Problem 1
Question
Let \(c \in[a, b]\) and \(f:[a, b] \rightarrow \mathbb{R}\) be given by
$$
f(x):=\left\\{\begin{array}{ll}
0 & \text { if } a \leq x \leq c \\
1 & \text { if } c
Step-by-Step Solution
Verified Answer
The given function \(f\) is integrable on the interval \([a, b]\) because the upper and lower sums, calculated as \(U(P, f) = m(b-c)\) and \(L(P, f) = 0\), converge to 0 as \(c\) approaches \(b\). Furthermore, this integrability can be proven using Proposition 6.10, which states that a function is integrable on an interval if for every \(\epsilon > 0\), there exists a partition such that the difference between the upper and lower sums is less than \(\epsilon\). In our case, we can choose a partition so that the upper and lower sums satisfy this condition, implying that the function \(f\) is indeed integrable on \([a, b]\).
1Step 1: Calculate the upper sum
To calculate the upper sum, find the supremum of the function \(f\) on each subinterval in the partition of \([a, b]\). Since the function is defined as 0 for \(a \leq x \leq c\) and 1 for \(c < x \leq b\), the upper sum is given by:
$$
U(P, f) = 1 * m(b-c) = m(b-c),
$$
where \(m\) is the length of the interval defined by the supremum value of 1, i.e., the interval \((c, b]\).
2Step 2: Calculate the lower sum
To calculate the lower sum, find the infimum of the function \(f\) on each subinterval in the partition of \([a, b]\). The lower sum is given by:
$$
L(P, f) = 0 * n(c-a) = 0,
$$
where \(n\) is the length of the interval defined by the infimum value of 0, i.e., the interval \([a, c]\).
3Step 3: Check if the upper and lower sums converge
Given that the supremum and infimum Definition of the Riemann Integral states that a function is integrable on a given interval if and only if the upper and lower sums can be made arbitrarily close, that is,
$$
\lim_{||P|| \to 0} U(P,f) = \lim_{||P|| \to 0} L(P,f) = I \in \mathbb{R}.
$$
In our case, the only way to make the upper and lower sums arbitrarily close is if the length \(m(b-c)\) approaches 0. This is only possible if \(c\) approaches \(b\). So we have:
$$
\lim_{c \to b} m(b-c) = 0.
$$
Since the lower sum is already 0, the function is integrable on \([a, b]\) as the upper and lower sums converge to 0.
Now let us prove this using Proposition 6.10.
4Step 4: Proposition 6.10
Proposition 6.10 states that a function \(f\) is integrable on \([a, b]\) if and only if for every \(\epsilon > 0\), there exists a partition \(P\) of \([a, b]\) such that:
$$
U(P, f) - L(P, f) < \epsilon.
$$
In our case, since \(L(P, f) = 0\) and \(U(P, f) = m(b-c)\), we need to find a partition such that:
$$
m(b-c) < \epsilon.
$$
Let us choose a partition in which \(c\) is very close to \(b\) or equivalently \(b-c\) is very small (say, \(b - c < \epsilon/2\)). In that case, we have:
$$
U(P, f) - L(P, f) = m(b-c) < m(\epsilon/2) = \epsilon/2 < \epsilon.
$$
Thus, for every \(\epsilon > 0\), we can find a partition \(P\) such that \(U(P, f) - L(P, f) < \epsilon\), which implies that \(f\) is integrable on \([a, b]\) by Proposition 6.10.
Key Concepts
Integrable FunctionsUpper Sum and Lower SumPartition of an Interval
Integrable Functions
An integrable function is one that can be summed up or measured over a certain interval. Specifically, when we talk about Riemann integrals, a function is considered integrable if there is a total area bounded by the function's graph and the x-axis across an interval \[a, b\]. This means that the function shouldn't have too wild of fluctuations on the interval. The classic example of an integrable function involves those that remain continuous throughout, or only have a finite number of discontinuities. Riemann integrability also requires the upper and lower sums to "meet" as we refine the partition of the interval, thereby enclosing a specific value which is the Riemann Integral. In simple words, for any given error range, there should be a way to define the partition such that the sum of the areas above and below the curve align closely around an accurate value of the integral.
Upper Sum and Lower Sum
The concept of "upper sum" and "lower sum" is central to understanding the Riemann integral. Imagine dividing the range \[a, b\] into smaller subintervals; the function's behavior varies over each of these intervals.
- **Upper Sum:** Determined by taking the highest value (supremum) within each subinterval and summing up the products of these maximum values with the subinterval lengths. This upper sum gives an approximation of the area above the curve.
- **Lower Sum:** Obtained by using the lowest value (infimum) within each subinterval and multiplying by the subinterval lengths. This represents the area beneath the curve.
Partition of an Interval
A partition of an interval is a way to break down that interval into smaller pieces or subintervals. Partitions are essential for calculating upper and lower sums in the process of finding a Riemann integral. Let's understand this more clearly:
- **Definition:** A partition of the interval \[a, b\] is a finite sequence of numbers starting from \a\ and ending at \b\ such that each number is greater than or equal to the previous one. In mathematical terms, this is represented as \{a = x_0, x_1, x_2, \.\.\., x_n = b\}\.
- **Role in Riemann Integral:** It divides the interval into sections which help calculate the upper and lower sums. A finer partition — one with more points — gives a more precise approximation of the integral.
Other exercises in this chapter
Problem 2
Let \(c \in(a, b)\) and \(f:[a, b] \rightarrow \mathbb{R}\) be given by $$ f(x):=\left\\{\begin{array}{ll} (x-c) /(a-c) & \text { if } a \leq x \leq c, \\ (x-c)
View solution Problem 3
Let \(f:[0,1] \rightarrow \mathbb{R}\) be given by $$ f(x):=\left\\{\begin{array}{ll} 1+x & \text { if } x \text { is rational } \\ 0 & \text { if } x \text { i
View solution Problem 4
Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable. Show that the Riemann integral of \(f\) is the unique real number \(r\) satisfying the following conditio
View solution