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Final

MA 141 Lecture 26: Area Between Two CurvesExam


Department
Mathematics
Course Code
MA 141
Professor
Kevin Flores
Study Guide
Final

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MA 141 Chapter 5
Section 5.1: Area Between Two Curves
In Chapter 4, we developed definite integrals to calculate the area "under" a curve - between
the graph of a function and the x-axis. We may want to find the area of a more complicated
shape, such as the area between two curves.
Figure 1
From the picture, we can see that the area under the curve y=f(x)is given by
Zb
a
f(x)dx =A1+A2
while the area under the curve y=g(x)is given by
Zb
a
g(x)dx =A2.
If we wanted to find just the area between the two curves, A1, we could do the following:
(A1+A2)A2=Zb
a
f(x)dx Zb
a
g(x)dx
=Zb
a
(f(x)g(x))dx
This yields the formula we will use to find the area "between two curves".
1

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Ex: Find the area of the region bounded by f(x) = x2+ 1,g(x) = x,x=1and x= 2.
For any area problem like this, we should graph the functions.
From the graph, we can see that f(x) = x2+ 1 is greater than g(x) = xon the interval
[1,2].
Then the area we want is given by
Z2
1
(f(x)g(x))dx =Z2
1
(x2+ 1 x)dx
=x3
3+xx2
2
2
1
=23
3+ 2 22
2(1)3
31(1)2
2
=8
3+ 2 2 + 1
3+ 1 + 1
2
=9
2
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Ex: Find the area of the region bounded by f(x) = 6 x2and g(x) = 2x+ 3.
Notice we are not given an interval here. Let’s graph to see if we can tell why.
The region "bounded" by the two curves is the region between their two intersection points.
So, we will find the intersection points and use those for our bounds.
6x2=2x+ 3
0 = x22x3 = (x3)(x+ 1)
Then the bounds of our interval are x=1and x= 3.
We also need to know which function has the bigger values on this interval. We can see from
the graph that f(x) = 6 x2is bigger than g(x) = 2x+ 3 on the interval [1,3].
Z3
1
(f(x)g(x))dx =Z3
1
(6 x2(2x+ 3))dx
=Z3
1
(x2+ 2x+ 3)dx
=x3
3+x2+ 3x
3
1
=33
3+ 32+ 3(3) (1)3
3+ (1)2+ 3(1)
=9 + 9 + 9 1
31 + 3
=32
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