Example.1
Find the area of the figure bounded by the parabola
(y-2)2 = x-1, the tangent to it at the point with ordinate 3,
and the x-axis.
Solution
The
equation (y – 2)2 = x – 1 represents a parabola having vertex at
A(1, 2) and axis as y =2 . The point P on this parabola with ordinate 3
has abscissa.
x = 1+ (y-2)2 = 2, then P º (2, 3)
(y - 2)2 = x -1
&⇒
2(y-2)
= 1
&⇒
Thus equation of tangent at P is given by
y - 3 = 
&⇒x
– 2y +4 = 0
The region, of which
area is to be found is shown by shaded portion. We have to find the
Area(EOBAPCE). C is the point of intersection of x-2y +4 =0 and y-axis
&⇒ C º (0, 2). Obviously .
D º (0, 3).
The required area
= area(OBAPDO) –
area (DPCD) + area(DEOC) 
= 
= 
sq. units
Example.2
Let f(x) = maximum {x2, (1-x)2,
2x(1-x)}. Determine the area of the region bounded by the curve y = f(x),
x-axis , x = 0 and x=1.
Solution
First we draw the graph of y = x2,
y = (1 - x)2 and y = 2x(1 - x) for
0 £ x £ 1. Obviously all these are parabolas.
Now graph of y = f(x) is
shown by dark curves. Hence C is the point of intersection of
y = x2 and y = (1-x)2 . So
C º
(1/2, ¼). B is the point of intersection of y =(1 – x)2,
y =2x(1-x). So B º(1/3, 4/9).
Similarly D º (2/3, 4/9)
The required area = Area (OPBDQAO)
= 
= 
= 
sq. units
Note: If f(x) = minimum {x2,
(1-x)2, 2x(1-x)}. , then the required area will be OCAO. .
Example.3
Find the area between the curves y = x2 + x
–2 and y = 2x, for which |x2 + x –2| + | 2x | = |x2
+ 3x –2| is satisfied.
Solution
y = x2 + x –2, y = 2x
| x2 + x –2 | + | 2x | = | x2 + 3x
–2 |
This means we have to find area in the region where x2
+ x –2 and 2x have same sign.
Thus, the required region is OABO and ECD.
\ x2 +x –2 = 2x
&⇒ x2 –x –2 = 0
&⇒ x = -1, x = 2
\ required area is,
A=Area
of OABO+ Area of ECD
\ 
= 
= 
= 
= 
= 
= 
Example.4
Let f(x) = min{ex, 3/2, 1+ e-x
| 0 £ x £ 1
}. Find the area bounded by
y = f(x) , x-axis, y-axis and the line x = 1 . .
Solution
f(x) = 
Let A be the required
area as shown in fig. 
A = 
= 
= 
= 2+ 
.
Example.5
Find a function ‘f’,
(x4 – 4x2 £
f(x) £ 2x2 – x3)
such that the area bounded by y = f(x), y =x4 –4x2,
the y-axis and the line x=t, (0 £
t £ 2) is k times the
area bounded by y = f(x), y = 2x2 – x3, y-axis and line
x =t ( 0£ t £ 2) . .
Solution
According to the
given condition 
Differentiate both sides w.r.t t , we get f(t) –(t4
– 4t2) = k(2t2 –t3 –f(t)).
&⇒
(1+k) f(t) = k 2t2 – kt3 + t4 – 4t2
&⇒ f(t) = 
Hence, the required function ‘f’ is given by
f(x) = 
(x4 – kx3 +2(k – 2)x2)
.
