Welcome, aspiring mathematicians! Today, we embark on a deep dive into one of the most fundamental and powerful concepts in Integral Calculus: the
Evaluation of Definite Integrals. While indefinite integrals give us a family of functions, definite integrals provide a specific numerical value, often representing quantities like area, volume, or total change. Our journey will begin with the bedrock theorem and then explore various techniques and properties that are indispensable for cracking complex problems, especially those encountered in JEE.
Let's begin!
### 1. The Fundamental Theorem of Calculus (Part 2: The Evaluation Theorem)
At the heart of evaluating definite integrals lies a profound connection between differentiation and integration, established by the
Fundamental Theorem of Calculus. Specifically, its second part (often called the Evaluation Theorem or Newton-Leibniz formula) provides a direct method for calculation.
Consider a continuous function $f(x)$ on an interval $[a, b]$. If $F(x)$ is *any* antiderivative of $f(x)$ (meaning $F'(x) = f(x)$), then the definite integral of $f(x)$ from $a$ to $b$ is given by:
$int_a^b f(x) dx = F(b) - F(a)$
Here, $a$ is called the
lower limit and $b$ is the
upper limit of integration. The constant of integration $C$ (which we usually add in indefinite integrals) cancels out because $(F(b) + C) - (F(a) + C) = F(b) - F(a)$. Hence, we don't need to include it when evaluating definite integrals.
Intuition: Think of $F(x)$ as a function that accumulates the "quantity" represented by $f(x)$ up to point $x$. Then $F(b) - F(a)$ represents the net change in this accumulated quantity as $x$ goes from $a$ to $b$. For instance, if $f(x)$ is a velocity function, $F(x)$ would be a position function, and $F(b) - F(a)$ would be the net displacement.
Example 1: Direct Application of FTC
Evaluate $int_1^2 (x^2 + 3) dx$.
Solution:
1. Find the antiderivative of $f(x) = x^2 + 3$.
$F(x) = frac{x^3}{3} + 3x$
2. Apply the FTC:
$int_1^2 (x^2 + 3) dx = F(2) - F(1)$
$= left(frac{2^3}{3} + 3(2)
ight) - left(frac{1^3}{3} + 3(1)
ight)$
$= left(frac{8}{3} + 6
ight) - left(frac{1}{3} + 3
ight)$
$= left(frac{8+18}{3}
ight) - left(frac{1+9}{3}
ight)$
$= frac{26}{3} - frac{10}{3} = frac{16}{3}$
### 2. Properties of Definite Integrals
Many definite integrals can be simplified or solved more easily by using their properties. These properties are critical for JEE problems.
Property No. |
Description |
Formula |
|---|
P1 |
Interchanging Limits |
$int_a^b f(x) dx = -int_b^a f(x) dx$ |
P2 |
Identical Limits |
$int_a^a f(x) dx = 0$ |
P3 |
Dummy Variable Property |
$int_a^b f(x) dx = int_a^b f(t) dt$ |
P4 |
Interval Splitting |
$int_a^b f(x) dx = int_a^c f(x) dx + int_c^b f(x) dx$ (where $a < c < b$) |
P5 |
King's Rule (Most Important) |
$int_a^b f(x) dx = int_a^b f(a+b-x) dx$ |
P6 |
Special Case of King's Rule |
$int_0^a f(x) dx = int_0^a f(a-x) dx$ |
P7 |
Interval Doubling / Symmetry (for $int_0^{2a} f(x) dx$) |
- If $f(2a-x) = f(x)$, then $int_0^{2a} f(x) dx = 2 int_0^a f(x) dx$
- If $f(2a-x) = -f(x)$, then $int_0^{2a} f(x) dx = 0$
|
P8 |
Even/Odd Functions (for $int_{-a}^a f(x) dx$) |
- If $f(x)$ is an even function ($f(-x) = f(x)$), then $int_{-a}^a f(x) dx = 2 int_0^a f(x) dx$
- If $f(x)$ is an odd function ($f(-x) = -f(x)$), then $int_{-a}^a f(x) dx = 0$
|
Let's delve deeper into some key properties with derivations and examples.
#### Derivation of King's Rule (P5): $int_a^b f(x) dx = int_a^b f(a+b-x) dx$
Let $I = int_a^b f(x) dx$.
