Finding the Area Between Curves
© 2005 by Karl Hahn
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Your instructor will, at some point, ask you to find the area between two
curves. Most of these problems are not that difficult, and once you understand
the attack, you should be able to do them fairly easily. So let's just dive
into a typical example. The problem is, find the area enclosed by the two
curves,
1
y = |
(x3 - 9x2 + 11x + 2) |
and
1
y = |
(-2x2 + 7) |
A plot of these two functions appears on the right. The first curve is shown
in blue, the second in green. The area to which the problem refers is pretty
clear because there is only one enclosed area -- that narrow strip that
runs diagonally down for a ways. Making even a crude plot of the functions
in question is always a good preliminary step because it gets you visually
in touch with what the problem is asking for.
So how do we establish the boundaries of the problem? Well, the area in question is bounded on the right and left by the intersections of the two curves. So a useful step would be to find where these two curves intersect. And that is the same as finding where their difference is zero (note that for the purpose of finding the intersections, we do not need to carry the 1/5 factor into the calculations below. BTW I put in 1/5 factor only so that the graphs of these functions would be manageable).
x3 - 9x2 + 11x + 2 - (-2x2 + 0x + 7)This is a cubic, but don't go rushing to look up the cubic formula. It can be factored easily. If you have forgotten the methods for factoring polynomials, then take a moment to review them now. You should see that if there are easy factors to this, the only choices for those factors are (x ± 1) and (x ± 5). From the plot you can see that the intersection points are to the right of the y axis, so the candidates are narrowed down to (x - 1) and (x - 5). And when you try polynomial long division with either of the candidates, you find that they both divide x3 - 7x2 + 11x - 5 evenly. When you work it all out you find thatx3 - 7x2 + 11x - 5
1 |
(x3 - 7x2 + 11x - 5) = |
1 |
(x - 1)(x - 1)(x - 5) |
So the difference between the two polynomials is zero at
x = 1 and x = 5. This means the two
polynomials, y = (1/5)(x3 - 9x2 + 11x + 2) and
y = (1/5)(-2x2 + 7) are equal to each other at those same
two values of x.
In the diagram on the right we have shown a vertical slice (in red) between the
two polynomials. Although the slice in not quite rectangular -- its top and bottom
edges are not horizontal straight lines -- as the width, dx, of the slice is made smaller
and smaller, the area of the slice becomes closer and closer to dx times the
difference between the two functions. Make sure you see why this is. If you add up all
the areas of all the slices that constitute the area between the two curves, you will
have the total area between the two polynomials. And with rectangular slices, you have
a Riemann sum. Hence the area between the curves is the definite integral of the difference
between the two polynomials taken from the left-hand point where they intersect
to the right-hand point where they intersect.
A = |
5 |
1 |
-2x2 + 7 |
- |
1 |
x3 - 9x2 + 11x + 2 |
dx |
1
Combining terms the above becomes
A = |
1 |
5 |
-x3 + 7x2 - 11x + 5 dx = |
1 |
- |
1 |
x4 + |
7 |
x3 - |
11 |
x2 + 5x |
5 |
1
= |
275 |
19 |
64 |
From the above example you should see that problems like this one, where you are to find the area enclosed between f(x) and g(x), are solved in just two steps:
Step 1: Find the values of x where f(x) - g(x) = 0.
Step 2: If the solutions to step 1 are x1 and x2, then find the definite integral of the difference between f(x) and g(x) from x1 to x2.
Exercise: Find the area between y = x2 - 3x + 3 and y = x.
Here's a variation on this type of problem that you might also encounter. Find the area bounded by the following:
| a) The x axis, b) the line, | y = |
3 |
1 |
and c) the curve, | y = -x2 + |
17 |
21 |
A plot of this problem is shown to the right. You can see that the area in question has
two straight-line sides on the top and bottom and two curved sides on the right and left.
