3. Area Between 2 Curves
by M. Bourne
We are trying to find the area between 2 curves, y1 = f1(x) and y2 = f2(x) and the lines x = a and x = b.
We see that if we subtract the area under lower curve
y1 = f1(x)
from the area under the upper curve
y2 = f2(x),
then we will find the required area. This can be achieved in one step:
Alternative Way to Find The Formula (First Principles)
Another way of deriving this formula is as follows (the thinking here is important for understanding how we develop the later formulas in this section).
Each "typical" rectangle indicated has width Δx and height y2 − y1, so its area is (y2 − y1)Δx.
If we add all these typical rectangles, starting from a and finishing at b, the area is approximately:
Now if we let Δx → 0, we can find the exact area by integration:
Likewise, we can sum vertically by re-expressing both functions so that they are functions of y and we find:
Example:
Need Graph Paper?
Find the area between the curves y = x2 + 5x and y = 3 − x2 between x = -2 and x = 0.
- Answer
-
Sketching first:
So we need to find:
Exercises
1. Find the area bounded by y = x3, x = 0 and y = 3.
- Answer
-
Sketch first:
We need to use:
and use horizontal elements.
In this case, c = 0 and d = 3.
We need to express x in terms of y:
y = x3 so x = y1/3
So
2. Find the area bounded by the curves
y = x2 + 5x and y = 3 − x2.
(This is an extension of the Example above.)
Here is the answer using LiveMath:
Normal answer:
- Answer
-
Sketch first:
We need to use:
We note that y = 3 − x2 is above y = x2+ 5x so we take
y2 = 3 − x2 and y1= x2 + 5x
Points of intersection occur where:
x2 + 5x = 3 − x2
2x2 + 5x − 3 = 0
(x + 3)(2x − 1) = 0
So x = -3 or x = 0.5
We take vertical elements.
So the area is given by:
3. Find the area bounded by the curves
y = x2, y = 2 − x and y = 1.
- Answer
-
Sketch first:
We need to take horizontal elements in this case.
So we need to solve y = x2 for x:
x = ±√y
We need the left hand portion, so x = − √y.
Notice that x = 2 − y is to the right of x = -√y so we choose
x2 = 2 − y and x1 = -√y.The intersection of the graphs occurs at (-2,4) and (1,1).
So we have: c = 1 and d = 4.
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