Brian E. Veitch
2
Applications of Integration
2.1
Area between curves
In this section we are going to find the area between curves. Recall that the integral can
represent the area between f (x) and the x-axis. And any area below the x-axis is considered
negative.
Z
b
f (x) dx
a
So the next question is, how do I find the area of the shaded region below?
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2.1 Area between curves
Brian E. Veitch
Notice that f (x) ≥ g(x) on the interval [a,b]. The formula for the area between f (x) and
g(x) is
Z
b
f (x) − g(x) dx
a
This should make sense. You find the area below f (x) and subtract the area under g(x),
which leaves just the area between the two functions.
Example 2.1. Find the area of the region bounded by f (x) = x2 + 1, g(x) = x, x = 0, and
x = 1.
Solution:
1. The first step would be to sketch and find the shaded region. We need to know which
function is on top and which is on the bottom.
2. Now that we see f (x) = x2 + 1 is on top, we can set up our integral.
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2.1 Area between curves
Brian E. Veitch
Z
1
(x2 + 1) − x dx
0
3. Now evaluate it,
Z
1
Z
2
1
x2 − x + 1 dx
0
1
1 3 1 2
=
x − x + x
3
2
0
5
=
−0
6
5
=
6
(x + 1) − x dx =
0
Example 2.2. Find the area of the region bounded by f (x) = 2x2 + 10 and g(x) = 4x + 16
We begin by sketching the graph. Notice that we don’t have any bounds and we don’t
know which function is on top.
1. The sketch is below
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2.1 Area between curves
Brian E. Veitch
2. You might be able to see that the two functions intersect at x = −1 and x = 3. I
want to verify this by doing some algebra. If we want to know where the two functions
intersect, then set them equal to each other.
2x2 + 10 = 4x + 16
2x2 − 4x − 6 = 0
2(x2 − 2x − 3) = 0
2(x − 3)(x + 1) = 0
Solving for x and we find x = −1 and x = 3.
3. We can see g(x) is above f (x). If you didn’t have a graph to verify that, then just plug
in any point between x = −1 and x = 3. Whichever has the larger y-value must be
the top function.
4. Set up the integral and evaluate
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2.1 Area between curves
Brian E. Veitch
Z
3
A =
top bottom
f unction − f unction dx
−1
3
Z
=
(4x + 16) − (2x2 + 10) dx
−1
3
Z
−2x2 + 4x + 6 dx
−1
3
2 3
2
= − x + 2x + 6x
3
−1
−10
= 18 −
3
64
=
3
=
What happens when f (x) ≥ g(x) is not always true in the interval [a,b]? Let’s take a
look at the following example.
Example 2.3. Find the area of the region bounded by the curves f (x) = sin(x), g(x) =
cos(x), x = 0, and x = π/2.
1. Let’s begin by sketching the graph
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2.1 Area between curves
Brian E. Veitch
Do you see that f (x) = sin(x) and g(x) = cos(x) switch being the top function somewhere in our interval [0, π/2]. So how do we find out where they switch?
We need to find the intersection. So set sin(x) = cos(x) and solve for x. Solving this
gets us x = π/4.
2. Now, we break the integral up so that each integral has the same top function in its
interval.
π/4
Z
Z
π/2
sin(x) − cos(x) dx
cos(x) − sin(x) dx +
A=
π/4
0
3. Integrate and Evaluate
Z
π/4
Z
π/2
sin(x) − cos(x) dx
cos(x) − sin(x) dx +
π/4
h
i
π/2
+
−
cos(x)
−
sin(x)|
= sin(x) + cos(x)|π/4
0
π/4
" √
# "
√ !
√
√ !#
2
2
2
2
=
+
− (1) + (−1) − −
−
2
2
2
2
√
= 2 2−2
A =
0
2.1.1
Steps to Finding the Area Between Functions of x
1. Sketch the graph (if anything, just to see what it looks like)
2. Identify the top and bottom function.
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2.1 Area between curves
Brian E. Veitch
If they switch, find out where they switch, by setting the functions equal to each other,
and then separate the integral.
