Problem 444 - Calculus Volume 1

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

Step-by-Step Solution

Answer

Calculate the Riemann sums numerically and compare with the integral of \(\ln(x^2+1)\) over the interval \([0, e]\).

Step 1: Define the Interval and Subdivision

We are given the interval \([0, e]\) with four subintervals. Therefore, the width of each subinterval \(\Delta x\) is \(\frac{b-a}{n} = \frac{e-0}{4} = \frac{e}{4}.\)

Step 2: Calculate Left Riemann Sum (L4)

For the left Riemann sum \(L_4\), we evaluate the function at the left endpoint of each subinterval. The left endpoints are \(x_0 = 0\), \(x_1 = \frac{e}{4}\), \(x_2 = \frac{2e}{4}\), and \(x_3 = \frac{3e}{4}\). \[L_4 = \Delta x \left[ f(x_0) + f(x_1) + f(x_2) + f(x_3) \right]\]Evaluate each: \[f(x_0) = \ln((0)^2 + 1) = \ln(1)=0\]\\[f\left(\frac{e}{4}\right)=\ln\left(\left(\frac{e}{4}\right)^2+1\right)\]\[f\left(\frac{2e}{4}\right)=\ln\left(\left(\frac{2e}{4}\right)^2+1\right)\]\[f\left(\frac{3e}{4}\right)=\ln\left(\left(\frac{3e}{4}\right)^2+1\right)\]Thus, \[L_4=\frac{e}{4}\left( f(x_0)+f(x_1)+f(x_2)+f(x_3) \right)\] Compute the sum numerically for the exact value.

Step 3: Calculate Right Riemann Sum (R4)

For the right Riemann sum \(R_4\), we evaluate the function at the right endpoint of each subinterval. The right endpoints are \(x_1 = \frac{e}{4}\), \(x_2 = \frac{2e}{4}\), \(x_3 = \frac{3e}{4}\), and \(x_4 = e\). \[R_4 = \Delta x \left[ f(x_1) + f(x_2) + f(x_3) + f(x_4) \right]\]Evaluate each: \[f(x_1) = \ln\left(\left(\frac{e}{4}\right)^2 + 1\right)\]\[f(x_2) = \ln\left(\left(\frac{2e}{4}\right)^2 + 1\right)\]\[f(x_3) = \ln\left(\left(\frac{3e}{4}\right)^2 + 1\right)\]\[f(x_4) = \ln(e^2 + 1)\]Thus, \[R_4=\frac{e}{4}\left( f(x_1)+f(x_2)+f(x_3)+f(x_4) \right)\] Compute the sum numerically for the exact value.

Step 4: Compare with Exact Integral

The exact integration of \(y = \ln(x^2 + 1)\) over \([0, e]\) can be computed using a calculator or integral table.Formally, evaluate \[\int_0^e \ln(x^2+1)\, dx\]Once you find the integral value, compare it with values of \(L_4\) and \(R_4\) calculated above.

Key Concepts

Left Riemann Sum

When trying to approximate the area under a curve using the Left Riemann Sum, you evaluate the function at the left endpoints of each subinterval. Imagine dividing a stretch of the graph into smaller sections. For each section, draw a rectangle where the height is determined by the value of the function at the beginning of that section. This approach may underestimate or overestimate the true area,which is why it's just an approximation.

In our exercise, we are looking at the function \( y = \ln(x^2 + 1) \) over the interval \([0, e]\). We divide this interval into four equal parts or subintervals.

• The subintervals are from \( x_0 = 0 \) to \( x_1 = \frac{e}{4} \), \( x_1 \) to \( x_2 = \frac{2e}{4} \), and so on.
• The step width \( \Delta x \) is \( \frac{e}{4} \).

For each subinterval, compute the function at the left endpoint and sum all these values multiplied by \( \Delta x \) to get the Left Riemann Sum, expressed as \( L_4 \). This approach gives us an approximation of the area under the curve over the given interval.

Right Riemann Sum

Similarly, the Right Riemann Sum offers another way to approximate the area under a curve, but this time by evaluating the function at the right endpoints of each subinterval. Visualize again dividing your interval into smaller stretches and drawing rectangles. The height of each rectangle comes from the function value at the end of each small section.

In our exercise scenario, we continue with the function \( y = \ln(x^2 + 1) \) over \([0, e]\), using the same subdivision for consistency in comparison. However, now:

• We evaluate at points such as \( x_1 = \frac{e}{4} \), \( x_2 = \frac{2e}{4} \), up to \( x_4 = e \).

The formula for the Right Riemann Sum \( R_4 \) is calculated by summing up the multiplied values of \( \Delta x \) and function values at these right endpoints.
Evaluating \( R_4 \) provides another estimation of the area beneath the curve, which will usually differ slightly from \( L_4 \), allowing you to see variations in approximation based on endpoint choices.

Integration

Integration fundamentally acts as the bridge to calculating the exact area under a curve, unlike the approximations from Riemann Sums. Through integration, you sum up infinitesimally small areas to find the entire area covered by the function over an interval.

In our example with \( y = \ln(x^2 + 1) \), integration mathematically gives us: \[\int_0^e \ln(x^2 + 1)\, dx\]This expression seeks the precise area under the curve from \( x = 0 \) to \( x = e \). You can use an integral table or a calculator to find this value accurately.After computing, contrast this with the Riemann Sums \( L_4 \) and \( R_4 \) from our previous steps to understand how closely these sums approximate integration results.
Integration serves as the gold standard when determining areas under curves for continuous functions. It's an essential mathematical tool in calculus that you can rely on for accurate assessments without relying solely on approximated sums.