Download The Fundamental Theorems of Calculus, Section 01, Spring 2009 and more Lecture notes Calculus in PDF only on Docsity!
The Fundamental Theorems of Calculus
Math 142 , Section 01 , Spring 2009 We now know enough about definite integrals to give precise formulations of the Fundamental Theorems of Calculus. We will also look at some basic examples of these theorems in this set of notes. The next set of notes will consider some applications of these theorems.
The First Fundamental Theorem of Calculus. If f (x) is continuous on [a, b] and F (x) is any anti- derivative of f (x) on [a, b], then ∫ (^) b
a
f (x) dx = F (b) − F (a).
In words: In order to compute a definite integral f (x), it suffices to find an antiderivative of f (x), then compute the difference of this antiderivative.
The Second Fundamental Theorem of Calculus. If f (x) is continuous on an interval and a is any number in this interval, then the function
A(x) =
∫ (^) x
a
f (t) dt
is an antiderivative of f (x). In other words,
d dx
[∫ (^) x
a
f (t) dt
]
In words: In order to constuct an antiderivative of a function f (x), it is enough to be able to compute the definite integral of f (x). Both of these theorems establish a relationship between antiderivatives (i.e., indefinite integrals) and net areas (i.e., definite integrals.) In effect, the First Fundamental Theorem says that definite integra- tion reverses differentiation, whereas the Second Fundamental Theorem says that differentiation reverses definite integration. Thus, the Fundamental Theorems say that definite integration is actually a sort of antidifferentiation. This explains why we use integration notation and terminology (indefinite integrals) for antiderivatives.
From a mathematical point of view, the Second Fundamental Theorem is the “true” fundamental the- orem of calculus. The First Fundamental Theorem is just a logical consequence of Second Fundamental Theorem. However, the First Fundamental Theorem is what most non-mathematicians think of as the fundamental theorem. This is because the First Fundamental Theorem is the one that appears in appli- cations in lots of different subjects. Thus, the First Fundamental Theorem is of practical interest whereas the Second Fundamental Theorem is primarily of theoretical interest, although it does have some practical applications.
- Using the First Fundamental Theorem of Calculus
Example 1 : Find ∫ 3 π 2 π 2
sin θ dθ.
Solution: We already know this integral is 0 by the previous set of notes. But lets compute this using the First Fundamental Theorem. First, we need to find an antiderivative of sin θ. But ∫ sin θ dθ = − cos θ + C, 1
so we’ll use F (θ) = − cos θ in the fundamental theorem. (The fundamental theorem only requires that we find some antiderivative, not the most general antiderivative, so we don’t need to use the +C.) So, ∫ 3 π 2 π 2
sin θ dθ = F (
3 π 2
) − F (
π 2
) = − cos (
3 π 2
) − (− cos (
π 2
consistent with the answer we obtained in the last set of notes.
We make two comments on the above example. First, what if we used a different antiderivative? Could that have changed our answer? Fortunately the answer is no. The most general antiderivative is F (θ) = − cos θ + C, so if we used this we would have had ∫ 32 π
π 2
sin θ dθ = F (
3 π 2
) − F (
π 2
) = (− cos (
3 π 2
) + C) − (− cos (
π 2
) + C) = −0 + C + 0 − C = 0.
The C’s will always cancel off like this, so we don’t need the +C when computing definite integrals.
Second, we will find it convenient to write F (x) |ba = F (b) − F (a).
Thus, we could write ∫ 32 π
π 2
sin θ dθ = F (θ) |
32 π π 2
Example 2 : Find (^) ∫ 4
1
x
dx.
Solution: First we find an antiderivative: ∫ 1 √ x
dx =
x−^1 /^2 dx = 2x^1 /^2 + C.
We’ll use 2
x as our antiderivative. Thus ∫ (^4)
1
x
dx = 2
x |^41 = 2
Example 3 : Find (^) ∫ 2
− 1
(x^3 + 5x) dx.
Solution: Often times, we will just compute the antiderivatives in our heads. Thus ∫ (^2)
− 1
(x^3 + 5x) dx = (
x^4 +
x^2 ) |^2 − 1 = (
· (−1)^4 +
· (−1)^2 ) = 11. 25
Example 4 : Things can go wrong if the integrand has a vertical asymptote. (Please note the Fundamental Theorem does require f (x) to be defined and continuous on the whole interval of integration.) For example, the following calculation is wrong. ∫ (^1)
− 1
x−^2 dx = −x−^1 | (^) −^11 = −(1)−^1 − (−(−1)−^1 ) = − 2.
We can tell immediately that something is wrong since
− 1 x
− (^2) dx is the net area of a figure that is always
above the x axis, and therefore the integral better be postive. Furthermore, using approximating sums, it looks as though the integral is infinite. Thus, while the fundamental theorem gives us a very powerful
(You could just as well have first computed the definite integral using the first Fundamental Theorem, then computed the derivative. Doing this should again give x^2 + 3x. But doing so misses the whole point of the Second Fundamental Theorem.)
