In Section 14.3, you will learn how to find the area between two curves. Suppose you have two functions f(x) and g(x). Also assume that the higher curve is f(x). We are interested in finding the area from a point x = a to x = b between the two curves. We can do this by finding the area below f(x) and above the x-axis,
$latex \displaystyle \int\limits_{a}^{b}{f(x),dx}$
and subtracting the area below g(x) and above the x-axis,
$latex \displaystyle \int\limits_{a}^{b}{g(x),dx}$
Alternately, we can subtract the functions first and then find the area,
$latex \displaystyle \int\limits_{a}^{b}{\left[ f(x)-g(x) \right],dx}$
On Monday, the face-to-face class worked several of these types of problems.
To start these problems, graph the different equations to see the region you are finding the area of.
Problem 1 Find the area between the curves $latex \displaystyle y={{x}^{2}}-30$ and $latex \displaystyle y=10-3x$.
The key to this problem is to first find where the curves intersect by setting the functions equal to each other…these solutions for the limits on the definite integral.
Problem 2 Find the area between the curves $latex \displaystyle y={{x}^{2}}-18$ and $latex \displaystyle y=x-6$.
The points of intersection may also be found using a graphing calculator.
In the next problem. the region that is enclosed is a little different since the curves cross. When this happens, the region needs to be broken into two parts.
Problem 3 Find the area of the region enclosed by x = 0, x = 6, y = 5x and y = 3x + 10.
The two line cross at x = 5. On the left side of this point y = 3x + 10 is higher than y = 5x. On the right side of x = 5, y = 5x is higher than y = 3x + 10. This means the order in which the functions are subtracted must change.
Problem 4 Find the area enclosed by x = -2, x = 1, y = 2x, and $latex \displaystyle y={{x}^{2}}-3$.