Apply a substitution: Let $t = a+b-x$.
Then $dt = -dx$, so $dx = -dt$.
When $x=a$, $t = a+b-a = b$.
When $x=b$, $t = a+b-b = a$.
Substituting these into the integral:
$I = int_b^a f(a+b-t) (-dt)$
$I = -int_b^a f(a+b-t) dt$
Using P1 ($int_a^b f(x) dx = -int_b^a f(x) dx$):
$I = int_a^b f(a+b-t) dt$
Using P3 (Dummy variable property: $int_a^b f(t) dt = int_a^b f(x) dx$):
$I = int_a^b f(a+b-x) dx$
Hence, the King's Rule is proven.
JEE Focus: King's Rule (and its special case P6) is arguably the most frequently used property in JEE definite integral problems. It is particularly effective for integrals involving trigonometric functions, logarithms, and powers, especially when the limits are symmetric or sum to a simple value (like $0$ to $pi/2$, or $0$ to $a$).
Example 2: Using King's Rule (P6)
Evaluate $int_0^{pi/2} frac{sin x}{sin x + cos x} dx$.
Solution:
Let $I = int_0^{pi/2} frac{sin x}{sin x + cos x} dx$ -------- (1)
Using P6: $int_0^a f(x) dx = int_0^a f(a-x) dx$. Here $a = pi/2$.
$I = int_0^{pi/2} frac{sin(pi/2 - x)}{sin(pi/2 - x) + cos(pi/2 - x)} dx$
$I = int_0^{pi/2} frac{cos x}{cos x + sin x} dx$ -------- (2)
Now, add (1) and (2):
$2I = int_0^{pi/2} frac{sin x}{sin x + cos x} dx + int_0^{pi/2} frac{cos x}{cos x + sin x} dx$
$2I = int_0^{pi/2} frac{sin x + cos x}{sin x + cos x} dx$
$2I = int_0^{pi/2} 1 dx$
$2I = [x]_0^{pi/2}$
$2I = frac{pi}{2} - 0$
$2I = frac{pi}{2}$
$I = frac{pi}{4}$
#### Derivation of Even/Odd Function Property (P8) for Symmetric Limits: $int_{-a}^a f(x) dx$
We start with P4: $int_a^b f(x) dx = int_a^c f(x) dx + int_c^b f(x) dx$.
For $int_{-a}^a f(x) dx$, we can split the interval at $c=0$:
$int_{-a}^a f(x) dx = int_{-a}^0 f(x) dx + int_0^a f(x) dx$
Consider the first integral, $int_{-a}^0 f(x) dx$.
Let $x = -t$. Then $dx = -dt$.
When $x = -a$, $t = a$.
When $x = 0$, $t = 0$.
So, $int_{-a}^0 f(x) dx = int_a^0 f(-t) (-dt)$
$= -int_a^0 f(-t) dt$
Using P1, this becomes $int_0^a f(-t) dt$.
Using P3 (dummy variable), this is $int_0^a f(-x) dx$.
Therefore, $int_{-a}^a f(x) dx = int_0^a f(-x) dx + int_0^a f(x) dx = int_0^a [f(x) + f(-x)] dx$.
Now, we check for even/odd functions:
1. If $f(x)$ is an
even function, then $f(-x) = f(x)$.
So, $int_0^a [f(x) + f(x)] dx = int_0^a 2f(x) dx = 2 int_0^a f(x) dx$.
2. If $f(x)$ is an
odd function, then $f(-x) = -f(x)$.
So, $int_0^a [f(x) - f(x)] dx = int_0^a 0 dx = 0$.
Hence, P8 is proven.
Example 3: Using Even/Odd Property
Evaluate $int_{-pi/2}^{pi/2} (sin^3 x + x^5 + cos x) dx$.
Solution:
Let $f(x) = sin^3 x + x^5 + cos x$.
We check if $f(x)$ is even or odd, or a sum of even and odd functions.
$f(-x) = sin^3 (-x) + (-x)^5 + cos (-x)$
$f(-x) = (-sin x)^3 - x^5 + cos x$
$f(-x) = -sin^3 x - x^5 + cos x$
We can write $f(x)$ as a sum of an odd part and an even part:
Let $g(x) = sin^3 x + x^5$. Then $g(-x) = -sin^3 x - x^5 = -g(x)$. So, $g(x)$ is an odd function.
