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Magazine Chapter 5: Problem 38
(a) Draw the rectangles that give the left-hand sum approximation to \(\int_{0}^{x} \sin x d x\) with \(n=2\) (b) Repeat part (a) for \(\int_{-\pi}^{0} \sin x d x\) (c) From your answers to parts (a) and (b), what is the value of the left-hand sum approximation to \(\int_{-\pi}^{\pi} \sin x d x\) with \(n=4 ?\)
Short Answer
Expert verified
The left-hand sum approximation for the integral from \(-\pi\) to \(\pi\) is 0.
Step by step solution
01
Understand the Problem
The problem asks us to find left-hand sum approximations for definite integrals involving \( \sin x \), with given intervals and subdivisions. The left-hand sum is a type of Riemann sum, which estimates the area under a curve using rectangles, where the height of each rectangle is determined by the function value at the left endpoint.
02
Set up Interval for Part (a)
We need to approximate \( \int_{0}^{x} \sin x \, dx \) with \( n = 2 \). Since the upper limit is \( x \), assume \( x = \pi \) for the endpoints to complete the calculation. We will divide the interval \([0, \pi]\) into 2 equal subintervals, so each subinterval has a width of \( \Delta x = \frac{\pi - 0}{2} = \frac{\pi}{2} \).
03
Calculate Left-Hand Sum for Part (a)
In the interval \([0, \pi]\), use \( x_0 = 0 \) and \( x_1 = \frac{\pi}{2} \) as the left endpoints of the subintervals. Calculate \( f(x_0) = \sin(0) = 0 \) and \( f(x_1) = \sin\left(\frac{\pi}{2}\right) = 1 \). The left-hand sum is given by:\[ R = f(x_0) \Delta x + f(x_1) \Delta x = 0 \cdot \frac{\pi}{2} + 1 \cdot \frac{\pi}{2} = \frac{\pi}{2} \]
04
Set up Interval for Part (b)
For \( \int_{-\pi}^{0} \sin x \, dx \), the interval is \([-\pi, 0]\), and we again divide this into 2 equal subintervals so \( \Delta x = \frac{0 + \pi}{2} = \frac{\pi}{2} \).
05
Calculate Left-Hand Sum for Part (b)
In the interval \([-\pi, 0]\), use \( x_0 = -\pi \) and \( x_1 = -\frac{\pi}{2} \) as the left endpoints of the subintervals. Calculate \( f(x_0) = \sin(-\pi) = 0 \) and \( f(x_1) = \sin\left(-\frac{\pi}{2}\right) = -1 \). The left-hand sum is:\[ R = f(x_0) \Delta x + f(x_1) \Delta x = 0 \cdot \frac{\pi}{2} + (-1) \cdot \frac{\pi}{2} = -\frac{\pi}{2} \]
06
Combine Results for Part (c)
For \( \int_{-\pi}^{\pi} \sin x \, dx \), use the results from parts (a) and (b) with \( n = 4 \) (2 subintervals on each half). The left-hand sum from \([-\pi, 0]\) is \(-\frac{\pi}{2}\) and from \([0, \pi]\) is \(\frac{\pi}{2}\). Adding these gives:\[ -\frac{\pi}{2} + \frac{\pi}{2} = 0 \]
07
Conclusion
The left-hand sum approximation to \( \int_{-\pi}^{\pi} \sin x \, dx \) with \( n = 4 \) is thus 0. This result is expected as \( \int_{-\pi}^{\pi} \sin x \, dx \) is zero because the positive area from \([0, \pi]\) cancels out the negative area from \([-\pi, 0]\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral is central in calculus and provides a way to calculate the exact area under a curve between two points on the x-axis. This is represented mathematically as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the function being integrated.
Definite integrals are used to find:
Definite integrals are used to find:
- The total accumulation of a quantity, such as distance, area, or volume, over a particular interval.
- Net change, by considering the positive and negative areas under a curve.
Left-Hand Sum
Left-hand sum is a method of estimating the value of a definite integral using rectangles. These rectangles approximate the area under a curve, and are a specific form of Riemann sum.
Here's how it works:
Here's how it works:
- The interval \([a, b]\) is divided into \(n\) equal subintervals, each with width \(\Delta x = \frac{b-a}{n}\).
- The height of each rectangle is determined by the function value at the left endpoint of the subinterval.
- For a function \(f(x)\), the left-hand sum \(R\) is calculated as:
\( R = \sum_{i=0}^{n-1} f(x_i) \Delta x \), where \(x_i\) are the left endpoints.
Trigonometric Integrals
Trigonometric integrals involve integrating functions that include trigonometric expressions like sine, cosine, and tangent. These are crucial for solving real-world problems in physics and engineering involving periodic phenomena.
With \( \sin x \), as used in this exercise:
With \( \sin x \), as used in this exercise:
- The integral \( \int \sin x \, dx \) results in \(-\cos x\), representing a shift in phase for the sine function.
- Trigonometric integrals require attention to symmetry and periodicity. For example, \( \int_{-\pi}^{\pi} \sin x \, dx \) naturally results in zero due to the symmetry of the sine function across the origin.
