Tuesday, October 15, 2013

Calculus Lesson: Area Between Curves

Area Between Two Curves

The general formula for finding the area between two curves is:

b
∫ T(x) - B(x) dx
a

I named the functions T(x) and B(x) specifically. T(x) represents the function "on top", while B(x) represents the function "on the bottom". In short, for x ∈ (a,b), T(x) ≥ B(x).

Here is an approach to use when finding areas between curves. A graphing calculator or mathematical software can be helpful in this procedure.

1. Draw the graphs. This is where a graphing calculator comes in handy.

2. Determine a and b, your limit points. Sometimes one or both of the limit points are found by finding the intersection of the two curves. Other problems will give you a third (or even fourth) condition. An example is x≥0, which makes one of our limit points x=0.

3. Determine which of the curves is T(x) and B(x). T(x) has greater numerical value than B(x) over the interval x ∈ (a, b).

4. Take the definite integral of T(x) - B(x). The result is your area.

b
∫ T(x) - B(x) dx = Area
a


About the examples and problem set: Please note that most examples in math textbooks, the numbers come out "nice and neat". Not so here, so I am using an HP Prime to demonstrate these examples and come up with answers. (any graphing calculator, HP, TI, Casio, all will do just fine!)

Example 1:

Find the area between:
y = -x^2 + 2
y = x^2 + 1

Here is a graph of the two curves:

Finding the limit points:

-x^2 + 2 = x^2 + 1
1 = 2*x^2
1/2 = x^2
Which implies x = -1/√2 and x = 1/√2

So our limit points are a = -1/√2 and b = 1/√2

From the graph we note that:
T(x) = -x^2 + 2
B(x) = x^2 + 1
and
T(x) - B(x) = -x^2 + 2 - (x^2 + 1) = -2*x^2 + 1

Taking the integral:

1/√2
∫ -2*x^2 + 1 dx ≈ 0.94281
-1/√2

So the area for Example 1 is about 0.94281.

Example 2:

Find the area between the two curves:
y = 2 cos x
y = x - 2
With the condition x≥0

First we graph the two curves:

The two graphs intersect when x>0. Since we are given the additional condition that x≥0, the lower limit for this problem is a=0.

To find the upper limit, we find intersection point:
2 cos x = x - 2
2 cos x - x + 2 = 0

Using a numeric solver on a calculator, we find that the intersection point is x ≈ 1.71419. (See, the numbers in these examples aren't always as "neat").

In the region x ∈ (0, 1.71419), we see that 2 cos x > x - 2. Hence:

T(x) = 2 cos x
B(x) = x - 2
and the area is approximately

1.71419
∫ 2 cos x - (x + 2) dx ≈ 3.93863
0

Problems to Try

Problem 1
Find the area between the curves of
y = x^2
y = 4

Problem 2

y = x^3
y = -3*x + 8
With the condition x≥0

Problem 3

y = 8*x^2 + 4*x - 6
y = x



Check your answers:
Areas are rounded to five decimal points:
1. 10.66667
2. 7.36017 (Hint: Upper Limit ≈ 1.51275)
3. 7.42101 (Hint: Use a calculator to find the limit points)


Any questions, please leave them in the comment box. Hope this is helpful and thanks!


Eddie


This blog is property of Edward Shore. 2013


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