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. \(y=\ln \left(x^{2}+1\right)\) over \([0, e]\)

Short answer

Calculate \(L_4\) and \(R_4\) using left and right endpoints, then compare with exact or numerically computed integral over \([0, e]\).

Step-by-step solution

Define the interval and subintervals

The interval is \([0, e]\). Since we are using 4 subintervals, the subinterval width, \(\Delta x\), is given by \(\Delta x = \frac{e-0}{4} = \frac{e}{4}\).

Evaluate Left Riemann Sum (\(L_4\))

The left Riemann sum is computed by evaluating the function at the left endpoints of each subinterval. These subintervals have endpoints at \(0, \frac{e}{4}, \frac{2e}{4}, \frac{3e}{4}, e\). Thus, \(L_4\) is:\[L_4 = \Delta x \cdot [f(0) + f(\frac{e}{4}) + f(\frac{2e}{4}) + f(\frac{3e}{4})]\]Substitute the function values:\[L_4 = \frac{e}{4} [\ln(0^2 + 1) + \ln((\frac{e}{4})^2 + 1) + \ln((\frac{2e}{4})^2 + 1) + \ln((\frac{3e}{4})^2 + 1)]\]

Evaluate Right Riemann Sum (\(R_4\))

The right Riemann sum is computed using the right endpoints of each subinterval. These endpoints are \(\frac{e}{4}, \frac{2e}{4}, \frac{3e}{4}, e\). Thus, \(R_4\) is:\[R_4 = \Delta x \cdot [f(\frac{e}{4}) + f(\frac{2e}{4}) + f(\frac{3e}{4}) + f(e)]\]Substitute the function values:\[R_4 = \frac{e}{4} [\ln((\frac{e}{4})^2 + 1) + \ln((\frac{2e}{4})^2 + 1) + \ln((\frac{3e}{4})^2 + 1) + \ln(e^2 + 1)]\]

Exact Evaluation

The exact integral of \(f(x) = \ln(x^2 + 1)\) over \([0, e]\) can be calculated (if needed) using a calculator or another method as the integral is not elementary. The calculated integral is:\[\int_0^e \ln(x^2 + 1)\, dx\]

Comparison and Calculation

Evaluate \(L_4\) and \(R_4\) using a calculator to get numeric approximations. Compare these with the exact integral value computed or approximated by numeric integration if an exact solution isn't directly feasible. This will illustrate how close the Riemann sums come to the actual area under the curve.

Key concepts

Left Riemann Sum

Understanding the Left Riemann sum involves approximating the area under a curve using rectangles. To calculate it, you find the height of each rectangle by evaluating the function at the left endpoint of each subinterval. This essentially means you take the value of the function at the beginning of each portion. Hence, for a function like \( y = \ln(x^2 + 1) \) in the interval \([0, e]\), you divide this range into four subintervals.

• The endpoints for these subintervals, marking the left side where the height is sampled, are \(0, \frac{e}{4}, \frac{2e}{4}, \frac{3e}{4}\).
• The width of each subinterval is \( \Delta x = \frac{e}{4} \).

You plug these values into the function to compute the sum: \[L_4 = \frac{e}{4} \left[ \ln(0^2 + 1) + \ln\left(\left(\frac{e}{4}\right)^2 + 1\right) + \ln\left(\left(\frac{2e}{4}\right)^2 + 1\right) + \ln\left(\left(\frac{3e}{4}\right)^2 + 1\right) \right]\] This aggregate represents the approximate area under the curve using the left corners. While simple, it typically tends to underestimate the actual integral if the function is increasing. You then evaluate this to get its numerical approximation.

Right Riemann Sum

Right Riemann sums, on the other hand, are similar to their left counterparts but instead use the right endpoint of each subinterval to determine the height of each rectangle. This change in evaluation point often fits the function better, especially when the function's incline increases over the interval.

• For the function \( y = \ln(x^2 + 1) \) on \([0, e]\), the subinterval widths and endpoints shift slightly to \( \frac{e}{4}, \frac{2e}{4}, \frac{3e}{4}, \) and \( e \).

You'll evaluate the function at these right endpoints: \[R_4 = \frac{e}{4} \left[ \ln\left(\left(\frac{e}{4}\right)^2 + 1\right) + \ln\left(\left(\frac{2e}{4}\right)^2 + 1\right) + \ln\left(\left(\frac{3e}{4}\right)^2 + 1\right) + \ln(e^2 + 1) \right]\] Right Riemann sums often yield a higher estimate if the curve is rising, compared to left Riemann sums. Using them helps students understand how sampling points affect integral approximations. To get an accurate number, calculators or computers are commonly employed to crunch the math.

Integral Approximation

In calculus, we often approximate an integral, especially when dealing with complex or non-elementary functions, like \( \ln(x^2 + 1) \). Integral approximation becomes necessary because some integrals cannot be solved exactly using standard antiderivative methods.

• Riemann sums (both left and right) are foundational techniques for these approximations, literally summing up small rectangles under a curve.
• These sums provide a numeric method to evaluate integrals by breaking the area into manageable sections.

The goal is to compare the Riemann sums with the exact integral value or its close numerical approximation if an exact solution is hard to find. The steps typically involve: - Calculating both \(L_4\) and \(R_4\), as demonstrated above.- Checking these against either the exact integral value or using numerical integration tools like calculators.Through this, students can see the connection between discrete sums and continuous areas, sharpening both their understanding and computational skills.

Logarithmic Functions

Logarithmic functions, such as \( y = \ln(x^2 + 1) \), play a crucial role in calculus and analysis. They are the inverse of exponential functions and grow at a slower rate as compared to polynomial or exponential functions, which gives them unique properties.

• In the context of Riemann sums and integrals, understanding how logarithmic functions behave is important. They often smooth out steep areas, affecting the convergence and behavior of these sums over an interval.
• The function \(\ln(x^2 + 1)\) specifically adds complexity due to the square inside the logarithm, meaning it represents a more complicated curvature.

Understanding these properties is vital for performing approximations accurately, as logarithmic functions can model natural phenomena and data trends, where continuous growth and decay are analyzed. As students work through exercises involving them, it emphasizes the need to grasp fundamental concepts such as growth patterns and the effect of mathematical transformations. Mastering these functions extends their capability to handle real-world analytical challenges.