Download The Fundamental Theorem of Calculus and Integration

The Fundamental Theorem of Calculus and
Integration
September 30, 2014
4.3
Areas and Definite Integrals
In the previous lecture, we learned about the concept of Riemann sums, and
how they can be used to approximate areas of shapes. For a function f (x),
a Riemann sum is a sum of the form
n
X
f (xi )∆x.
i=1
Here, n is the number of rectangles used in the approximation, xi is the xvalue at the left-hand edge of each rectangle, and ∆x is the width of each
rectangle. We concluded the lecture by saying that, in practice, we never
compute this limit directly (though in some cases it is possible, and not
too hard). The reason why is because we have a theorem, known as the
Fundamental Theorem of Calculus, which tells us exactly what that limit
will be. Before we can state it, we need the following definition.
Definition 1. If f (x) is continuous on [a, b] and F (x) is the antiderivative
of f (x), then we define the definite integral, denoted by
Z b
f (x)dx
a
as
Z
b
f (x) dx = F (b) − F (a).
a
The numbers a and b are known as the limits of integration.
1
b
Remark 1. We often use the notation F (x) to represent F (b) − F (a).
a
With this definition, we can state the Fundamental Theorem of Calculus,
which has many implications in areas well outside of mathematics.
Theorem 1. (Fundamental Theorem of Calculus) If f is continuous on [a, b]
and has antiderivative F , then
Z b
n
X
f (x) dx = F (b) − F (a).
lim
f (xi )∆x =
∆x→0
a
i=1
The thing about this theorem that makes it so important is that it relates
areas and derivative/antiderivatives with a nice simple equation.
Example 1. Compute the following integrals.
Z 2
1.
x2 dx
1
Solution: By the FTC, we have
Z 2
1 3 2
2
x dx = x 3 1
1
8 1
= −
3 3
7
= .
3
Z 1
2.
ex dx
0
Solution: By the FTC, we have
Z 1
1
x
x
e dx = e 0
0
1
= e − e0
= e − 1.
One thing you should note about the FTC is that it tells you what the
limit of the Riemann sum is as ∆x → 0. It doesn’t actually say anything
about areas of shapes. With this in mind, consider the following example.
2
Example 2. Evaluate the following integral.
Z 0
x3 dx
−1
Solution: By the FTC, we have
Z 0
1 4 0
x dx = x 4 −1
−1
1
=0−
4
1
=− .
4
3
The point of this problem is that while we like to think of areas as positive
quantities, definite integrals can give you negative answers. The reason for
this comes from the definition we used in the Riemann sums:
n
X
f (xi )∆x.
i=1
In the expression above, there is no reason why f (xi ) must always be positive.
If a function is always negative, then the Riemann sum will be negative.
Thus, we can think of the FTC as saying two different things:
• If f (x) ≥ 0, then the integral gives you the area.
• If f (x) ≤ 0, then the integral gives you the negative of the area.
Suppose now that we have a function f (x) ≥ 0 on some interval [a, b]. For
any x between a and b, we can define a function, which we will call the area
function and denote it by A(x), as the area under the curve between a and
x. Based on the above considerations, we must have
Z x
A(x) =
f (s) ds.
a
Based on the FTC, A(x) has the following property.
Theorem 2. If f (x) ≥ 0 on [a, b], and a < x < b, then the area function
A(x) satisfies A0 (x) = f (x).
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Example 3. Differentiate the following functions.
Z x
sin(s) ds
1.
0
Solution: By the previous theorem, A0 (x) = sin(x).
Z x
2.
s sin(s) ln(s) ds
1
Solution: By the previous theorem, A0 (x) = x sin(x) ln(x).
Remark 2. Note that in these examples, we did not have to evaluate the
integrals at all.
4.4
Properties of Definite Integrals
In the previous section, we discussed the problem of computing areas, and
what to do in the two cases where a function is either positive or negative.
This leaves a third case: a function which is sometimes positive, sometimes
negative. To be able to handle such cases, we need the following theorem.
Theorem 3. If a < b < c, then
Z c
Z b
Z c
f (x) dx =
f (x) dx +
f (x) dx.
a
a
b
What this theorem is saying is that an integral can be chopped up into
pieces, and the sum of all the pieces is equal to the original integral.
Example 4. If f (x) is given by
(
9
f (x) =
x2
if x < 3
if x ≥ 3,
compute
Z
5
f (x) dx.
−4
Solution: According to the Fundamental Theorem of Calculus, we should
find the antiderivative of f (x). The problem is that since the function has
different definitions, we will have different expressions for the antiderivative.
4
Thus, we will need to use Theorem 3 to break up the integral into pieces
where each piece has only one definition. For the above example, we can
break it up in the following way:
Z 3
Z 5
Z 5
f (x) dx =
f (x) dx +
f (x) dx
−4
−4
3
Z 5
Z 3
x2 dx
9 dx +
=
3
−4
3
1 3 5
= 9x + x 3 3
−4
98
= 63 + .
3
Now, we would like to go back to the area problem for functions which
are both positive and negative. Consider the function f (x) = cos(x) on the
interval [0, π]. The graph looks like this:
As we can see, the function is both positive and negative within the interval
[0, π]. However, we can also see that it is positive on [0, π/2] and negative on
[π/2, π]. Based on the discussion after Example 2 and on Theorem 3, we can
break up the region into areas where the function is positive and negative.
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Thus
Z
π
2
π
Z
− cos(x) dx
cos(x) dx +
A=
π
2
0
Z
=
π
2
π
Z
cos(x) dx −
cos(x) dx
π
2
0
π
π
2
= sin(x) − sin(x) π
0
2
= 1 − 0 − (0 − 1) = 2.
You should compare this with
Z π
π
cos(x) dx = sin(x) = 0 − 0 = 0.
0
0
Areas Between Curves
Up until now, we discussed the problem of finding areas. However, the
areas we discussed had a very specific form: they we formed by vertical lines,
the x-axis, and the upper boundary was the graph of a function f (x). What
if we now want the area for a shape where, say, both the top and bottom
boundaries are given by functions? For example, suppose we wanted the area
of the following shape:
We can think of this area as the area of this graph:
6
with the following area removed:
Mathematically, this is simply
Z b
Z b
A=
f (x) dx −
g(x) dx
a
a
Recalling the properties of integrals, we can simplify this to
Z b
A=
f (x) − g(x) dx.
a
We summarize this in the following theorem.
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Theorem 4. Let f and g be continuous functions with g(x) ≤ f (x) on [a, b].
Then the area between the two curves from x = a to x = b is
Z b
f (x) − g(x) dx.
a
Example 5. Find the area between the curves f (x) = x2 and g(x) = 9
between x = 0 and x = 2.
Solution: Note that in this example, f (x) ≤ g(x). Thus, by Theorem 4, we
have
Z 2
9 − x2 dx
A=
0
1 2
= 9x − x3 3 0
8
= 18 − − (0 − 0)
3
46
= .
3
8