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CALCULUS 1 PROBLEMS & SOLUTIONS

6.3
The Fundamental Theorem Of Calculus


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Review


1. Notes

Integrals Are Antiderivatives

The graph of the function y = f(x) = (1/3) x, which is a straight line, is shown in Fig. 1. For any x ³ 0
let A be the area of the region under the graph of f, over the x-axis, and between the y-axis and the
vertical line at x; it's shaded in Fig. 1. So A is a function of x, thus denoted A(x). We have A(x) =
(1/2) x f(x) = (1/2) x (1/3) x = (1/6) x². Hence, A'(x) = (1/6) (2x) = (1/3) x = f(x). Therefore, the
derivative of the area under f over the x-axis is f itself.

Fig. 1 · f(x) = (1/3) x. · Area A(x) = (1/6) x^2. · A'(x) = (1/3) x = f(x).

Now let f be a continuous non-negative-valued function on (a, b) and a a given point in (a, b), as
shown in Fig. 2. For any x in (a, b) consider the area A of the region under the graph of f over

Fig. 2

[a, x]. In Fig. 2, it's the shaded region. Clearly A depends on x, and only on x. Thus, A is a function of
x, hence called the area function and denoted A(x). In term of integral we have:



Let's not be confused with the appearance of the letter t. Here, it's a dummy variable (see Section
6.2.1 Remarks 2). We would really be confused if we use the same letter x for both the variable of integration and the upper limit of integration.


(But that's not the case for A(x), because an area is ³ 0, and thus, for example, cannot decrease as x
increases away from a both where f(x) ³ 0 and where f(x) < 0, hence its derivative cannot be < 0.
Areas and integrals are not always the same thing. See Section 6.2.1 Review Part 3.)


Computations Of Definite Integrals Using Antiderivatives

Fig. 3

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2. The Fundamental Theorem Of Calculus

Theorem 1 – The Fundamental Theorem Of Calculus
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Proof
i. As f is continuous on any closed subinterval of I, it's integrable there (see Section 6.2.2 Theorem 1).



EOP

Fig. 4 · U1( f, x, x + h) = area of rectangle ABEF. · L1( f, x, x + h) = area of rectangle CDEF. Note the "1" in U1 and L1. We "partition" [x, x + h] into 1 subinterval only: just [x, x+ h] itself.

Fig. 5 · U1( f, x, x + h) = – (area of rectangle ABCD). · L1( f, x, x + h) = – (area of rectangle ABEF).

Remarks 1

i. The hypothesis of the fundamental theorem doesn't require that f be non-negative-valued on its
domain. Clearly the proof doesn't need that condition. The theorem applies to any continuous
function.

ii. The fundamental theorem asserts that any integral function of f is an antiderivative of f, or, in other
words, that f is the derivative of any of its integral functions. The theorem establishes the
relationship between the derivative and the integral, hence the adjective "fundamental" in its name.

iii. It follows directly from the fundamental theorem that every continuous function is antidifferentiable
(ie, has an antiderivative). Recall that not every continuous function is differentiable.


Evaluation Notation



Note that we don't need to add a constant C to the antiderivative. Any one of the following facts is a
complete reason:
· Any antiderivative can be used in the evaluation.
· The constant C cancels out in the subtraction.

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Problems & Solutions


1. Evaluate the following definite integrals.



Solution



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Solution





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3. Find the following derivatives.



Solution



Note



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4. Consider the function y = 1/x², whose graph is shown in the figure below.





Solution

The function y = 1/x² is discontinuous at x = 0. So it's discontinuous on any open interval containing
x = –1 and x = 1. Thus, the fundamental theorem doesn't apply to it.

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5. Prove this limit:



Solution

Let B_n be the sum in brackets. Then:



Let:


So we've got a regular partition {x₀ = 0, x₁, x₂, ..., x_n = 1} of order n of the interval [0, 1]. Now, f is
continuous everywhere on R, and in particular on [0, 1]. So f is integrable on [0, 1] (see Section 6.2.2
Theorem 1). Thus:



Note

We have f '(x) = – (2x)/(1 + x²)² < 0 for all x > 0. So f is decreasing on [0, 1]. Consequently, on
each subinterval [x_i–1, x_i], f(x_i) is the minimum of f on that subinterval. Thus, (1/n)(f(x₁) + f(x₂) +
f(x₃) + ... + f(x_n)) is actually a lower Riemann sum L_n ( f, 0, 1). But we don't have to worry about it,
because f is continuous (see Section 6.2.2 Theorem 1). It's enough to refer simply to the general
Riemann sum R_n ( f, 0, 1).

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