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Mathematical Expression Editor
Here we’ll practice finding area between curves.
Find the area of the region bounded by the curves and .
First we find the points of intersections.
By graphing the two functions, we can see that is always greater than on the
interval .
Thus the area of this region is
Find the area of the region bounded by the curves and in the first quadrant.
First we find the point of intersections.
is greater than on the interval
Thus the area of this region is
Since this region is a triangle with base of length on the -axis, and height , one could
also have calculated this area geometrically.
Find the area of the region bounded by two consecutive intersections of the curves
and .
We can use any two consecutive intersections, but the first two positive intersections
are convenient.
These occur at and
is greater than on the interval
Thus the area of this region is
Find the area of the region bounded by the vertical lines and , and the curves and
.
There are no intersection points between the curves on this interval, and is always
above on this interval.
Thus the area of this region is
Find the area of the region bounded by the curves and , and .
First sketch a picture of the region
Next we find the -coordinates of the relevant points of intersection.
Let be the point of intersection between and , between and , and between and
.
Then
Thus , , and (make sure you understand why and are not the correct
intersections).
Thus the area of this region is
The first summand is
The second summand is
Note that these two area equal, which we could have also discovered by observing
the symmetry of the figure about the line .
Find the area of the region bounded by the curves and and above the -axis. Try to
solve this problem using horizontal rectangles, i.e. by integration with respect to
.
First sketch a picture of the region
Rather than using two different regions of integration, this problem is a great case
opportunity to use horizontal rectangles.