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Area Enclosed Between the Curves


Area between curves y = Φ(x) and y = Ψ(x) and ordinates x = x1 and x = x2

img7   

(i) To determine the area between curves, first find out the points of intersection of the two curves. (See figure)

img8 

(ii) If in the domain common to both (i.e. the domain given by the points of intersection) the curves lie above x-axis, then area is
img16 
Note: If however one part of one or both the curve lies below x-axis, then the individual integral must be evaluated according to the case considered in the last topic.

3. Area between curve y = f(x) and y-axis. 
To obtain the area between the curve and the y-axis, the function must be written in y. (see figure A and B) i.e. y = f(x) must be inverted to x = g(y) (where g(x) = f–1 (x)) and the integral to be evaluated is ∫y1y2x dy or ∫y1y2g(y) dy.

img9


Similarly the area bounded between the y-axis & the curves y = f(x) and y = g(x) can be determined. In general, to find the area of the region one must draw the curve and locate the region. The limits and sign of different definite integral are determined accordingly.

Illustration: 
Suppose we have to calculate the area bounded by y2 = 4x and 2x = y. 
Solution: 
First we should calculate the points of intersection. 
4x2 – 4x = 0   ⇒ x = 0, x = 1 
                                 img10
The shaded portion gives the required area. 
This area is given by 
  img17
Illustration: 
Find the area bounded by y = x |sin x| and the x-axis between x = 0, x = 2Π. Solution: 
img18 

img11


Hence the required area
img19

Illustration: 
Find the area bounded by the curve |x| + y = 1 and the x-axis. 
Solution: 
The given curve is |x| + y = 1 …… (1)
i.e. x + y = 1, when x > 0
and –x + y = 1, when x < 0.


img12


The required area 
= area (CAOC) + area (OABO) 
    img20 
    = 1 sq. unit. 
Note:    Obviously, y = 1 – [x] is an even function. Hence graph of y = 1 – |x| is symmetrical about the y-axis. Thus the required area = 2 ∫01(1-x)dx. 

Illustration: 
       
Calculate the area bounded by the curve y = x(3 – x)2, the x-axis and the ordinates of the maximum and minimum points of the curve. 
Solution: 
       Since y = x(3 – x)2. Now for points of maxima or minima, we have dy/dx = 0 
       ⇒ (3 – x)2 – 2x(3 – x) = 0 
       ⇒ (3 – x)(3 – x – 2x) = 0          ∴ x = 1, 3 


img13

img24
img21
Illustration: 
Let f be a real valued function satisfying f(x/y) = f(x) – f(y) andimg27  then find the area bounded by y = f(x), y-axis and line y = 3.
Solution:
f(x/y) = f(x) – f(y) 
 Putting x = y = 1, we get f(1) = 0
img25
   
Putting x = 1, we get c = 0 
f(x) = 3 lnx 
Hence required area = ∫3 e y/3 dy = 3e sq. units. 

Illustration: 
Find the area between curves y = exlnx and y = lnx/ex. 
Solution: 
The two curves intersect where
img26
⇒ x = 1/e or x = 1 (lnx is not defined for x = –1/e) 
At x = 1/e or ex = 1 
lnx = –1, y = –1


img14

⇒ (1/e, –1) is one point of intersection and at x = 1, ln 1 = 0, y = 0 
⇒ (1, 0) is the other common point of the curves. 
The required area = ∫1/e1  (y1-y2) dx
img22

Illustration:
Let f(x) = minimum (x + 1, √(1-x)) for all x < 1. Find the area bounded by y = f(x) and the x-axis. 
Solution: 
 Required area = Area ABCA
img23
img15

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