Solution to Exercise 12.3-0© 2005 by Karl Hahn |
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The problem was to find the area between the curve, x2 - 3x + 3, and the line, y = x. Since having a graph for problems like this is always useful, such a graph is shown here on the right. The graph suggests where the intersection points are, but still we should do the first step of the procedure, which is to find where the difference of the two functions is zero.
x2 - 3x + 3
x
x2 - 4x + 3 = (x - 1)(x - 3)
You should be able to factor this difference by inspection. The factorization confirms
what the graph suggests, that the intersection points are at x = 1
and x = 3.
The second step is to take the definite integral of the difference from the first intersection point to the second.
A = |
3 |
x2 - 4x + 3 dx = |
x3 |
- 2x2 + 3x |
3 |
= 0 - |
4 |
4 |
1
1
So why, you might ask, does the area end up being negative? Look at the graph and then look at how we took the difference between the curve and the line. In the region of the enclosed area, the line is consistently greater than the curve. Yet we took the curve minus the line. That difference must be negative throughout the region. Don't let that throw you. When you get a negative area in a problem like this, simply look at the polarity of how you took the difference. Once you can see that you took the negative version of the difference, simply take the absolute value of the result of the definite integral, and that will be the correct area. So the correct area in this problem is 4/3.
Supplement: If you plot the line, y = 3 - x, over the curve in
this problem, you can see by symmetry that the resulting area enclosed between the
curve and this new line ought to be the same. Use the method outlined here to confirm
that.