3. Find the limits of integration. They may tell you this, or you have to find them. Looking at the graph would help very much.
4. Set up the integral and evaluate
Z b top bottom A=
f unction − f unction dx
a
2.1.2
Finding the Area Between Functions of y
We’re moving on to a variation of finding the area between curves. All the functions we’ve
dealt with so far have been in terms of x. Now we have to find the area between curves when
they are in terms of y.
Formula for the area between curves
Z
A=
d
lef t f unction − f unction dy
right
c
or
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2.1 Area between curves
Brian E. Veitch
Z
d
f (y) − g(y) dy
A=
c
Since your functions are in terms of y, we integrate along the y-axis, from c to d.
Example 2.4. Find the area enclosed by the line y = x − 1 and y 2 = 2x + 6.
Solution: So why can’t we just solve this like the others? Well for one thing, y 2 = 2x + 6
is not in the form y = f (x). Let’s take a look at the graph.
Notice that there are parts of the graph where y 2 = 2x + 6 is both the top and bottom
function. This is why it’s not a good idea to try integrating with respect to x. But actually,
I’m getting a bit ahead of myself. Graphing calculators don’t graph in terms of y. So how
did I graph this?
2.1.3
Graphing Functions in terms of y
1. If you need to graph in terms of y, set up your equations in the form x = f (y).
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2.1 Area between curves
Brian E. Veitch
IMPORTANT: If you also have functions that are in terms of x and can be easily
graphed, wait until after you finish graphing the functions in terms of y.
1
Rewrite y 2 = 2x + 6 as x = (y 2 − 6)
2
2. Swap x and y.
1
y = (x2 − 6)
2
It’s now a function of x and can graph it. You can graph this by hand or use a graphing
calculator. Either way, you should get this
3. You need to rotate your graph along the line y = x. It’s hard to demonstrate that on
paper (much better in person), so instead here’s another way.
(a) Flip your graph over the y-axis. You’re graph should now look like this,
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2.1 Area between curves
Brian E. Veitch
You probably can’t tell a difference since the function is symmetric about the
y-axis. But you can see the x-axis is now pointing to the left.
(b) Rotate your graph 90◦ (90 degrees) to the right.
(c) You’re done graphing y 2 = 2x + 6. Now just re-label the axes.
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2.1 Area between curves
Brian E. Veitch
4. Since y = x − 1 isn’t a function of y, I didn’t use the method described above. If
it’s a function of x, wait until you’re done graphing all functions of y, then just graph
y = x − 1 normally.
5. We shade the area enclosed by the two functions.
6. Since we are integrating with respect to y, we need to find our endpoints. At what
y-value does the shaded region begin and at what y-value does the shaded region end?
y = −2 and y = 4
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2.1 Area between curves
Brian E. Veitch
How do you find this algebraically? You’re looking for where the functions intersect.
So set them equal to each other. BUT! you need to make sure both functions are in
terms of the same variable. Rewrite them as
1
x = y + 1 and x = (y 2 − 6)
2
and solve
1
y + 1 = (y 2 − 6)
2
I’ll leave that to you. You should still get y = −2 and y = 4.
7. Now set up the integral and evaluate. At this point all functions have to be in terms
of the same variable (in our case, y).
A =
=
=
=
=
4
right lef t f unction − f unction dy
−2
Z 4
1 2
(y + 1) −
y − 3 dy
2
−2
Z 4
1
− y 2 + y + 4 dy
2
−2
4
1 3 1 2
− y + y + 4y 6
2
−2
18
Z
Wasn’t that fun?!? Let’s do another!
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2.1 Area between curves
Brian E. Veitch
Example 2.5. Find the area bounded by x = −y 2 + 10 and x = (y − 2)2 .
Solution: Since both of these functions are in terms of y (i.e, written as x = f (y)), we
use the method from the last example to graph them.
1. Graph x = −y 2 + 10 and x = (y − 2)2 .
(a) Write the functions in terms of y. Check!
