Download The Fundamental Theorem of Calculus [1]

The Fundamental Theorem of Calculus [1]
Guangliang Zhao
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Review
(i) Definite Integral: Limit of Riemann Sum. Three steps: Divide-Sum-Limit (DSL). §5.2, p372.
∫x
2
2
2
Ex: 0 tdt = x2 − 02 = x2 .
(ii) Derivative: f ′ (a) is the instantaneous rate of change of y = f (x) with respect to x when
x = a. (TANGENT). §2.7, p148.
Ex: ( x2 )′ = x.
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(iii) Antiderivative F of f is defined by: F ′ = f .
Q: Definite integral arose from the area problem, and Derivative arose from the tangent problem. What’s the deep relationship between definite integral and derivative?
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2.1
The Fundamental Theorem of Calculus (FTC)
Part 1
Suppose f is a continuous function on [a, b], x varies between a and b. FTC deals with the function
∫ x
f (t)dt,
g(x) =
a
which can be interpreted as the area under the graph of f from a to x, and the blue part in Figure
1. Here in the graph f happens to be a positive function.
To compute g ′ (x), let’s use the definition of a derivative. For h > 0, g(x + h) − g(x) is obtained
by subtracting areas, so it is the area under the graph of f from x to x + h (the red area in Figure
1). For small h, the area is approximately equal to the area of the rectangle with height f (x) and
width h:
g(x + h) − g(x) ≈ hf (x)
so
g(x + h) − g(x)
≈ f (x).
h
Therefore we expect that
g ′ (x) = lim
h→0
g(x + h) − g(x)
= f (x).
h
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Figure 1:
Theorem 2.1 (The Fundamental Theorem of Calculus, Part 1.) If f is continuous on [a, b],
then the function g defined by
∫ x
g(x) =
f (t)dt a ≤ x ≤ b
a
is continuous on [a, b] and differentiable on (a, b), and g ′ (x) = f (x).
Proof. If x and x + h are in (a, b), then
∫
∫ x
f (t)dt −
f (t)dt
) ∫ x
(a∫ x
∫ ax+h
f (t)dt
f (t)dt −
=
f (t)dt +
a
x
a
∫ x+h
f (t)dt
=
g(x + h) − g(x) =
x+h
(by Property 5 of integrals)
x
so for h ̸= 0,
g(x + h) − g(x)
1
=
h
h
∫
x+h
f (t)dt.
x
Now let’s assume that h > 0. Since f is continuous on [x, x+h], by Extreme Value Theorem, there
are numbers u and v in [x, x + h] such that f (u) = m and f (v) = M , where m and M are the
absolute minimun and maximum values of f on [x, x+h]. By Property 8 of integrals, we have
∫
x+h
mh ≤
f (t)dt ≤ M h
x
that is
∫
f (u)h ≤
x+h
f (t)dt ≤ f (v)h
x
Since h > 0, we can divide this inequality by h:
1
f (u) ≤
h
∫
x+h
x
2
f (t)dt ≤ f (v)
replace the middle part by g, we get
f (u) ≤
g(x + h) − g(x)
≤ f (v)
h
(2.1)
Letting h → 0. Then u → x and v → x, since u and v lie between x and x + h. Therefore
lim f (u) = lim f (u) = f (x)
u→x
h→0
and
lim f (v) = lim f (v) = f (x)
h→0
v→x
because f is continuous at x. By Squeeze Theorem and (2.1) we conclude that
g ′ (x) = lim
h→0
g(x + h) − g(x)
= f (x).
h
(2.2)
If x = a or b, then (2.2) can be interpreted as a one-sided limit. Then Theorem 2.8.4 shows
that g is continuous on [a, b].
Remark 2.2 This theorem first tells us that definite integral of f is one of it’s infinitely many
antiderivatives. That is because g is a definite integral by definition, and the conclusion g ′ = f
indicates that g is an antiderivative of f .
Remark 2.3 Using Leibniz notion for derivatives, we can write FTC1 as
∫ x
d
f (t)dt = f (x)
dx a
(2.3)
when f is continuous. Roughly speaking, 2.3 says that if we first integrate f and then differentiate
the result, we get back to the original function f . This shows that an antiderivative can be reversed by
a differentiation, and it also guarantees the existence, continuity, differentiability of antiderivatives
∫x
d
) of the area ( a ),
for continuous functions. In other words, the instantaneous rate of change ( dx
which located under the curve f from a to x as shown in Figure 1, evaluated at x, is exactly f
evaluated at point x.
∫x√
Example 2.4 Find the derivative of the function g(x) = 0 1 + t2 dt.
Solution. Since f (t) =
Example 2.5 Find
d
dx
√
1 + t2 is continuous, by FTC part 1,
√
g ′ (x) = 1 + x2 .
∫ x4
1
sec tdt.
Solution. Let u = x4 . Then
∫ u
∫ x4
d
d
sec tdt
dx 1 sec tdt = dx
1
[∫
]
u
du
d
sec tdt
(by the Chain Rule)
=
du 1
dx
du
(by FTC1)
= sec u
dx
4
= sec(x ) · 4x3
References
[1] James SteWart, Calculus, Early Transcendentals, 7th ed., Brooks/Cole.
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