where f (xi) is the height and 
is the base of the ith rectangle. The Fundamental Theorem of Calculus tells us that this area can be found by evaluating the definite integral 
where a and b are the limits of integration.
| AREA |
Area Under a Curve
In the lesson on Riemann sums, we learned that the area between a curve and the x-axis is found by evaluating where f (xi) is the height and is the base of the ith rectangle. The Fundamental Theorem of Calculus tells us that this area can be found by evaluating the definite integral where a and b are the limits of integration.
If the curve is always above the x-axis, we simply integrate from a to b. However, if the curve of the function crosses the x-axis, we must break the area up into sections in order to evaluate it. For this reason, it is always best to make a diagram of the situation.
Example 1: Find the area between f (x) = x² + 3 and the x-axis from x = 0 to x = 4.
Since the curve doesn't cross the x-axis, we simply evaluate the integral from x = 0 to x = 4.
So, the area is 
Since the value = 0 at x = 0, we substitute x = 4
The area under this curve from x = 0 to x = 4 is 48 units².
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Example 2: In this one, we have to break the area up into 2 sections since the curve crosses the x-axis on the interval of integration.
Find the area between the x-axis and the curve f (x) = x² – 4 from x = 0 to x = 4.
Here's the picture
Because this curve crosses the x-axis at x = 2, making the y-values negative, we must integrate the negative of f (x) from 0 to 2 and the function f (x) from 2 to 4.
In this case, 
This is really 
Once we integrate (antidifferentiate), we get
again, at x = 0 everything disappears so,
The area between this curve and the x-axis from x = 0 to x = 4 is 16 units².
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Area Between Two Curves
When we have to find the area between two curves, we must pay attention to which curve forms the top or ceiling and which forms the bottom or floor of the region between them. We must also find their points of intersection to determine the limits of integration.
Example 3: Evaluate the area between y = 9 – 2x² and y = – x².
Setting y = y, we learn that the points of intersection are at x = ± 3.
Here's the picture
Since the curves are symmetric to the y-axis, we need only find the area on one side and then multiply it by 2. We'll take twice the area on the right side of the axis so we'll integrate from x = 0 to x = 3.
The blue curve, y = 9 – 2x² is the ceiling and
the red curve y = – x² is the floor, so our area is:
Once we antidifferentiate and evaluate, we get that the Area = 48 units².
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Area of a Region Defined by More Than Two Curves
Now let's do an example where we're given 3 different functions and we want to find the area of the region enclosed between them.
Example 4: Find the area of the region defined by y = x, y = 3x, and y = – x + 4.
Here's the picture:
The line y = x is always the floor.
At first, the ceiling is y = 3x from x = 0 to x = 1
right of x = 1, y = – x + 4 is the ceiling up to x = 2.
Therefore, the area here is found by evaluating
the area is 2 square units.
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Sometimes, the functions are such that it is impossible to use the x-axis as reference to complete the question because the top and bottom of the region are on the same curve. In such a case, we use x = f (y) and y-values (instead of y = f (x) and x-values) to evaluate the integral that represents the area of the region between the curves. With these questions, the rectangles that make up the area are horizontal rather than vertical, so each base is a chunk of the y-axis instead of the x-axis. The "top" or "ceiling" is now the curve that is furthest right and the "bottom" or "floor" is formed by the curve that is furthest left.
Example 5: Find the area of the region bounded by x = y² and x = – y² + 8.
Here's the picture:
When we set x = x, we get that y = ± 2, and since everything is symmetric to the x-axis, we will evaluate twice the area above the x-axis. So instead of integrating from y = – 2 to y = 2, we will integrate from y = 0 to y = 2 and then we'll double it.
The "ceiling" is the blue curve, the "floor" is the red one.
This becomes 
the area is 64/3 square units.
Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.
Practice
Sketch the region bounded by the graphs of these functions, then find the area of the region.
| 1) y = 1/x² | y = – x² | from x = 1 to x = 2 |
2)  |
y = – x | from x = 1 to x = 4 |
| 3) y² = – x | x – y = 4 | from y = – 1 to y = 2 |
| 4) y² = x | y – x = 2 | from y = – 2 to y = 3 |
| 5) x – y + 1 = 0 | 7x – y – 17 = 0 |
2x + y + 2 = 0 |
Solutions
1) y = 1/x², ................ y = – x², .......... x = 1, .......... and .......... x = 2
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2) 
, .............. y = – x, ........... x = 1, ............ and .......... x = 4
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3) y² = – x , ................x – y = 4 ......... y = – 1, .......... and .......... y = 2
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4) y² = x , ..................y – x = 2 ......... y = – 2, .......... and .......... y = 3
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5) x – y + 1 = 0 ........ 7x – y – 17 = 0 .......... 2x + y + 2 = 0
When we put these equations in standard form we get:
y = x + 1........ y = 7x – 17.......... and y = – 2x – 2
( Calculus II MathRoom Index )
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