1. 4.6 Area Between Curves
We applied the notion of the integral to calculate areas of only one type: the area under a curve bounded by the x-axis. Now, we expand this technique to the area between any curves.
To refresh the technique:
Example: find the area bounded by the curve y=x2, the verticals x=2 and x=3, and the x-axis.
Solution: Form the integral that represents the area....
To understand this in more detail, draw a picture that shows the area..
The expression under the integral, ydx, represents the area of the typical element. Then, the typical element itself is a rectangle of the height y and base dx.
Draw such a rectangle…. 1

2. Area Between two curves:
Now, we can easily extend this technique to the case where the curve bounding the area from below is not the x-axis (which is the line y=0) but any curve:
In the general case of the curve f(x) bounding the area from above, and g(x) bounding the area from below, the height of the typical element is [f(x)-g(x)], and the base is the same, dx.
The area of this element is the product of those two: [f(x)-g(x)]dx.
Finally, we need to sum these areas over the interval of integration [a,b]: 2

3. Note on the sign of integral:
Using the linearity of the integral, the above expression can be rewritten as
If both functions are above the x-axis, this can be interpreted as the difference between two areas: the area under f(x) bounded by the x-axis and the area under g(x) bounded by the same axis.
In the case when g(x) is under the axis, its values are negative, and the area of the region between the curves must be grater than the area under f(x) bounded by the x-axis, , because the integrand, that is the height of the typical element, is greater:
So, in the above difference of integrals, the second must be negative 3

4. Conclusions on the sign of integral:
1. The negative sign of the integral shows that the curve bounding the figure lies below the x-axis. The area of the figure in this case is the absolute value of the integral
2. If the curve f(x) bounding a region intersects the x-axis at some point c inside the interval of integration [a,b], then the area bounded by this curve and the x-axis must be calculated as a sum of the areas of the figures to the left and to the right from the point of the intercept:
Exercise: Find the area of the region bounded by y=x3, x=-2, x=1and the x-axis.
Hint: find the x-intercept of the graph of the given function. 4

5. Example: Find the area between curves:
Solution:
1. Graph the functions and find the intersections of the graphs…
2. Chose a typical element as a thin vertical rectangle if the curves are given as functions of x (horizontal rectangle if the curves are given as functions of y).
3. Define the height of this rectangle (which curve bounds form below, which from above?): [________]
4. The area of the rectangle is [_________]dx
5. The area of the region between the curves is the integral of the above areas in the limits of integration defined by the intersections of the graphs: 5

6. Example: Find the area between curves:
Solution:
1. Graph the function as x=y2-y. The other function is x=0 since this is the formula that gives the y-axis. Intersections between them:
2. Chose a typical element as a thin horizontal rectangle.
3. The length of this rectangle (which curve bounds form the right, which from the left?): [0– (y2–y)]= [y–y2]
4. The area of the rectangle is [y–y2]dy
5. The limits of integration are defined by the intersections of the graphs: 6

7. Exercise: Find the area between curves:7

8. Homework:
Section 4.6: 1,3,5,9,11,13,21,23,29,31. 8