Question

(a) If \(\mathrm{n}\) is a positive integer, prove that \(\int_{0}^{n}[t] d t=n(n-1) / 2\) (b) If \(\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}}[\mathrm{t}] \mathrm{dt}\) for \(\mathrm{x} \geq 0\), draw the graph of fover the interval \([0,4]\).

Short answer

Based on the given step by step solution, answer the following question: Question: Prove that for any positive integer n, \(\int_{0}^{n}[t] dt = n(n-1)/2\), and sketch the graph of the function \(f(x) = \int_{0}^{x}[t]dt\) over the interval \([0, 4]\). Answer: To prove that \(\int_{0}^{n}[t] dt = n(n-1)/2\), we break the integral into smaller sub-intervals \([0, 1), [1, 2), ... , [n-1, n)\) and integrate in each sub-interval, where the floor function \([t]\) takes a constant value of k. The sum of these integrals for all sub-intervals is \(\sum_{k=0}^{n-1} k = n(n-1)/2\). To sketch the graph of \(f(x) = \int_{0}^{x}[t]dt\) over the interval \([0, 4]\), we find the function values at integer points using the derived formula and connect them with straight lines for non-integer intervals. The graph resembles a stair with steps at integer points and sloping lines in between the steps.

Step-by-step solution

(Step 1: Breaking down the integral)

(Since the floor function, \([t]\), is discontinuous in the interval \(\left[0,n\right]\), we will break it into \(n\) smaller sub-intervals, \([0, 1), [1, 2), ... , [n-1, n)\). In each interval, the floor function takes a constant value, so we can replace it with this constant and easily integrate.)

(Step 2: Calculate integral in each sub-interval)

(For each sub-interval \([k, k+1)\) for \(k = 0, 1, 2, ..., n-1\), the floor function \([t]\) takes the value of \(k\). So we can integrate the function in these sub-intervals as follows: \(\int_{k}^{k+1}[t] dt = \int_{k}^{k+1}k dt = k((k+1)-k) = k\). Now, we can add up these integrals to get the integral over the whole interval \([0, n]\).)

(Step 3: Add up the integrals in all sub-intervals)

(Add up the integrals for all sub-intervals: \(\int_{0}^{n}[t] dt = \sum_{k=0}^{n-1} k = \frac{n(n-1)}{2}\). So, we have proven that \(\int_{0}^{n}[t] dt = n(n-1)/2\). )

(Step 4: Finding f(x) for different x values)

(To graph the function \(f(x) = \int_{0}^{x}[t]dt\) over the interval \([0, 4]\), we need to find the function values at different points in this interval. We can simply use the formula we derived in Step 3 and plug in the integer values of \(x\), and use the linearity of the integral for the non-integer values.)

(Step 5: Plot the graph of f(x))

(Finally, we can plot the graph of \(f(x)\) over the interval \([0, 4]\). The function is constant over each integer interval \([k, k+1)\), and it changes when \(x\) crosses an integer value. The function takes the values \(f(0) = 0, f(1) = 0, f(2) = 1, f(3) = 3\). For \(x\) in the non-integer intervals, the function is linear. So we can connect the function values at the integer points with straight lines. The graph of \(f(x)\) over the interval \([0, 4]\) will look like a stair with steps at the integer points and sloping lines in between the steps.)

Key concepts

Floor Function Integration

The floor function, denoted by \[t\], represents the greatest integer less than or equal to \(t\). Integrating the floor function can seem challenging due to its piecewise nature; it is discontinuous at all integer points. However, a strategic approach can simplify the task considerably.

To perform the integration of the floor function over an interval, one effective strategy is to divide the range of integration into sub-intervals where the function is continuous and essentially constant. For instance, the integral \(\int_{0}^{n}[t] dt\), where \(n\) is a positive integer, can be split up into \(n\) separate integrals over the intervals \([0, 1), [1, 2), ..., [n-1, n)\).

In each of these intervals, the floor function equals the integer at the lower bound, making it easy to integrate. Adding up these integrals then gives us the total area under the floor function curve from 0 to \(n\). This sum can be calculated using arithmetic series formulas, yielding the result \(\frac{n(n-1)}{2}\), proving the integral in question.

Grasping this technique not only aids in solving the immediate problem but also builds a deeper understanding of how to integrate functions that exhibit discontinuities due to their piecewise-defined nature.

Definite Integrals

Definite integrals are a cornerstone of calculus, providing a way to calculate the accumulation of quantities and the area under curves. They are expressed as the integral of a function \(f(x)\) with respect to \(x\) over a closed interval \([a, b]\). The basic notation is \(\int_{a}^{b} f(x) dx\), and the result is a number that quantifies the net area between the function and the \(x\)-axis.

In the context of the floor function, the definite integral accumulates the constant values taken by the floor function in each sub-interval. When the intervals are integer bounded, the exercise simplifies dramatically due to the nature of \([t]\), which is equal to an integer within each sub-interval. By assessing the function piece by piece and summing the areas, one can compute the total value of the definite integral.

Understanding definite integrals is also essential when dealing with piecewise functions. Piecewise functions are split into distinct sections, each of which may require a separate integration process. The computation becomes an exercise in segmenting the function, integrating each segment individually, and then combining the results to obtain the final integral value. Thus, mastering definite integrals is invaluable for solving a broad range of calculus problems.

Graphing Piecewise Functions

Piecewise functions are defined by different expressions for different portions of their domain. A graph of a piecewise function can show abrupt changes in direction, which correspond to the points at which the function's formula changes.

In the case of \(f(x) = \int_{0}^{x}[t]dt\), the function is built upon the integral of a piecewise constant function - the floor function. The nature of \(f(x)\) is such that it becomes stair-stepped, where each step corresponds to a sub-interval where the floor function maintains a constant value.

To graph \(f(x)\) over a specific interval, like \([0, 4]\), one would evaluate the function at various points within the interval, particularly at points of discontinuity (in this case, the integers). Between these points, \(f(x)\) will have linear segments whose slopes depend on the values taken by the floor function. The final graph would resemble a staircase, with flat sections corresponding to integer intervals and sloped lines connecting these flats.

Graphing piecewise functions therefore requires attention to the intervals where the function's rules change. By carefully plotting the points and connecting them with the appropriate lines, one can accurately represent the function's behavior visually.