as limit of a sum. Evaluation of simple integrals: Fundamental Theorem of Calculus. Properties of definite integrals, evaluation of definite integrals, determining areas of the regions bounded by simple curves in standard form.
IITian Academy Notes for IIT JEE (Main) Mathematics – definite integrals
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