Thursday, 4 April 2013

Areas of Curves Given by Cartesian Equations


Let, y = f (x) be a finite and continuous function in the interval [a,b]


Let the variable are GLCP bounded by the curve y = f (x), the x-axis, the fixed ordinate GL and the variable ordinate PC is A. Thus it is clear that A is a function of x and has a unique value corresponding to each of value of x.

When x is increased by an amount x( =CD), Area assumes an increment A = the trapezoidal area PQDC.

Clearly this trapezoidal area PQDC lies in between the area of the inscribed and circumscribed rectangles PFDC and EQDC.

OL = a, OM = b and OC = x

Ar. (PFDC) < Ar. (PQDC) < Ar. (EQDC)

f (x).x <A < f (x + x).(x)

f ( x) < A / x < f (x + x)

Taking the lime x -> 0, we have

dA / dx = f (x)

By definition dA = f (x) dx.

i.e A = ∫ f(x) dx = Φ (x) + C, where C is an arbitarary constant of integration and Φ (x) is an indefinite integral of f (x).

Now, when x = OM = b, the Area A becomes the required Area A1.
 0 = Φ (a) + C and A1 = Φ (b) + C



Corollary : 
In the same way it can be shown that the area bounded by any curve x = g (y), y-axis and two abscissae y = c and y = d is



The Sign Convention for Finding the Area Using Integration
We can tactfully apply the concept of definite integration to find the area enclosed between the curves but then we must be very careful, while applying the discussed sign convention we will discuss the three cases:


for all a≤ x ≤ b, then this integration will give the area enclosed between the curve, f (x), x-axis and the line x =a and x = b which is positive. No need of any modification.



f (x) dx if b>a and f(x) < 0 for all a≤ x ≤ b, then this integration will calculate to be negative. But the numerical or the absolute value is to be taken to mean the area enclosed between the curve y = f (x), x-axis and the lines x = a and x = b.


f (x) dx where b > a but f(x) change its sign a numbers of times in the interval a≤ x ≤ b, then we must divide the region [a, b] in such a way that we clearly get the points lying between [a, b] where (x) change its sign. For the region and then add the absolute value of the integration calculated in the region where f (x) < 0 to get the desired area between the curve y = f(x), x-axis and the line x = a and x = b.


Hence, if f (x) is as shown in above figure, the area enclosed by y = f (x); x-axis and the line x = a and x = b is given by



Example 1:

Find the area bounded by y = cos x, x = 2
 and the x - axis.

Solution

The required area

=2 + 2 +1 = 5 sq units

Example 2:

Find the area bounded by y = x |sin x| and the x-axis between x = 0, x =2

Solution

y = {x sin x, if sin x≥ 0 ie 0 ≤ x ≤ , - x sin x, if sin x < 0 ie  < x ≤ 2 }

=4 sq units

Example 3:

Find the area bounded by the curve |x| + y = 1 and the x - axis.

Solution

The given curve is |x| + y = 1 (i)

ie, x + y = 1, when x≥ 0

and - x + y = 1, when x< 0


Area Between the given Line
Area bounded by the curve y = f (x), y = g(x) and the lines x= a and y = b. Let the curves y = f (x) and y = g (x) be represented by AB and CD respectively. We assume that the two curves do not intersect each other in the interval [a, b].

Thus the shaded area = Area of curvilinear trapezoid APQB - Area of curvilinear trapezoid CPQD.

Now consider the case when f (x) and g (x) intersect each other in the interval [a, b].

First of all we should find the intersection point of y = f (x) and y = g (x). Let the root is x = c. (We consider only one intersection point to illustrate the phenomenon)

Which gives the required area.

Area between two curves y = f (x), y = g (x) and the lines x = a and x = b is always given by 
provided f 9x) > g (x) in [a, b];

the position of the graph in immaterial. As shown in given figure.



Example 4:

Find the area included between the curves y =sin-1 x, y = cos-1 x and and x-axis.

Solution

Clearly, we have to find the area of the shaded region OPBO. The point P of intersection of y = sin-1 x and y = cos-1 x is obtained by solving sin-1 x = cos-1


= >  / 2 - cos-1 x = cos-1 x

= > 2cos-1 x =  / 2 => cos-1 x =  / 4

= > x = 1 / √ 2


Example 5:

Compute the area of the region bounded by the straight lines x = 0, x = 2 and the curves y =2x, y = 2x - x2

Solution

Since the maximum of the function y = 2x - x2 is attained at the point x = 1 and is equal to 1, and the function y = 2x ≥ 1 on the interval [0, 2], we have

2x > 2 x - x2, for all x Є |0, 2|.



Example 6:

Find the area bounded by the curves x = y2 and x = 3 - 2y2

Solution

The two curves represent parabolas with vertices at (0, 0) and (3,0). They interesect at (1, 1) and (1, -1), so Area of curve OPMQO = 2 ( area of curve OPMO )