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Chapter 5: Problem 8

Does a left Riemann sum underestimate or overestimate the area of the region under the graph of a function that is positive and increasing on an interval \([a, b] ?\) Explain.

Short Answer

Expert verified
Answer: A left Riemann sum underestimates the area under the graph of a positive increasing function on the interval [a, b].

Step by step solution

01

Understanding the Left Riemann Sum

A left Riemann sum is an approximation of the area under the curve of a function by selecting the left endpoint of each subinterval as the sample point. It is calculated by dividing the interval \([a, b]\) into equal subintervals and drawing rectangles with a height that corresponds to the function value at the left endpoint of each subinterval. For the interval \([a, b]\), we can divide it into \(n\) equal-width subintervals, each with a width \(\Delta x = \frac{b-a}{n}\). Starting from the left, the \(i\)-th subinterval would have a left endpoint at \(x = a + (i-1) \Delta x.\) The left Riemann sum is given by the expression $$L_n = \sum_{i=1}^{n} f(a + (i-1) \Delta x) \Delta x$$
02

Compare Function Values on the Interval \([a, b]\)

Since the function is increasing on the interval \([a, b]\), for any two points, \(x_1\) and \(x_2\) in the interval such that \(x_1 < x_2\), we know that \(f(x_1) < f(x_2)\). Recall that by selecting the left endpoint as a sample point, we base the height of the rectangle on the function value at the left endpoint of each subinterval. This means that for every subinterval, we will compare the height of the rectangle (function value at the left endpoint) with the function value at the right endpoint.
03

Determine if Underestimate or Overestimate

Since \(f(x_1) < f(x_2)\) for \(x_1 < x_2\) and we are using the left Riemann sum, we will constantly miss out on the additional area between the top of our rectangle and the curve. This shows that the left Riemann sum underestimates the actual area under the graph of the increasing function.

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Most popular questions from this chapter

Use a change of variables to evaluate the following definite integrals. $$\int_{1 / 3}^{1 / \sqrt{3}} \frac{4}{9 x^{2}+1} d x$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume that \(f, f^{\prime},\) and \(f^{\prime \prime}\) are continuous functions for all real numbers. a. \(\int f(x) f^{\prime}(x) d x=\frac{1}{2}(f(x))^{2}+C\) b. \(\int(f(x))^{n} f^{\prime}(x) d x=\frac{1}{n+1}(f(x))^{n+1}+C, n \neq-1\) c. \(\int \sin 2 x d x=2 \int \sin x d x\) d. \(\int\left(x^{2}+1\right)^{9} d x=\frac{\left(x^{2}+1\right)^{10}}{10}+C\) e. \(\int_{a}^{b} f^{\prime}(x) f^{\prime \prime}(x) d x=f^{\prime}(b)-f^{\prime}(a)\)

Consider the integral \(I=\int_{0}^{\pi / 2} \sin x d x\), a. Write the left Riemann sum for \(I\) with \(n\) subintervals. b. Show that \(\lim _{\theta \rightarrow 0} \theta\left(\frac{\cos \theta+\sin \theta-1}{2(1-\cos \theta)}\right)=1\). c. It is a fact that \(\sum_{k=0}^{n-1} \sin \left(\frac{\pi k}{2 n}\right)=\frac{\cos \left(\frac{\pi}{2 n}\right)+\sin \left(\frac{\pi}{2 n}\right)-1}{2\left(1-\cos \left(\frac{\pi}{2 n}\right)\right)}\). Use this fact and part (b) to evaluate \(I\) by taking the limit of the Riemann sum as \(n \rightarrow \infty\).

Evaluate the following integrals. $$\int_{0}^{\pi / 6} \frac{\sin 2 y}{\sin ^{2} y+2} d y(\text {Hint}: \sin 2 y=2 \sin y \cos y .)$$

Use a change of variables to evaluate the following definite integrals. $$\int_{0}^{\pi / 4} \frac{\sin x}{\cos ^{2} x} d x$$
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