Using Integration to Find Areas
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As you know from the end of Calculus I, the definite integral a f (x)dx represents the area between the function f (x) and the x-axis along the interval
[a, b]. In this chapter we will expand upon this idea. In this section specifically, we will revisit talking about area under and between curves.
Objectives
• Be able to draw regions in the plane
• Be able to calculate area under and between curves
• Be able to calculate the area bound by multiple curves or piecewise
defined functions
General Strategies
Area underneath f (x) Area underneath g(x)
along [a, b]
along [a, b]
g
g
f(x)
f(x)
a
b
Z
b
fx
a
b
Z
f (x)dx
a
b
Area between f (x)
and g(x) along [a, b]
g
f(x)
fx
a
Z
a
1
fx
b
[f (x) − g(x)] dx
g(x)dx
a
b
Examples
Example 1 Find the area bounded by f (x) =
interval [1, 3]
x2
2
and the x-axis along the
Solution.
f
f(x)
1
3
Z
Area =
1
3
3
x
x2
dx
2
x 3
|
6 1
27 1
=
−
6
6
26
=
6
13
=
3
=
Example 2 Find the area of the region bounded by the positive y-axis, y = 8,
and f (x) = x3 .
2
Solution.
f
f(x)
y=8
x
Notice that we weren’t given the interval over which to integrate. We were
only given what the region looks like. How do we know what our limits
of integration should be? We need to find the points of intersection which
begin and end our region. We know that we are bounded by the positive
y-axis, which means we should begin from x = 0 (you can see this from our
drawing of the desired region). We find the second endpoint by seeing where
f (x) = x3 intersects the horizontal line y = 8.
x3 = 8
=⇒ x = 2
So, we will integrate over the interval [0, 2]. This gives
Z 2
Area =
8 − x3 dx
0
(Remember that when finding the area between two curves we subtract the
bottom function from the top function within the integral)
= 8x −
3
x4 2
|
4 0
= (16 − 4) − (0 − 0)
= 12
Example 3 Find the area bounded by the curves f (x) = 2−x2 and g(x) = x
Solution.
f(x)
g
x
f
Firstly, we need to identify which is the top function and which is the bottom.
From our picture, it should be clear that f (x) = 2 − x2 is the top function
and g(x) = x is the bottom function.
Next we need to determine the limits of integration. That is, we need to
know when these two functions intersect.
2 − x2 = x
=⇒ x2 + x − 2 = 0
=⇒ (x + 2)(x − 1) = 0
=⇒ x = −2 and x = 1
4
Now, we have found that we should integrate f (x) − g(x) along the interval
[−2, 1], to find the desired area.
Z 1
(2 − x2 ) − x dx
Area =
−2
Z
1
=
2
−x − x + 2 dx
−2
−x3 x2
−
+ 2x|1−2
3
2
−1 1
8
=(
− + 2) − ( − 2 − 4)
3
2
3
−9
1
=
+8−
3
2
1
= −3 + 8 −
2
10 1
−
=
2
2
9
=
2
=
Example 4 Find the area of the region between the graphs of f (x) = 3x3 −
x2 − 10x and g(x) = −x3 + 2x
Solution.
5
f(x)
f
x
g
What do we do to find this area? Well, what does it look like? It looks
like the area comes in two pieces, so maybe we should find the area of each
individual piece (that is, use two separate integrals). Again, we need to find
the limits of integration as well as the integrand (what should go inside of
our integral).
Let’s find where the two functions intersect:
3x3 − x2 − 10x = −x2 + 2x
=⇒ 3x3 − 12x = 0
=⇒ 3x(x2 − 4) = 0
=⇒ 3x(x + 2)(x − 2) = 0
=⇒ x = −2, 0, and 2
6
f
f(x)
R1
(0, 0)
(2, 0)x
g
R2
(−2, −8)
Area = Area(R1 ) + Area(R2 )
Now, we need to set up our integrals. Note that while f (x) is the top function
for R1 , g(x) is the bottom function for R2 . Keep this in mind when setting
up your integrals.
Z 0
Z 2
3
2
2
Area =
3x − x − 10x − −x + 2x dx+
−x2 + 2x − 3x3 − x2 − 10x dx
−2
0
Z
0
3
3x − 12x dx +
=
−2
Z
2
12x − 3x3 dx
0
3x
3x4 2
2 0
2
=
− 6x |−2 + 6x −
|
4
4 0
= [0 − (12 − 24)] + [(24 − 12) − 0]
4
= 24
7
Example 5 Find the area of the region bounded by the curves y = x − 1 and
x = 3 − y2.
Solution.
y
y =x−1
x = 3 − y2
x
Trying to integrate this poses some problems. For example, try drawing your
vertical rectangle slices near x = 2.5. What do you notice?
Okay, so vertical slices pose a problem-what about horizontal slices? The
horizontal slices seem to work without a problem. In fact, if you were to turn
your paper sideways, this region might look similar to one we have already
looked at.
A few questions. If we use horizontal rectangles...which variable is the
width of each rectangle in terms of (x or y)? Since the width is in terms of
y, we should be integrating in terms of y as well. Now, which function is the
‘top’ function here? Turning your paper sideways (making sure the positive
x-axis is pointing up) makes the answer obvous...it is x = 3 − y 2 . Now, lastly,
since we are integrating with respect to y, then both of our functions should
be written in terms of y:
y =x−1
⇐⇒ x = y + 1
Now we just need to find our limits of integration and we will be on our way:
y + 1 = 3 − y2
8
=⇒ y 2 + y − 2 = 0
=⇒ (y − 1)(y + 2) = 0
=⇒ y = −2 and 1
And so we integrate over [−2, 1] (as an interval of y-values):
Z 1
(3 − y 2 ) − (y + 1) dy
Area =
−2
Z
1
=
2 − y 2 − ydy
−2
y3 y2 1
− |−2
3
2
1 1
−8
= (2 − − ) − (−4 −
− 2)
3 2
3
9
=
2
= 2y −
9