Download integration_class_12th_CBSE.pdf and more Lecture notes Mathematics in PDF only on Docsity!
Chapter: Integration
Integral Calculus
1 Introduction
Integration is one of the two fundamental operations of calculus, the other being differ-
entiation. While differentiation deals with the rate of change of a quantity, integration
deals with accumulation. Historically, integration arose from problems of finding areas,
volumes, arc lengths, and physical quantities such as work and mass.
Mathematically, integration is the inverse process of differentiation.
2 Indefinite Integral
2.1 Definition
Let f (x) be a real-valued function defined on an interval I. If there exists a function
F (x) such that d dx
F (x) = f (x),
then F (x) is called an antiderivative of f (x).
The collection of all antiderivatives of f (x) is called the indefinite integral and is
denoted by (^) Z
f (x) dx = F (x) + C,
where C is an arbitrary constant.
2.2 Basic Properties
- Linearity: (^) Z (af (x) + bg(x)) dx = a
Z
f (x) dx + b
Z
g(x) dx
- Constant of integration must always be included.
2.3 Solved Example
Example: Evaluate
R
5 x^4 dx. Solution: (^) Z 5 x^4 dx = 5 · x^5 5
3 Standard Integrals
Some important standard integrals are listed below:
Z xn^ dx = xn+ n + 1
Z 1 x dx = ln |x| + C
Z ex^ dx = ex^ + C
Z ax^ dx = ax ln a
Z sin x dx = − cos x + C
Z cos x dx = sin x + C
Z sec^2 x dx = tan x + C
4 Methods of Integration
4.1 Integration by Substitution
This method is based on reversing the chain rule.
4.1.1 Formula
If (^) Z
f (g(x))g′(x) dx,
then put u = g(x), so that du = g′(x) dx and
Z f (u) du
4.1.2 Example
Evaluate
R
2 x cos(x^2 ) dx. Solution: Let u = x^2 , then du = 2x dx.
5.3 Example
Evaluate
R 1
0 x (^2) dx.
Z (^1)
0
x^2 dx =
x^3 3
0
6 Fundamental Theorem of Calculus
6.1 Statement
If f (x) is continuous on [a, b] and F ′(x) = f (x), then
Z (^) b
a
f (x) dx = F (b) − F (a)
This theorem establishes the connection between differentiation and integration.
7 Applications of Integration
7.1 Area Under a Curve
The area bounded by the curve y = f (x), the x-axis, and the vertical lines x = a and
x = b is
A =
Z (^) b
a
f (x) dx
7.1.1 Example
Find the area under y = x from x = 0 to x = 2.
A =
Z 2
0
x dx =
x^2 2
0
8 Improper Integrals
An integral is called improper if either the interval is infinite or the integrand becomes
infinite.
8.1 Example
Z ∞
1
x^2 dx = lim b→∞
Z (^) b
1
x^2 dx
= lim b→∞
x
b
1
9 Integration Using Partial Fractions
Partial fraction decomposition is a powerful technique used to integrate rational functions,
that is, functions of the form P (x) Q(x)
where P (x) and Q(x) are polynomials and deg P (x) < deg Q(x).
9.1 Basic Idea
The method consists of expressing a rational function as a sum of simpler rational func-
tions whose denominators are linear or quadratic factors.
9.2 Case I: Distinct Linear Factors
If the denominator factors into distinct linear factors, i.e.,
P (x) (x − a)(x − b)
then it can be written as
P (x) (x − a)(x − b)
A
x − a
B
x − b
9.2.1 Example
Evaluate (^) Z 3 x + 5 (x − 1)(x + 2)
dx.
Solution: Assume 3 x + 5 (x − 1)(x + 2)
A
x − 1
B
x + 2
Multiplying both sides by (x − 1)(x + 2),
3 x + 5 = A(x + 2) + B(x − 1).
ln |x − 1 | −
2(x − 1)
ln |x + 1| + C.
9.4 Case III: Irreducible Quadratic Factors
If the denominator contains an irreducible quadratic factor,
P (x) (x^2 + ax + b)
then the decomposition takes the form
Ax + B x^2 + ax + b
9.4.1 Example
Evaluate (^) Z x x^2 + 1 dx.
Solution: Let (^) Z x x^2 + 1 dx.
Put u = x^2 + 1, then du = 2x dx. Z x x^2 + 1 dx =
Z
du u
ln
x^2 + 1
+ C.
