Question

Evaluate the Riemann sums \(L_{4}\) and \(R_{4}\) for the following functions over the specified interval. Compare your answer with the exact answer, when possible, or use a calculator to determine the answer. \(y=\sqrt{x}+\frac{1}{x}\) over \([1,4]\)

Short answer

L4 ≈ 5.576, R4 ≈ 6.901, and the exact integral ≈ 6.430.

Step-by-step solution

Determine the Interval and Number of Subintervals

The given interval is \([1,4]\)\. We are instructed to use four subintervals, so each subinterval will have a width of \( \Delta x = \frac{4 - 1}{4} = \frac{3}{4} \). Thus, the subintervals are \([1, 1.75], [1.75, 2.5], [2.5, 3.25], [3.25, 4]\).

Calculate the Left Riemann Sum

To find the left Riemann sum \(L_4\), we use the left endpoints of each subinterval: \(x_0 = 1\), \(x_1 = 1.75\), \(x_2 = 2.5\), and \(x_3 = 3.25\). We evaluate the function at these points and multiply each by \(\Delta x\):\\[\L_4 = \left( \sqrt{1} + \frac{1}{1} \right) \cdot \frac{3}{4} + \left( \sqrt{1.75} + \frac{1}{1.75} \right) \cdot \frac{3}{4} \+ \left( \sqrt{2.5} + \frac{1}{2.5} \right) \cdot \frac{3}{4} + \left( \sqrt{3.25} + \frac{1}{3.25} \right) \cdot \frac{3}{4}\]\Evaluating these terms gives \(L_4 \approx 5.576\).

Calculate the Right Riemann Sum

For the right Riemann sum \(R_4\), use the right endpoints: \(x_1 = 1.75\), \(x_2 = 2.5\), \(x_3 = 3.25\), and \(x_4 = 4\). Compute the function, multiplying each result by \(\Delta x\):\\[\R_4 = \left( \sqrt{1.75} + \frac{1}{1.75} \right) \cdot \frac{3}{4} + \left( \sqrt{2.5} + \frac{1}{2.5} \right) \cdot \frac{3}{4} \+ \left( \sqrt{3.25} + \frac{1}{3.25} \right) \cdot \frac{3}{4} + \left( \sqrt{4} + \frac{1}{4} \right) \cdot \frac{3}{4}\]\Evaluating these gives \(R_4 \approx 6.901\).

Compute the Exact Integral

The integral of \(y=\sqrt{x}+\frac{1}{x}\) from \(1\) to \(4\) is calculated by evaluating \[ \int_{1}^{4} \left( \sqrt{x} + \frac{1}{x} \right) \, dx = \left[ \frac{2}{3}x^{3/2} \right]_1^4 + [\ln|x|]_1^4\]. \ Computing the result, we have: \[ \left( \frac{2}{3}(4^{3/2}) + \ln(4) \right) - \left( \frac{2}{3}(1^{3/2}) + \ln(1) \right) = \frac{16}{3} + \ln(4) - \frac{2}{3}\]. \" \This equals approximately \(6.430\).

Compare Riemann Sums to Exact Value

Now compare the calculated Riemann sums and the exact integral value: \- \(L_4 \approx 5.576\) \- \(R_4 \approx 6.901\) \- Exact value \(\approx 6.430\) \ \(R_4\) is closer to the exact value than \(L_4\).

Key concepts

Integral Calculus

Integral calculus is a fundamental branch of calculus focusing on accumulating quantities and finding areas under curves. It essentially deals with integrals, which are the inverse operations of derivatives. Integrals can be thought of as the accumulation of small quantities to determine a larger total. For example, if you want to find the area under a curve from one point to another on a graph, you'd use integral calculus.
One of the key applications of integral calculus is to compute areas, volumes, and other related quantities. It also allows us to model physical situations involving continuous change, like in physics and engineering. In summary, integral calculus helps us understand and calculate the cumulative effect of changes within a particular interval.

Numerical Integration

Numerical integration is a technique to approximate the value of integrals, especially when they cannot be evaluated analytically. This approach comes in handy when dealing with complex functions lacking elementary antiderivatives. Riemann sums are among the simplest numerical methods used for integration.
There are various approaches for numerical integration, including:

• Trapezoidal Rule
• Simpson's Rule
• Midpoint Rule
• Riemann Sum methods

Each method has its advantages and trade-offs, and choosing the appropriate method depends on the function's behavior and the desired level of accuracy.

Definite Integral

A definite integral is an integral represented with upper and lower limits, defining where the function starts and ends. It is written as \( \int_a^b f(x) \, dx \) where \( f(x) \) is the function to be integrated from \( a \) to \( b \). The definite integral represents the net area between the graph of \( f(x) \) and the x-axis from \( a \) to \( b \).
In definite integrals, the integration leads to a number describing the accumulated quantity, such as the total area under a curve. It's important in finding quantities like mass, displacement, and even computing workloads in various fields.
A definite integral provides a precise calculation, but in cases where it's difficult to find by hand, numerical methods such as Riemann sums are employed as viable alternatives.

Approximation Methods

Approximation methods are valuable tools in calculus for estimating mathematical values rather than computing them exactly. They are particularly useful when dealing with functions that are difficult to integrate analytically. With Riemann sums being a common approximation method, they estimate the area under a curve by dividing the interval into small sections (subintervals).
Riemann sums come in different flavors:

• Left Riemann Sum: Uses the left endpoint of each subinterval.
• Right Riemann Sum: Takes the right endpoint.
• Midpoint Riemann Sum: Uses the midpoint of each subinterval.

These methods work by approximating the area under a curve as a series of rectangles, with each rectangle's height determined by the function's value at a specific point. While these approximations become more accurate with more subintervals, they are most useful for estimating when exact calculations are complex or impossible.