Example 6 : Find d dx
[∫ (^) x
0
sin (θ^2 ) dθ
]
Solution: A few notes back we mentioned that sin (θ^2 ) has no antiderivative in terms of “simple” functions, so it is impossible to compute this by first computing the definite integral, then computing the derivative. But this doesn’t bother us since we know the Second Fundamental Theorem:
d dx
[∫ (^) x
0
sin (θ^2 ) dθ
]
= sin (x^2 ).
Matters are a bit more complicated if the limits of integration are not simply x. The most general situation allows both the upper and lower limits of integration to be functions of x.
The Second Fundamental Theorem of Calculus with arbitrary limits of integration. If f (x) is continuous and u(x) and `(x) are differentiable functions, then
d dx
[∫
u(x)
`(x)
f (t) dt
]
= u′(x)f (u(x)) − ′(x)f ((x)).
Example 7 : According to this form of the Fundamental Theorem,
d dx
[∫ (^) x
−x
e−t^ dt
]
= (x)′^ · ex^ − (−x)′e−x^ = ex^ + e−x.
This has a nice geometric interpretation. As x increases, the area represented by
∫ (^) x −x e
t (^) dt expands both
to the left and to the right. On the right we add a little rectangle of height ex, on the left we add a little rectangle of height e−x.
Example 8 :
d dx
[∫
x^2
1
cos t dt
]
= (x^2 )′^ · cos (x^2 ) − (1^2 )′^ · cos (1^2 ) = 2x · cos (x^2 ),
Example 9 : d dx
[∫ (^) ex
3 x
t^3 dt
]
= (ex)′^ · (ex)^3 − (3x)′^ · (3x)^3 = e^4 x^ − 81 x^3.
Example 10 : Let G(x) =
∫ (^) ln x 0 √^1 2 π e
−t^2 / (^2) dt. Find G′(x).
Solution:
G′(x) =
d dx
[∫ (^) ln x
0
2 π
e−t
(^2) / 2 dt
]
= (ln x)′^
2 π
e−(ln^ x)
(^2) / 2
x
2 π
· e−(ln^ x)
(^2) / 2 .
The lower limit of integration is constant, so the −(0)′^ · √^12 π e−^0
(^2) / 2 term is just 0, so we did not include
that term.
We’ll see lots of applications of the First Fundamental Theorem of Calculus in the next set of notes. Direct applications of the Second Fundamental Theorem are a bit harder to come by, but applications do exist for it. In particular, the last example is a typical sort of calculation you might do in a probability class (specifically, this example relates the probability densities of the normal distribution and the log-normal distribution.)
- Theory
We look at some of the more theoretical aspects of the Fundamental Theorems in this section. We begin by outlining a proof of the fundamental theorems. Actually, this argument should be very familiar to you by now. We’ve gone through this argument in several specific cases already.
Even though I’m taking the time to go through the proofs of these theorems, you will not be responsible for either of these proofs. Being able to do calculutions like the examples at the beginning of the set of notes on Areas and Antiderivatives is far more important than memorizing this proof. After you’ve done enough examples like those you should see that the proof of the fundamental theorem is the exact same argument just written out in abstract generality.
We begin with the Second Fundamental Theorem: We suppose f (x) is continuous on an interval, a is in the interval, and we need to show that the function
A(x) =
∫ (^) x
a
f (t) dt
is an antiderivative of f (t). In other words, we need to show that A′(x) = f (x), or
d dx
[∫ (^) x
a
f (t) dt
]
= f (x).
We don’t have an explicit formula for the integral, so our only hope for computing the derivative is to write out the definition of derivative:
d dx
[∫ (^) x
a
f (t) dt
]
= lim h→ 0
h
[∫ (^) x+h
a
f (t) dt −
∫ (^) x
a
f (t) dt
]
The numerator is the difference between two net areas, the first from a to x + h, the second from a to x. Thus, the difference is the net area from x to x + h, i.e., ∫ (^) x+h
a
f (t) dt −
∫ (^) x
a
f (t) dt =
∫ (^) x+h
x
f (t) dt.
If h is very small then this last integral is the net area of a figure that is essentially a rectangle with width h and height f (x), so (^) ∫ x+h
x
f (t) dt ≈ f (x) · h.
Piecing together everything,
d dx
[∫ (^) x
a
f (t) dt
]
= lim h→ 0
h
[∫ (^) x+h
a
f (t) dt −
∫ (^) x
a
f (t) dt
]
= lim h→ 0
h
[∫ (^) x+h
x
f (t) dt
]
≈ lim h→ 0
f (x) · h h
= f (x).
This only shows d dx
[∫ (^) x
a
f (t) dt
]
≈ f (x),
so technically this is not a proof of the second Fundamental Theorem. To get a rigorous proof we would need to quantify just how good the approximation
∫ (^) x+h x f^ (t)^ dt^ ≈^ f^ (x)^ ·^ h^ is.^ Requiring^ f^ (x) to be continuous is enough to guarantee that this approximation is good enough. You should look back at the first two examples in the the notes on Areas and Antiderivatives. The proof of the Second Fundamental Theorem of Calculus is exactly the same argument we used in those specific examples.
