Question

Prove using the concept of definite integral as limit of sum that \(\frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=1}^{\mathrm{n}} \cos \frac{\mathrm{ka}}{\mathrm{n}}<\sin \mathrm{a}<\frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=0}^{\mathrm{n}-1} \cos \frac{\mathrm{ka}}{\mathrm{n}}, 0<\mathrm{a}<\frac{\pi}{2}\)

Short answer

Question: Prove that, for \(0<a<\frac{\pi}{2}\), the inequality \(\frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=1}^{\mathrm{n}} \cos\frac{\mathrm{ka}}{\mathrm{n}}<\sin \mathrm{a}<\frac{\mathrm{a}}{\mathrm{n}}\sum_{\mathrm{k}=0}^{\mathrm{n}-1} \cos \frac{\mathrm{ka}}{\mathrm{n}}\) holds true. Solution: The inequality holds true because, for \(0<a<\frac{\pi}{2}\), the function \(\cos x\) is positive and decreasing over the interval \(0 \leq x \leq a\). Therefore, the left-endpoint sum is an underestimate and the midpoint sum is an overestimate of the area under the curve. As the number of rectangles (n) approaches infinity, both sums converge to the definite integral of \(\cos x\) over the interval \([0, a]\), which is \(\sin a\). Therefore, the inequality holds true.

Step-by-step solution

Identify the function and Riemann sum

We will first identify the function whose definite integral we want to approximate. For this problem, the function is \(f(x) = \cos x\). We can observe that the sum of \(\mathrm{a}/\mathrm{n}\) coeffficients in both inequalities are approximating the integral of \(\cos x\) over the interval \([0, a]\) which is equal to \([\sin(a) - \sin(0)] =\sin a\). Now, we will rewrite the inequality in terms of the Riemann sum. We will first rewrite it in the midpoint and left-endpoint forms: \(\frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=1}^{\mathrm{n}} f\left(\frac{\mathrm{ka}}{\mathrm{n}}\right) < \sin a < \frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=0}^{\mathrm{n}-1} f\left(\frac{\mathrm{ka}}{\mathrm{n}}\right)\).

Geometric interpretation

Now, let's think about the geometric interpretation of the integral. The definite integral of a function over an interval can be interpreted as the area under the curve of the function over the interval. The Riemann sum is essentially an approximation of the area under the curve, using rectangles. Since \(0<a<\frac{\pi}{2}\), the function \(f(x) = \cos x\) is positive and decreasing over the interval \(0 \leq x \leq a\). Therefore, the left-endpoint sum (first sum in the inequality) will be an underestimate, and the midpoint sum (second sum in the inequality) will be an overestimate of the area under the curve (integral).

Limiting process and prove the inequality

Now, we will take the limit as n approaches infinity to obtain the definite integral and prove the inequality: \(\lim_{n \to \infty} \frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=1}^{\mathrm{n}} f\left(\frac{\mathrm{ka}}{\mathrm{n}}\right) \leq \int_0^a \cos x \, dx \leq \lim_{n \to \infty} \frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=0}^{\mathrm{n}-1} f\left(\frac{\mathrm{ka}}{\mathrm{n}}\right)\). We know that \(\int_0^a \cos x\,dx = [\sin x]_0^a =\sin a\). Hence, we have: \(\lim_{n \to \infty} \frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=1}^{\mathrm{n}} f\left(\frac{\mathrm{ka}}{\mathrm{n}}\right) \leq \sin a \leq \lim_{n \to \infty} \frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=0}^{\mathrm{n}-1} f\left(\frac{\mathrm{ka}}{\mathrm{n}}\right)\). As \(n \to \infty\), both sums converge to the definite integral of \(\cos x\) over the interval \([0, a]\), which is \(\sin a\). Therefore, this inequality holds true, and we have proved that: \(\frac{\mathrm{a}}{\mathrm{n}} \sum_{\mathrm{k}=1}^{\mathrm{n}} \cos\frac{\mathrm{ka}}{\mathrm{n}}<\sin \mathrm{a}<\frac{\mathrm{a}}{\mathrm{n}}\sum_{\mathrm{k}=0}^{\mathrm{n}-1} \cos \frac{\mathrm{ka}}{\mathrm{n}},0<\mathrm{a}<\frac{\pi}{2}\).

Key concepts

Understanding the Riemann Sum

The Riemann sum is a fundamental concept in calculus that serves as a building block for understanding definite integrals. Imagine you have a smooth curve on a graph, and you want to find the area under the curve between two points on the x-axis. The Riemann sum helps us approximate this area by dividing the region into a finite number of rectangles, calculating the area of each rectangle, and summing them up.

The accuracy of the approximation improves as the number of rectangles increases because the rectangles can fit more snugly under the curve. In mathematical terms, we express a Riemann sum for a function, like our cosine function, as \(\sum_{k=1}^n f(x_k^*)\Delta x\), where \(x_k^*\) represents the sample points, \(\Delta x\) is the width of each rectangle, and the function \(f\) gives the height. Depending on where the sample point is taken on the interval, there can be different types of Riemann sums, such as the left-endpoint, right-endpoint, and midpoint sums.

Left-Endpoint and Midpoint Sums

In the case of the given exercise, the left-endpoint sum underestimates the actual area since \(\cos x\) is decreasing on the interval \(0
Using Riemann sums, calculus allows us to understand not only geometric shapes but also various physical phenomena and statistical data, by modeling and approximation, which is a powerful tool in applied mathematics and engineering fields.

The Integral as an Approximation

The concept of an integral is tightly linked with the idea of approximation. When we deal with definite integrals, we essentially look at the exact area under a curve within specified bounds. This precision is achieved in the limit as the Riemann sums approach infinity, that is, when the number of rectangles becomes infinitely large and their width infinitely small.

Integrals provide not just an approximation but the exact value of the area, given the form \(\int_{a}^{b} f(x) dx\), where \(f(x)\) is our function, and \(a\) and \(b\) are the lower and upper bounds respectively. In our exercise, we are investigating the integral of the cosine function, \(\int_0^a \cos x\,dx\), where the process is to approximate the integral using finite sums and then to take the limit of these sums as \(n\) approaches infinity, showcasing the integral as the limit of a sum.

The beauty of integral calculus is how it provides an exact counterpart to the approximation methods of Riemann sums. It allows us to analyze and compute real-world quantities precisely, such as distance, volume, work, and many other important concepts that rely on accumulation and continuous change.

The Sin and Cos Integral Relationship

Trigonometric functions like sine and cosine are often studied for their unique relationship in calculus, particularly when integrated. The exercise explores this by showing the relationship between the definite integral of the cosine function and the sine function.

Mathematically, when we integrate \(\cos x\) with respect to \(x\), we obtain the sine function \(\sin x\): \(\int \cos x\,dx = \sin x + C\), where \(C\) is the constant of integration. When we evaluate this over an interval [0, a], we apply the Fundamental Theorem of Calculus to find the exact change in the sine function over that interval, which is \(\sin a - \sin 0 = \sin a\), as the sine of zero is zero.

Understanding this integral relationship is not just academic; it reflects the periodic nature of these functions and is essential in areas like signal processing, wave mechanics, and alternating current (AC) circuit analysis. The exercise uses this relationship to provide a concrete example of how integrals capture this periodicity by comparing the sum of function evaluations at discrete points to the smooth transition of the sine function over the interval.