Area Between Curves

Area Between Curves

Using integration we can find the area between a function and the x-axis, which allows us to find the area of many different shapes. This method has one serious limitation, however - it can only be used to find the area of an object when one side of the object is flat. In general, we might want to find the area in an arbitrary shape. How can we propose doing so? We can define each edge of the object as a different function, and find the area between the two functions. In doing so, we will always be considering conventional area.

Suppose we want to find the area between sin(x) and - sin(x) over the interval [0, ]. In this case some of the space is above the x-axis, and some is below. If we want to find the conventional area between these two functions, then we must count the area above and below the x-axis both as positive. Using the fact that

2

sin(x) = 2

0

and recognizing the area below the x-axis is exactly the same, we find the total area to be 4. The next question is how to solve such a problem more generally, without needing to rely on symmetry considerations.

Let us suppose we have functions f and g, and we want to find the area between them over some interval of interest, where f (x) g(x). Suppose we transform both f and g in an identical way, for instance, by adding 2 to both functions. In doing so both functions will shift up by 2, but the distance between the two functions will not be altered at all. Thus, the area between the two functions will remain the same as before the transformation. Keeping this in mind, we can go one step further, and think of deforming both functions in a way so that the bottom curve g is fit to the x-axis. Since both functions are being deformed identically, the area between them will not change. Now the area between f and g is defined by the distance between a deformed version of f and the x-axis, which is an area we know how to calculate. We simply need to integrate the deformed f over the interval of interest, and we will have the area between f and g.

It may appear that there is one difficult thing which we still have not accounted for - how do find such a deformation? In fact, it is quite an easy obstacle to overcome. If we subtract g(x) from g(x) at all points, the result will be a function that is identically 0. We simply perform the same deformation to f , so the function we are interested in integrating is f - g. In conclusion, if we have two functions f and g and over some interval [a, b], and f (x) g(x), the area between f and g is given by

b

A = (f (x) - g(x))dx

a

When faced with a specific example of finding the area between two functions, we simply need to identify on which intervals one function is greater than or equal to the other, and evaluate the appropriate integrals. One additional subltety to be wary of is finding the area

enclosed between two curves. In this situation we will only be interested intervals that have endpoints where the functions f and g are equal, so that the area will form a closed region. In contrast, if we were trying to find the area enclosed between x and -x the answer would be 0, because these functions only intersect at one point, not forming a closed shape.

Example 1 Find the area enclosed by 2 - x2 and -x. Solution The first consideration we need to make is to find where the two curves intersect, so that we can determine which regions are enclosed by the two functions. To do so we set the two functions equal, and solve.

2 - x2 = -x x2 - x - 2 = 0 (x + 1)(x - 2) = 0

So the region of interest is [-1, 2]. Next, we need to identify which function is greater on

this interval. To do so, we only need to evaluate a convenient sample point, because our

functions are continuous, and only intersect at -1 and 2, so they cannot cross one another in this interval. Choosing the convenient point x = 0, we see that 2 - 02 > -0, so the parabola is our top function. Now we subtract -x from 2 - x2 and integrate, yielding

2 (2 - x2 + x)dx = 2x - x3 + x2 2

84

11 9

= (4 - + ) - (-2 + + ) =

-1

3 x -1

32

32 2

Example 2 Find the area enclosed between x from above and the x-axis and y = x - 2

from below.

Solution The first thing to recognize is that we can rewrite the lower bound of the x-axis

and the line y = x - 2 as a single function, defined piecewise. At first the x-axis is above

the line, so it acts as our lower bound, but after they intersect at x = 2, the line is higher

and becomes our new lower bound. Thus, we can define a new function

f (x) =

0 0x2 x-2 x>2

and recognize that the original problem was really just to find the area between x and f (x). Now we note that these two functions intersect at x = 0 (where x stems from the x-axis), and x = 4, because 4 = 2 = 4 - 2. Finally, we integrate

4

4

4

4

2

4

A = ( x - f (x))dx =

xdx - f (x)dx =

xdx - 0dx - (x - 2)dx

0

0

0

0

0

2

=

3 x3/2

4

+

x2 (

-

2x)

4

=

10

2 02

23

Above we could have just as well split the integral from 0 to 2 and 2 to 4 for both functions, but we saved ourselves one integration of x by evaluating the integral as we did.

If we think about the previous problem is a slightly different way, we can much more easily find the area. Rather than thinking of both curves as functions of x, we can think of them as

functions of y, and integrate along y. In doing so we avoid dealing with piecewise functions, which simplifies the calculation. To find the area between two functions of y, where f is to the right of g, we simply evaluate

d

A = (f (y) - g(y))dy

c

where we are interested in the vertical region c to d.

Example 3 Find the area enclosed between x from above and the x-axis and y = x - 2 from below, by integrating with respect to y. Solution The first thing is to write our functions in terms of y, so we have x = y2, and x = y +2. The functions still intersect at the same points (0, 0 and (4, 2), but now y +2 > y2. Rather than integrating from 0 to 4 along x, we integrate from 0 to 2 along y, and look at the difference of these two functions of y.

A=

2

(y + 2 - y2)dy

=

y2

+ 2y -

y3

2

=2+4-

8

=

10

0

2

30

33

In general, it is helpful to draw a picture of the situation to decide which variable to integrate along, and determine which function is above the other (or to the right of the other).

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