9.5 Case IV: Product of Linear and Quadratic Factors
If the denominator is a product of linear and quadratic factors, the decomposition is a
combination of previous cases.
9.5.1 Example
Evaluate (^) Z 2 x + 3 (x − 1)(x^2 + 1)
dx.
Solution: Assume 2 x + 3 (x − 1)(x^2 + 1)
A
x − 1
Bx + C x^2 + 1
Multiplying both sides,
2 x + 3 = A(x^2 + 1) + (Bx + C)(x − 1).
Comparing coefficients and solving,
A = 1, B = 1, C = 1.
Thus, (^) Z 1 x − 1
x + 1 x^2 + 1
dx
= ln |x − 1 | +
ln
x^2 + 1
10 Definite Integral
10.1 Definition
Let f (x) be a real-valued function defined and continuous on the closed interval [a, b].
The definite integral of f (x) from a to b is defined by
Z (^) b
a
f (x) dx = F (b) − F (a),
where F (x) is any antiderivative of f (x), that is,
F ′(x) = f (x).
Geometrically, the definite integral represents the signed area bounded by the curve
y = f (x), the x-axis, and the vertical lines x = a and x = b.
10.2 Fundamental Properties
Let f (x) and g(x) be integrable functions on [a, b] and c be a constant. Then:
- Zero limit: (^) Z (^) a
a
f (x) dx = 0
- Change of limits: (^) Z b a
f (x) dx = −
Z (^) a
b
f (x) dx
- Linearity: (^) Z b a
[cf (x) + g(x)] dx = c
Z (^) b
a
f (x) dx +
Z (^) b
a
g(x) dx
Example 3: Using Properties
Evaluate (^) Z (^) π
0
sin x dx.
Solution: Z (^) π
0
sin x dx = [− cos x]π 0
= (− cos π) − (− cos 0) = 2.
Example 4: Symmetry Property
Evaluate (^) Z (^) a
−a
x^3 dx.
Solution: Since x^3 is an odd function, (^) Z (^) a
−a
x^3 dx = 0.
10.4 Important Symmetry Results
- If f (x) is even, then (^) Z (^) a
−a
f (x) dx = 2
Z (^) a
0
f (x) dx
- If f (x) is odd, then (^) Z a −a
f (x) dx = 0
11 Area Under a Curve
One of the most important applications of definite integrals is the calculation of area
bounded by curves.
11.1 Area Between a Curve and the x-Axis
Let y = f (x) be a continuous function on the interval [a, b].
- If f (x) ≥ 0 on [a, b], then the area bounded by the curve y = f (x), the x-axis, and the lines x = a and x = b is A =
Z (^) b
a
f (x) dx.
- If f (x) ≤ 0 on [a, b], then the area is
A =
Z (^) b
a
|f (x)| dx = −
Z (^) b
a
f (x) dx.
11.2 Solved Example 1
Find the area under the curve y = x^2 from x = 0 to x = 2.
Solution: Since x^2 ≥ 0 on [0, 2], A =
Z 2
0
x^2 dx =
x^3 3
0
11.3 Area When the Curve Cuts the x-Axis
If the curve intersects the x-axis between a and b, the interval must be divided at the
points of intersection.
11.4 Solved Example 2
Find the area bounded by the curve y = x^2 − 4 and the x-axis.
Solution: First find the points where the curve meets the x-axis:
x^2 − 4 = 0 =⇒ x = ± 2.
On [− 2 , 2], x^2 − 4 ≤ 0. Hence, area is
A =
Z 2
− 2
|x^2 − 4 | dx =
Z 2
− 2
(4 − x^2 ) dx.
Using symmetry, A = 2
Z 2
0
(4 − x^2 ) dx.
4 x − x^3 3
0
11.5 Area Between Two Curves
Let y = f (x) and y = g(x) be two continuous curves on [a, b] such that f (x) ≥ g(x).
The area bounded by the two curves is given by
A =
Z (^) b
a
[f (x) − g(x)] dx.
11.10 Important Remarks
- Ensure that the degree of numerator is less than the degree of denominator.
- If not, perform polynomial division first.
- Partial fractions reduce complicated rational functions into integrable standard forms.
12 Summary
In this chapter, we studied:
- Indefinite and definite integrals
- Standard formulas
- Methods of integration
- Fundamental Theorem of Calculus
- Applications and improper integrals
Integration plays a crucial role in higher mathematics, physics, engineering, and ap-
plied sciences.