(b) Swap x and y
y = −x2 + 10 and y = (x − 2)2
(c) Graph these like you normally graph functions of x.
(d) Flip horizontally (over the y axis) and rotate right 90◦ .
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2.1 Area between curves
Brian E. Veitch
(e) Relabel and shade the bounded region
2. Identify the left and right functions.
Right Function: x = −y 2 + 10
Left Function: x = (y − 2)2
3. Identify the bounds. You can set the equations together and solve for y or you can
just look at the graph. At what y-value does the shaded region begin and end?
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2.1 Area between curves
Brian E. Veitch
y = −1 to y = 3
4. Set up the integral and evaluate
Z
3
A =
(−y 2 + 10) − (y − 2)2 dy
−1
3
Z
−2y 2 + 4y + 6 dy
−1
3
2 3
2
= − y + 2y + 6y 3
−1
64
=
3
=
Let’s try one more.
Example 2.6. Find the area of the region bounded by y = x3 − x and y = −0.5 + x − x3 .
1. Let’s begin by graphing these two functions. Both are in terms of x.
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2.1 Area between curves
Brian E. Veitch
2. Notice that we are going to have to break this up into two integrals since the top
function changes. There are three places the functions intersect. Let’s find those.
x3 − x = −.5 + x − x3
2x3 − 2x + .5 = 0
Using a calculator, the solutions are x = −1.107, x = 0.27, and x = 0.84.
3. Set up the integrals and evaluate
Z
0.27
3
Z
0.84
(x − x) − (−.5 + x − x ) dx +
(−.5 + x − x3 ) − (x3 − x) dx
−1.107
0.27
Z 0.27
Z 0.84
=
2x3 − 2x + .5 dx +
−2x3 + 2x − .5 dx
0.27
"−1.107
0.27 # "
0.84 #
1 4
1
+ − x4 + x2 − .5x
x − x2 + .5x
=
2
2
−1.107
0.27
A =
3
= 1.19427
I’ll leave the final evaluation to you, mainly because I don’t want to type it.
Ok, ok.. one more problem. This one is from the textbook. I like it because the integration requires u-substitution and a bit of other stuff. If you can’t follow along, check out the
solution in the textbook.
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2.1 Area between curves
Brian E. Veitch
Example 2.7. Find the bounded region by y = √
x
and y = x4 − x.
+1
x2
1. As always, let’s begin by graphing.
2. Next, we need to find the bounds. They’re not easily done algebraically, so we’ll use
the Internet or a graphing calculator.
The bounds are x = 0 to x = 1.18.
3. Identify the top and bottom function.
x
Top: y = √
x2 + 1
Bottom: y = x4 − x
4. Set up the integral and evaluate
Z
A =
0
1.18
x
√
x2 + 1
27
− (x4 − x) dx
2.1 Area between curves
Brian E. Veitch
So we have a bit of a problem. We can’t simplify the integrand. Plus the first part
x
√
requires u-substitution. The second part is just a basic integral. Let’s break
2
x +1
them up into two separate integrals. You may recall this is one of the properties of
integrals.
Z
A =
0
1.18
x
√
dx −
x2 + 1
Z
1.18
x4 − x dx
0
5. Let’s integrate these separately (less messy that way).
Z
0
1.18
√
x
dx:
x2 + 1
(a) Let u = x2 + 1
(b) du = 2x dx →
du
= x dx
2
(c) Change the bounds
If x = 0 → u = 1
if x = 1.18 → u = 2.39
(d) Substitute
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2.1 Area between curves
Brian E. Veitch
A =
=
=
=
Z
1 2.39 1
√ du
2 1
u
Z 2.39
1
u−1/2 du
2 1
2.39
u1/2 1
√
2.39 − 1
= 0.5460
Z
1.18
x4 − x dx;
0
Z
1.18
x4 − x dx
0
1.18
1 5 1 2 =
x − x
5
2 0
= −0.2386 − 0
A =
= −0.2386
6. Final Answer:
A = 0.5460 − (−0.2386) = 0.7846
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