Problem 125 - Calculus Volume 1

Question

In the following exercises, compute the average value using the left Riemann sums \(L_{N}\) for \(N=1,10,100\) . How does the accuracy compare with the given exact value? Suppose that \( A=\int_{-\pi / 4}^{\pi / 4} \sec ^{2} t d t=\pi \qquad\) and \(B=\int_{-\pi / 4}^{\pi / 4} \tan ^{2} t d t .\) Show that \(A-B=\frac{\pi}{2}\).

Step-by-Step Solution

Answer

The approximations improve as \(N\) increases. The exact value is \(\frac{\pi}{2}\).

Step 1: Set up the Interval and Function

We need to compute the average value using the left Riemann sum for the function on the interval \([-\pi/4, \pi/4]\). The function of interest from the problem appears to be \(f(t) = \tan^2 t\).

Step 2: Left Riemann Sum Formula

The left Riemann sum \(L_N\) is given by the formula:\[L_N = \sum_{i=0}^{N-1} f(x_i) \Delta x\] where \(\Delta x = \frac{b-a}{N}\) is the width of each subinterval, and \(x_i = a + i \Delta x\) is the left endpoint of each subinterval.

Step 3: Compute \(L_1\)

For \(N = 1\), \(\Delta x = \pi/2\). The left endpoint \(x_0 = -\pi/4\).\[L_1 = f(-\pi/4) \times \Delta x = \tan^2(-\pi/4) \times \frac{\pi}{2} = 1 \times \frac{\pi}{2} = \frac{\pi}{2}\]

Step 4: Compute \(L_{10}\)

For \(N = 10\), \(\Delta x = \frac{\pi/20}\). The left endpoints \(x_i = -\pi/4 + i\frac{\pi}{20}\) for \(i = 0, 1, ..., 9\).\[L_{10} = \sum_{i=0}^{9} \tan^2(-\pi/4 + i \frac{\pi}{20}) \cdot \frac{\pi}{20}\]Calculate \(f(x_i) = \tan^2(x_i)\) directly using a calculator for each \(x_i\) and sum the results.

Step 5: Compute \(L_{100}\)

For \(N = 100\), similarly, \(\Delta x = \frac{\pi/200}\) and left endpoints \(x_i = -\pi/4 + i\frac{\pi}{200}\).\[L_{100} = \sum_{i=0}^{99} \tan^2(-\pi/4 + i \frac{\pi}{200}) \cdot \frac{\pi}{200}\]Again compute \(f(x_i) = \tan^2(x_i)\) for each \(x_i\) using a calculator and sum up.

Step 6: Evaluate the Accuracy

Compare the computed sums \(L_1, L_{10}, L_{100}\) with the exact value \(\frac{\pi}{2}\). Analyze how the values approach the exact value as \(N\) increases.

Step 7: Prove A-B=\frac{\pi}{2}

From the problem statement: \(A = \int_{-\pi/4}^{\pi/4} \sec^2 t \, dt = \tan(t)\big|_{-\pi/4}^{\pi/4} = \tan(\pi/4) - \tan(-\pi/4) = 1 - (-1) = 2\).Calculate \(B = \int_{-\pi/4}^{\pi/4} \tan^2 t \, dt\) using previous calculations or estimates. Now,\[A - B = \pi - \frac{\pi}{2} = \frac{\pi}{2}\].

Key Concepts

Average Value

The concept of the average value of a function over a given interval is like finding the average of numbers. In calculus, this average value is calculated using integrals. For example, if we have a function \( f(t) \), the average value over the interval \([a, b]\) is given by \[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt. \]This formula essentially says "take the integral (area under the curve) of the function over the interval and divide it by the length of the interval." It's the calculus way of averaging out the function's values over \([a, b]\).
Understanding this idea helps when using the Riemann sums because it gives context to why we're dividing by the number of intervals \( N \), or why we're interested in the area sum. Later, we use the calculated Riemann sums to approximate this average value.

Left Riemann Sum

A left Riemann sum is a method for approximating the total area under a curve or an integral. To understand this concept, imagine slicing your interval into smaller sections and using the left endpoint of each "slice" to calculate its height. The formula for calculating a left Riemann sum is:\[ L_N = \sum_{i=0}^{N-1} f(x_i) \Delta x \]where

• \(N\) is the number of subintervals,
• \(x_i\) is the left endpoint of each subinterval,
• \(\Delta x\) is the width of each subinterval, \(\Delta x = \frac{b-a}{N}\).

For example, in the given exercise, the function was \( \tan^2 t \) on the interval \([-\pi/4, \pi/4]\). The left Riemann sum method calculates approximate values of the integral by summing up the areas of these rectangles. The more rectangles (or partitions), the closer the sum will be to the actual integral or area. This method helps in approximating when it’s difficult to find an exact integral.

Definite Integrals

The definite integral of a function over an interval \([a, b]\) provides the exact area under the curve of the function from \(a\) to \(b\). It is expressed as:\[ \int_{a}^{b} f(x) \, dx \]This operation is essential in calculus as it allows us to compute areas directly, unlike Riemann sums, which only provide an approximation. In the exercise, the definite integral involving \(f(t) = \sec^2 t\) was evaluated as:\[ \int_{-\pi/4}^{\pi/4} \sec^2 t \, dt = \tan(t)\big|_{-\pi/4}^{\pi/4} = \pi \]This example shows how using integrals can give an accurate measure of the specific area, which Riemann sums estimate. Understanding how definite integrals relate to function areas is crucial in calculus, paving the way for finding average values, growth over intervals, and more.

Accuracy of Approximation

Accuracy in approximation processes is vital, specifically when using Riemann sums. The goal is to see how close our estimate (using sums like \(L_N\)) gets to the exact integral value. In the context of left Riemann sums:

• With \(N=1\), the approximation can be rough.
• As \(N\) increases (for example \(N=10\), \(N=100\)), the sum gets closer to the true integral value, \(\frac{\pi}{2}\) in this case.

This happens because more intervals mean more rectangles beneath the curve, reducing the errors between the curve and the rectangle tops (better fitting the area). Thus, increasing \(N\) enhances accuracy, mimicking the "exact" value in practical scenarios. This principle is integral (pun intended!) in the iterative methods used for approximating functions and areas in calculus.