Let $h(x) = cos x$. Then $h(-x) = cos x = h(x)$. So, $h(x)$ is an even function.
So, $int_{-pi/2}^{pi/2} (sin^3 x + x^5 + cos x) dx = int_{-pi/2}^{pi/2} (sin^3 x + x^5) dx + int_{-pi/2}^{pi/2} cos x dx$.
Using P8:
For the odd function part: $int_{-pi/2}^{pi/2} (sin^3 x + x^5) dx = 0$.
For the even function part: $int_{-pi/2}^{pi/2} cos x dx = 2 int_0^{pi/2} cos x dx$.
$2 int_0^{pi/2} cos x dx = 2 [sin x]_0^{pi/2}$
$= 2 (sin(pi/2) - sin(0))$
$= 2 (1 - 0) = 2$.
Therefore, the original integral evaluates to $0 + 2 = 2$.
### 3. Techniques for Evaluation (Beyond Direct Application)
While properties simplify many integrals, sometimes a change of variable or integration method is required.
#### 3.1 Substitution Method
When using substitution in definite integrals, remember to change the limits of integration according to the new variable.
Example 4: Substitution Method
Evaluate $int_0^1 x sqrt{1-x^2} dx$.
Solution:
Let $u = 1-x^2$.
Then $du = -2x dx$, so $x dx = -frac{1}{2} du$.
Now, change the limits:
When $x=0$, $u = 1-(0)^2 = 1$.
When $x=1$, $u = 1-(1)^2 = 0$.
Substitute these into the integral:
$int_0^1 x sqrt{1-x^2} dx = int_1^0 sqrt{u} left(-frac{1}{2} du
ight)$
$= -frac{1}{2} int_1^0 u^{1/2} du$
Using P1, we can flip the limits and change the sign:
$= frac{1}{2} int_0^1 u^{1/2} du$
$= frac{1}{2} left[frac{u^{3/2}}{3/2}
ight]_0^1$
$= frac{1}{2} imes frac{2}{3} [u^{3/2}]_0^1$
$= frac{1}{3} (1^{3/2} - 0^{3/2})$
$= frac{1}{3} (1 - 0) = frac{1}{3}$.
#### 3.2 Integration by Parts
The integration by parts formula for definite integrals is:
$int_a^b u , dv = [uv]_a^b - int_a^b v , du$
Example 5: Integration by Parts
Evaluate $int_0^{pi/2} x cos x dx$.
Solution:
We use the ILATE rule for choosing $u$ and $dv$. Here, $u=x$ (algebraic) and $dv = cos x dx$ (trigonometric).
$u = x implies du = dx$
$dv = cos x dx implies v = int cos x dx = sin x$
Apply the integration by parts formula:
$int_0^{pi/2} x cos x dx = [x sin x]_0^{pi/2} - int_0^{pi/2} sin x dx$
$= left(frac{pi}{2} sin(frac{pi}{2}) - 0 sin(0)
ight) - [-cos x]_0^{pi/2}$
$= left(frac{pi}{2} imes 1 - 0
ight) - (-cos(frac{pi}{2}) - (-cos(0)))$
$= frac{pi}{2} - (0 - (-1))$
$= frac{pi}{2} - 1$.
#### 3.3 Integrals Involving Modulus Functions
When an integrand contains a modulus function, say $|g(x)|$, we need to split the integral at points where $g(x)$ changes sign (i.e., where $g(x) = 0$). This is an application of Property P4.
Example 6: Modulus Function
Evaluate $int_{-1}^2 |x^2 - x| dx$.
Solution:
First, analyze the expression inside the modulus: $x^2 - x = x(x-1)$.
The expression $x(x-1)$ changes sign at $x=0$ and $x=1$.
These points are within our integration interval $[-1, 2]$.
We need to split the integral based on the sign of $x(x-1)$:
* For $x in [-1, 0]$, $x < 0$ and $x-1 < 0$, so $x(x-1) > 0$. Thus, $|x^2 - x| = x^2 - x$.
