Problem 36 - Thomas Calculus

Question

For the functions in Exercises \(35-40\) find a formula for the upper sum obtained by dividing the interval \([a, b]\) into \(n\) equal subintervals. Then take a limit of these sums as \(n \rightarrow \infty\) to calculate the area under the curve over \([a, b]\). $$ f(x)=2 x \text { over the interval }[0,3] $$

Step-by-Step Solution

Answer

The area under the curve is 9.

Step 1: Divide the Interval

First, divide the interval \( [0, 3] \) into \( n \) equal subintervals. Each subinterval will have a width of \( \Delta x = \frac{3-0}{n} = \frac{3}{n} \).

Step 2: Determine the Subinterval Endpoints

The endpoints of the subintervals are \( x_0 = 0, x_1 = \frac{3}{n}, x_2 = \frac{6}{n}, \ldots, x_n = 3 \). For the \( k^{th} \) subinterval, the right endpoint is \( x_k = \frac{3k}{n} \) where \( k = 1, 2, \ldots, n \).

Step 3: Find the Maximum Function Value for Each Subinterval

Since \( f(x) = 2x \) is increasing over \( [0, 3] \), the maximum value on the \( k^{th} \) subinterval is \( f(x_k) = 2 \left( \frac{3k}{n} \right) = \frac{6k}{n} \).

Step 4: Calculate the Upper Sum

The upper sum is calculated by summing up the products of the maximum value from each subinterval and the width of the subinterval:\[ U_n = \sum_{k=1}^{n} f(x_k) \Delta x = \sum_{k=1}^{n} \frac{6k}{n} \cdot \frac{3}{n} = \sum_{k=1}^{n} \frac{18k}{n^2}. \]

Step 5: Simplify the Sum

The expression for the upper sum simplifies to:\[ U_n = \frac{18}{n^2} \sum_{k=1}^{n} k. \]The sum of the first \( n \) integers is given by \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \). Thus, the sum becomes:\[ U_n = \frac{18}{n^2} \cdot \frac{n(n+1)}{2} = \frac{18n(n+1)}{2n^2} = \frac{9(n+1)}{n}. \]

Step 6: Take the Limit as n Approaches Infinity

Compute the limit of \( U_n \) as \( n \rightarrow \infty \):\[ \lim_{n \to \infty} \frac{9(n+1)}{n} = \lim_{n \to \infty} \left(9 + \frac{9}{n}\right) = 9. \]

Step 7: Conclusion: Area Under the Curve

The area under the curve \( f(x) = 2x \) over the interval \( [0, 3] \) is \ 9 \. This is the result of the limit of the upper sums as \( n \rightarrow \infty \).

Key Concepts

Upper Sum

The concept of the Upper Sum is a fundamental step in understanding Riemann Sums and ultimately Integral Calculus. When we talk about the Upper Sum, we're essentially summing up the upper bounds of the subintervals of a function's graph to approximate the area under the curve.

Here's how it works:

• For a given interval, it’s divided into smaller subintervals.
• In an increasing function like our example, the maximum value at each subinterval is at the right endpoint.
• The function value at this point is multiplied by the width of the subinterval.

These products are summed to approximate the area under the curve. For instance, with the function \( f(x) = 2x \) over \([0, 3] \), the maximum values are multiplied by the width \( \Delta x = \frac{3}{n} \). When summed, these products give us the Upper Sum.

Limit

When we calculate the area under a curve using Riemann Sums, the number of subintervals plays a crucial role. The process involves taking these subintervals to the limit, specifically, as their number \( n \) approaches infinity.

By doing so, we refine our approximation of the area:

• The subintervals become infinitesimally thin.
• The Upper Sum moves closer to the actual area under the curve.
• The error in our approximation decreases, ultimately converging on the true area.

In our exercise, as \( n \rightarrow \infty \), the term \( \frac{9}{n} \) becomes insignificant, allowing us to conclude that the limit of the Upper Sum equals 9. Thus, it gives an accurate depiction of the area under the curve.

Area Under the Curve

The "Area Under the Curve" is a crucial concept, bridging our understanding of geometry and calculus. When we discuss this in context, we're essentially looking to find the exact space between a curve of a function and the x-axis over a specified interval.

In this exercise, the goal is to compute this area for the function \( f(x) = 2x \) over the interval \([0, 3] \). By employing the limit of the Riemann Upper Sum process, we get closer to finding this exact area. Ultimately, through the calculation, the area is revealed to be 9 square units.

This concept lays the foundational block for Integral Calculus, further connecting the area under the curve to the integral of functions.

Integral Calculus

Integral Calculus is one of the two main branches of calculus, with the other being Differential Calculus. It is fundamentally about finding the area under a curve, along with solving a wide variety of other problems.

In this discipline, integrals are used to accumulate quantities, such as area, volume, and other sum-based measurements. For the function \( f(x) = 2x \), the area we calculated using limits and upper sums becomes the integral of the function over \([0, 3] \).

Integral Calculus allows:

• The determination of accumulated change, such as total distance or total area.
• The calculation of areas and volumes of irregular shapes.
• Understanding of more complex relationships between varying quantities.

Thus, by taking the limit of the Riemann Upper Sum, we showcased a key application of Integral Calculus, where these abstract calculations come together to form a real-world measurement of area.