Question

Riemann sums for constant functions Let \(f(x)=c,\) where \(c>0,\) be a constant function on \([a, b] .\) Prove that any Riemann sum for any value of \(n\) gives the exact area of the region between the graph of \(f\) and the \(x\) -axis on \([a, b]\).

Short answer

Question: Prove that any Riemann sum for the constant function \(f(x) = c\), where \(c > 0\), gives the exact area of the region between the graph of \(f\) and the \(x\)-axis on the interval \([a, b]\). Answer: We have shown that for the constant function \(f(x) = c\), any Riemann sum \(R_n = c\sum_{i = 1}^{n} \Delta x_i\) evaluates to \(R_n = c(b-a)\). The exact area of the region between the graph of \(f\) and the \(x\)-axis on the interval \([a, b]\) is also given by \(A = c(b-a)\). Since both the Riemann sum and the exact area have the same value, it proves that any Riemann sum for the constant function \(f(x) = c\) gives the exact area of the region between the graph of \(f\) and the \(x\)-axis on the interval \([a, b]\).

Step-by-step solution

Writing the Riemann sum

Before we prove the given statement, we should first write the general form of the Riemann sum for the function \(f(x) = c\) on the interval \([a, b]\). The Riemann sum is given by: $$ R_n = \sum_{i = 1}^{n} f(x_i) \Delta x_i $$ Where \(n\) is the number of divisions of the interval \([a, b]\), and \(\Delta x_i = x_{i+1} - x_i\) is the width of the subinterval.

Evaluating the function at the subintervals

Since \(f(x) = c\) is a constant function, it means that the value of the function will be the same for all values of \(x\) within the interval \([a, b]\). Therefore, we can rewrite the Riemann sum as follows: $$ R_n = \sum_{i = 1}^{n} c \Delta x_i $$

Combining the constant term with the sum

Now we will combine the constant term with the sum. Since the value of \(c\) does not change for each subinterval, it can be factored out of the sum: $$ R_n = c\sum_{i = 1}^{n} \Delta x_i $$

Summing the widths of the subintervals

The next step is to evaluate the sum of the widths of the subintervals. Since the interval \([a, b]\) is divided into \(n\) equal subintervals, the width of each subinterval is \(\Delta x = \frac{b-a}{n}\). Thus, the sum of all the widths is: $$ \sum_{i = 1}^{n} \Delta x_i = n\Delta x = n\left(\frac{b-a}{n}\right) = (b-a) $$

Calculating the Riemann sum

Now, we can substitute the sum of the widths of the subintervals back into the Riemann sum: $$ R_n = c\sum_{i = 1}^{n} \Delta x_i = c(b-a) $$

Comparing the Riemann sum with the exact area

We can now compare the value of the Riemann sum with the exact area of the region between the graph of \(f\) and the \(x\)-axis on the interval \([a, b]\). The exact area is given by the constant value of the function \(c\) multiplied by the length of the interval \((b-a)\). This gives us: $$ A = c(b-a) $$ Since both the Riemann sum and the exact area have the same value, it proves that any Riemann sum for the constant function \(f(x) = c\) gives the exact area of the region between the graph of \(f\) and the \(x\)-axis on the interval \([a, b]\).

Key concepts

Constant Functions

A fundamental concept in calculus is that of \textbf{constant functions}. These are functions where the output value is the same regardless of the input. That is, for a constant function given by \(f(x) = c\), the value of \(c\) remains unchanged for all \(x\) within the domain of \(f\).

In the context of Riemann sums and integration, constant functions play a special role because they greatly simplify the process of finding the area under the curve. Since the function does not vary, the height of each rectangle in the Riemann sum is consistent. This constant height can be factored out of the summation, reducing the problem to a simple multiplication of this constant value by the width of the interval over which the area is being calculated. Therefore, when working with constant functions, the Riemann sum conveniently coincides with the exact area under the curve.

Exact Area Calculation

The \textbf{exact area calculation} is a precise determination of the area bounded by the graph of a function and the x-axis over a given interval. In general, calculating the exact area under a curve can be quite intricate, especially for functions that exhibit complex behavior. However, for constant functions, this calculation simplifies considerably.

As outlined in the exercise solution, we can quickly find that the area under a constant function \(f(x) = c\) over an interval \([a, b]\) is \(A = c(b-a)\). This formula employs the most fundamental principle of area calculation: area equals height (the value of the constant function) times width (the length of the interval). It's a direct application of the geometrical concept of the area of a rectangle, extended to the calculus setting. Therefore, understanding the exact area calculation for constant functions provides a solid foundation for grasping more complex integration techniques.

Integration Fundamentals

Delving into \textbf{integration fundamentals} can initially seem daunting, but they are crucial for understanding the basic premise behind finding areas under curves. Integration is the process of summing up infinitesimally small pieces to determine a whole. A Riemann sum is a stepping stone towards understanding integration, as it approximates the area using a finite number of rectangles.

However, when dealing with constant functions, as seen in the exercise, the Riemann sum quickly yields the exact area, thereby serving as a gentle introduction to the concept of definite integration. Typically, to find the exact area under a curve for non-constant functions, you would take the limit of the Riemann sums as the number of rectangles approaches infinity, leading to the definite integral. In the simplest form, the definite integral of a constant function over an interval \([a, b]\) is simply \(c \times (b-a)\), mirroring our findings from the exercise. As you progress in calculus, you will encounter many more sophisticated and complex functions, but the basic understanding of integration starts here, with these simple but powerful fundamentals.