* For $x in [0, 1]$, $x > 0$ and $x-1 < 0$, so $x(x-1) < 0$. Thus, $|x^2 - x| = -(x^2 - x) = x - x^2$.
* For $x in [1, 2]$, $x > 0$ and $x-1 > 0$, so $x(x-1) > 0$. Thus, $|x^2 - x| = x^2 - x$.
Now, split the integral using P4:
$int_{-1}^2 |x^2 - x| dx = int_{-1}^0 (x^2 - x) dx + int_0^1 (x - x^2) dx + int_1^2 (x^2 - x) dx$
Evaluate each part:
1. $int_{-1}^0 (x^2 - x) dx = left[frac{x^3}{3} - frac{x^2}{2}
ight]_{-1}^0$
$= left(0 - 0
ight) - left(frac{(-1)^3}{3} - frac{(-1)^2}{2}
ight)$
$= 0 - left(-frac{1}{3} - frac{1}{2}
ight) = - left(-frac{2+3}{6}
ight) = frac{5}{6}$
2. $int_0^1 (x - x^2) dx = left[frac{x^2}{2} - frac{x^3}{3}
ight]_0^1$
$= left(frac{1^2}{2} - frac{1^3}{3}
ight) - (0 - 0)$
$= frac{1}{2} - frac{1}{3} = frac{3-2}{6} = frac{1}{6}$
3. $int_1^2 (x^2 - x) dx = left[frac{x^3}{3} - frac{x^2}{2}
ight]_1^2$
$= left(frac{2^3}{3} - frac{2^2}{2}
ight) - left(frac{1^3}{3} - frac{1^2}{2}
ight)$
$= left(frac{8}{3} - frac{4}{2}
ight) - left(frac{1}{3} - frac{1}{2}
ight)$
$= left(frac{8}{3} - 2
ight) - left(frac{2-3}{6}
ight)$
$= left(frac{8-6}{3}
ight) - left(-frac{1}{6}
ight) = frac{2}{3} + frac{1}{6} = frac{4+1}{6} = frac{5}{6}$
Summing the three parts:
Total Integral $= frac{5}{6} + frac{1}{6} + frac{5}{6} = frac{11}{6}$.
#### 3.4 Integrals Involving Greatest Integer Function (GIF)
The greatest integer function, $[x]$ or $lfloor x
floor$, is piecewise constant. To integrate it, we split the interval of integration at the integer points where the value of $[x]$ changes.
Example 7: Greatest Integer Function
Evaluate $int_0^3 [x] dx$.
Solution:
The value of $[x]$ changes at $x=1$ and $x=2$ within the interval $[0, 3]$.
* For $x in [0, 1)$, $[x] = 0$.
* For $x in [1, 2)$, $[x] = 1$.
* For $x in [2, 3)$, $[x] = 2$.
* At $x=3$, $[x] = 3$, but the integral considers the limit from the left. We can treat $x=3$ as the upper limit for the last interval.
Split the integral using P4:
$int_0^3 [x] dx = int_0^1 [x] dx + int_1^2 [x] dx + int_2^3 [x] dx$
$= int_0^1 0 , dx + int_1^2 1 , dx + int_2^3 2 , dx$
$= [C]_0^1 + [x]_1^2 + [2x]_2^3$
$= 0 + (2-1) + (2 imes 3 - 2 imes 2)$
$= 0 + 1 + (6 - 4)$
$= 1 + 2 = 3$.
### Conclusion
Evaluating definite integrals is a cornerstone of integral calculus. Mastering the
Fundamental Theorem of Calculus is essential, but equally important is a strong grasp of the various
properties of definite integrals. Techniques like
substitution and
integration by parts extend our ability to tackle more complex functions. Furthermore, special functions like
modulus functions and the
greatest integer function demand careful interval splitting based on their definitions.
For JEE, always remember to:
*
Check for properties first: P5 (King's Rule) and P8 (Even/Odd) are incredibly powerful.
*
When using substitution, *always* change the limits of integration.
*
For modulus and greatest integer functions, accurately identify the critical points to split the integral.
*
Simplify the integrand before integrating whenever possible.
With consistent practice and a clear understanding of these concepts, you'll be well-equipped to ace definite integral problems in your exams! Keep practicing, and you'll build the intuition to spot the right approach for any given integral.