There are two ways you can work this. One is to do three separate integrals and add
the resulting areas. Starting from the left, you would integrate the curve from its left-hand
intersection with the x axis to its left-hand intersection with the sloped line.
Next you would
integrate the sloped line from its left-hand intersection with the curve to its right-hand
intersection with the curve. Finally you would integrate the curve from its right-hand
intersection with the sloped line to its right-hand intersection with the x axis.
The other way of doing it is to use the method we've already discussed to establish the
area enclosed by the sloped line and the curve and subtract that area from the area enclosed
by the curve and the x axis. Either of the two ways you do it, you must know
both points of intersection between the sloped line and the curve and both points where
the curve intersects the x axis.
You find the points where the curve intersects the x axis simply by applying the quadratic formula to the quadratic equation given for the curve. You find that these points are at x = 3/2 and at x = 7. To find the points where the curve intersects the sloped line, take the equation of the curve minus the equation of the line, which is -x2 + 7x - 10. When you apply the quadratic formula to that you get x = 2 and x = 5.
Using the second method outlined above:
A = |
7 |
-x2 + |
17 |
x - |
21 |
dx - |
5 |
-x2 + 7x - 10 dx |
A = |
- |
x3 |
17x2 |
- |
21x |
7 | - |
- |
x3 |
7x2 |
- 10x |
5 |
= |
1331 |
- |
9 |
= |
1115 |
I leave it as an exercise to you to use the first method outlined for this problem and come up with the same answer.
Now here's an example of one last variation on this that you might encounter. Find
the area between the curves y = x and y = sin(2x). This
is done exactly the same as finding the area between the parabola and the line except
for one new wrinkle. Finding one of the points of intersection is not as easy. Clearly
the left-hand point of intersection is x = 0. But when you go trying
to solve sin(2x) - x = 0, you find that no matter how much algebra
and trig manipulations you do, you cannot get a closed form solution to this equation
other than x = 0. The right-hand solution does exist. You just can't
solve for it. Because of that we say that the right-hand intersection point is
transcendental (math-speak for you-can't-solve-for-it). So instead we do
Newton-Raphson iteration to approximate the solution.
Recall that Newton-Raphson finds the solution to f(x) = 0 by iterating xn+1 = xn - f(xn)/f'(xn). In this case f(x) = sin(2x) - x and f'(x) = 2cos(2x) - 1. So we iterate
xn+1 = xn - |
sin(2xn) - xn |
The only question is what do we use for our first guess, x0?
Since this equation does have two solutions, we don't want the iteration to converge on the
solution we already know, that is we don't want it converging to x = 0.
So let's start with
x0 past the "hump" in sin(2x). Since the
hump happens at x = π/4 ≈ 0.785, let's start with
x0 = 1. Using a 15-digit calculator, I find:
x0 = 1.000000000000000 x1 = 0.950497797101954 x2 = 0.947755822689797 x3 = 0.947747133604357 x4 = 0.947747133516990 x5 = 0.947747133516990So you would find
A = |
0.947747133516990
|
sin(2x) - x dx |
= |
- |
1 |
cos(2x) - |
1 |
x2 |
0.947747133516990 |
The evaluation of the above is left as an exercise.
Exercises
1) Find the area between the curve y = x4 - 3x3 + 3x2 - 2x + 2 and the line, y = x. (Note that the difference between the line and the fourth degree polynomial is easily factored).
2) Find the area bounded by the y-axis, the curve, y = ex, and the curve, y = 2e-x. (Note that you do not need to use Newton-Raphson to find where the two curves intersect. Just set the function for one of the curves equal to the function for the other, take the natural log of both sides -- remembering the log identity, ln(ab) = ln(a) + ln(b), and solve for x).
3) Find the area bounded by the y-axis, the curve, y = cos(x), and the line, y = x. (Note that you will have to use Newton-Raphson to find the intersection point for this).
4) Find the total area enclosed between the curves,
y = |
9 |
and | y = |
13 |
x2 |
Note that two curves can enclose more than one area.