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An explanation of how to calculate the area between two curves using the definite integral. It covers the concept of integrals as a summing mechanism, setting up integrals for area calculations, and using the fundamental theorem of calculus or approximation. Examples and exercises.
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MATH 116 ACTIVITY 3: Area between curves and the definite integral
WHY: Area between curves gives a nice “practice” application of the definite integral (some setup, but not too complex)
but also serves as a model of the total effect of a difference over time – total change in a population will be the
difference between the effect of birth rate and death rate, for example [so that total change would be represented
by the area between the birth curve and the death curve )
with all the results in the recorder's report.
We can use the definite integral to calculate the area between two graphs in much the same way we use it to calculate the
area under the graph of a positive function. As before, this gives a representation of the total change - but this time we
allow for two rates of change interacting. We still want to multiply the base by height to get area – but this time the height
depends on two graphs (one at the bottom of the area, one at the top) rather than just one.
the rule is that the area between two graphs, from x = a to x = b, is given by
( y ( top ) − y ( bottom )) dx
a
b
up rectangles for approximation, the base for each is x and the height is (top – bottom).
If the graphs do not cross, this is straightforward – we use
f
top
( x ) − f
bottom
( x ) dx
a
b
If the graphs cross, then the formula for y(top) sometimes comes from one function, sometimes from the other, and we
have to split up the integral using the
f ( x ) dx =
a
b
f ( x ) dx +
a
c
f ( x ) dx
c
b
property of the definite integral.
Usually we need to sketch the graphs (take advantage of our calculator] to see whether the graphs cross and which graph is
on top where.
Example 1: Area between the graphs of
1
2
2
between x = 0 and x = 2.
Looking at the graphs, we see that the graph of f 2
is above the graph of f 1
(f 2
(x) is larger than f 1
(x) ) for all x values
between 0 and 2, so y(top) is f 2
(x) and y(bottom) is f 1
(x). this tells us area =
f
2
( x ) − f
1
( x ) dx =
0
2
2 x + 1 ( )
− x
2
dx
0
2
€
= 2 x − x
2
0
2
= x
2
−
x
3
3
0
2
= 8 −
8
3
=
16
3
Example 2: Area between the graphs of
1
2
and
2
between x = -1 and x = 2.
Looking at the graphs, we see that these graphs cross twice between x = 1 and x = 2 – solving the equation
1
2
2
] tells us they cross at x = 0 and at x = 1. For x < 0, the graph of
1
2
is on
top; from x = 0 to x = 1, it is that is on top, and for x > 1,
1
2
is again on top. We split up the integral so that
we can write
( y ( top ) − y ( bottom )) dx
a
b
(and know what formula is “top” and which is “bottom”) and get
Area =
2 x
2
− 2 x ( )
dx +
− 1
0
2 x ( )
− 2 x
2
dx
0
1
2
− 2 x ( )
dx
1
2
3
2
− 1
0
2
2
0
1
3
2
1
2
3
2
3
2
2
3
2
3
3
2
3
2
[I suggest using a calculator to sketch graphs of these, before diving into calculations]
from x = 0 to x = 4 Make sure you
show the integral (or integrals) involved here.
3
2
and
2
from x
= 0 to x = 5
3 This table gives a set of values for two functions f and g at various values of x. Use the midpoint approximation, with n
= 5, to approximate the area between the graphs from x = 2.0 to x = 3.0 [there is enough information here – in fact,
there’s more than you need -- to calculate this approximation). What is x? what are the subintervals? What are the
midpoints? What is the value of (top – bottom) at the midpoints?