Summary - Applied Mathematics

Additional Topics
Applied Mathematics-XII
(M.L. Aggarwal and Neeraj Raj Jain)
Additional matter related to Chapter 7 and Chapter 12
as per syllabus released by CBSE on 25th March 2025 for 2026 examinations

Chapter 7: Integrals
Applications of Integrals
In geometry, we learnt some formulae to calculate areas of various geometrical figures such as
triangles, quadrilaterals, polygons and circles etc., which are very useful in many real life problems.
However, by these formulae of geometry, we can calculate areas of only simple figures. They are
inadequate for calculating the areas enclosed by closed curves. For that we need definite integrals.
In fact, definite integrals have a wide range of applications.

Areas of Bounded Regions
If the function f is continuous and non-negative in the closed interval [a, b], then the area of the
region below the curve y = f (x), above the x-axis and between the ordinates x = a and x = b or briefly
the area of the region bounded by the curve y = f (x), the x-axis and the ordinates x = a, x = b is given
b b
by ∫ f (x) dx or ∫ y dx.
a a

Proof. Let AB be the curve y = f (x) between AC(x = a) and BD(x = b), then the required area is
the area of the shaded region ACDB.
Y
Let P(x, y) be a point on the curve y = f (x) and Q(x + δx,
y + δy) be a neighbouring point on the curve, then MP = y, B

NQ = y + δy and MN = δx. Let A be the area of the region S Q
ACMP and A + δA be the area of the region ACNQ, then A P R
δA = area of region PMNQ.
Area of rectangle PMNR = yδx and area of rectangle
x=b
x=a




SMNQ = (y + δy)δx.
From fig. S-1, area of rectangle PMNR ≤ area of region O MN
C D X
PMNQ ≤ area of rectangle SMNQ
⇒ yδx ≤ δA ≤ (y + δy)δx Fig. S-1
δA
⇒ y≤ ≤ y + δy …(i)
δx
δA dA
When P → Q, δx → 0, δy → 0 and → .
δx dx
δA
From (i), Lt y ≤ Lt ≤ Lt ( y + δy )
δx → 0 δx → 0 δx δx → 0

dA dA
⇒ y≤ ≤y⇒y= .
dx dx
Integrating both sides w.r.t. x between the limits a to b, we get
b b
dA
∫ ∫ dx = [A ]a
b
y dx =
dx
a a

= (value of area A when x = b) – (value of area A when x = a)
= area ACDB – 0 = area ACDB.
Additional Topics S-1

, If a function f is continuous and non-positive in the closed Y
interval [a, b], then the curve y = f (x) lies below the x-axis and
b M δx N
the definite integral ∫ f (x) dx is negative. Since the area of a O X




x=b
a y




x=a
region is always non-negative, the area of the region bounded
by the curve y = f (x), the x-axis and the ordinates x = a,
b b P(x, y) Q
x = b is given by ∫ f ( x ) dx or ∫ y dx .
Fig. S-2
a a

Hence, if the curve y = f (x) is continuous and does not cross the x-axis, then the area of the
region bounded by the curve y = f (x), the x-axis and the ordinates x = a and x = b is given by

b b Y

∫ f ( x ) dx or ∫ y dx . y=d
a a


Similarly, if the curve x = g (y) is continuous and does δy
not cross the y-axis, then the area of the region bounded by P(x, y)




x = g(y)
the curve x = g (y), the y-axis and the abscissae y = c, y = d is
given by
d d
O X
∫ g ( y ) dy or ∫ x dy . y=c
c c
Fig. S-3

Remark:
b
It may be noted that when sign of f (x) is not known, then ∫ f (x) dx may not represent the area
a
b
enclosed between the curve y = f (x), the x-axis and the ordinates x = a and x = b, whereas ∫ |f (x)| dx
a

equals the area enclosed between the graph of the curve y = f (x), the x-axis and the ordinates x = a
and x = b.


1 1
For example, let us consider the integrals ∫ x dx and ∫ | x | dx.
−1 −1

1 1
 x2  1 2
First integral = ∫ x dx = 
  −1 2
2
2
 = (1 – (– 1) ) = 0, whereas second integral
−1

1 0 1
= ∫ | x | dx = ∫ (– x) dx + ∫ x dx
−1 −1 0

(Common sense suggests this division as | x | = – x in [– 1, 0] and | x | = x in [0, 1])
0 1
 x2   x2  1 1
= −  +   = − (0 – 1) + (1 – 0) = 1.
 2 2
 −1   0 2 2




S-2 Applied Mathematics-XII

, Y Y

x
= 0
y >




y
x




=
x,




x,
–1 =




x
y




<
O 1 X




0
–1 O 1 X



Fig. S-4
Clearly, the area enclosed between y = x, the x-axis and the ordinates x = – 1 and x = 1 is not zero.
It follows that if the graph of a function f is continuous in [a, b] and crosses the x-axis
at finitely many points in [a, b], then the area enclosed between the graph of the curve
b b
y = f (x), the x-axis and the ordinates x = a, x = b is given by ∫ |f (x)|dx or ∫ |y|dx.
a a
Area between Two Curves Y
If f (x), g (x) are both continuous in [a, b] and 0 ≤ g (x) ≤ f (x) y = f (x)
for all x ∈ [a, b], then the area of the region between the


x=a
graphs of y = f (x), y = g (x) and the ordinates x = a, x = b is
given by




x=b
b b

∫ f (x) dx – ∫ g (x) dx y = g (x)
a a
MO N X
b Fig. S-5
= ∫ ( f (x) – g (x)) dx. Y
a
y=d

Similarly, the area of the region between the graphs of
x = f (y), x = g (y) and the abscissae y = c, y = d is given by
x = f (y)
x = g (y)




d

∫ ( f (y) – g (y)) dy.
c


O X
y=c
Remarks: Fig. S-6
1. If f (x), g (x) are both continuous in [a, b] and g (x) ≤ f (x) for all x ∈ [a, b], then the above formula
also holds when one or both of the curves y = f (x) and y = g (x) lie partially or completely below
the x-axis.
2. If the graphs of the curves y = f (x) and y = g (x) cross each other at finitely many points,
then the area enclosed between the graphs of the two curves and the ordinates x = a and
b
x = b is given by ∫ |f (x) – g (x)| dx.
a
3. Similarly, the area of the region between the graphs of x = f (y), x = g (y) and the abscissae y = c,
b
y = d is given by ∫ |f (y) – g (y)| dy.
c
4. In case of symmetrical closed area, find the area of the smallest part and multiply the result by
the number of symmetrical parts.


Additional Topics S-3