Solution to Problem 12.3-4© 2005 by Karl Hahn |
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The problem is to find the area enclosed by the curves,
y = |
9 |
and | y = |
13 |
x2 |
Before looking at the graph, we will find points of intersection by solving for when the difference between the two functions is zero.
x2 |
13 |
9 |
Multiplying by 4x2 gives
x4 - 13x2 + 36 = 0which is a quartic (i.e., a fourth degree polynomial). But before you go looking up the quartic formula, observe that all the exponents are even. So you can substitute u = x2 to convert it to a quadratic in u.
u2 - 13u + 36 = 0You can apply the quadratic formula to solve this, but you shouldn't have to. It is easily factored into (u - 4)(u - 9). If you have u = 4 and u = 9 and u = x2, then x = ±2 and x = ±3. More than one pair of solutions means more than one region of enclosed area. But with four solutions for the intersection points, how do you pair them to find the regions of enclosed area? Observe that y = 9/x2 has an undefined point at x = 0. So no region of enclosed area can include x = 0. That means you must pair x = -2 with x = -3 and x = 2 with x = 3.
Observe that the functions in the graph are symmetric about the y-axis. This is the case because all the exponents are even. It means that the two enclosed regions have equal areas. So we only have to integrate over one of the two regions, then multiply the result by 2 to get the total area.
To avoid getting a negative result, we take into account that parabola (that is, the blue function) is consistently greater than the y = 9/x2 function in the regions of enclosed area. This clues us as to how we take the difference in the integral.
A = 2 |
3 |
13 |
x2 |
9 |
dx = |
1 |
3 |
13 - x2 - |
36 |
dx |
2
A = |
1 |
13x - |
x3 |
36 |
3 |
= 21 - |
62 |
1 |
2