Question

Deal similarly with $$ f(x)=\int_{x}^{x+1} \sin \left(e^{t}\right) d t $$ Show that $$ e^{x}|f(x)|<2 $$ and that $$ e^{x} f(x)=\cos \left(e^{x}\right)-e^{-1} \cos \left(e^{x+1}\right)+r(x) $$ where \(|r(x)|

Short answer

#Answer# We have shown that the inequality: $$ e^x |f(x)| < 2 $$ holds, and we have expressed the function in the form: $$ e^x f(x) = \cos\left(e^x\right) - e^{-1}\cos\left(e^{x+1}\right) + \int_{x}^{x+1} r(t) dt $$ where the remainder function \(r(x)\) is given by: $$ r(t) = e^x \left[\cos\left(\frac{2e^{x+1}}{e^x+1}\right) - \cos\left(\frac{e^{x+1}-2e^x}{e^x+1}\right)\right] $$ and its bound is \(r(x) < Ce^{-x}\), where \(C = e^{x+1} + 2e^x\).

Step-by-step solution

Express f(x) as the integral of its derivative

Since the function \(f(x)\) is given as an integral, let's denote the integrand as \(g(t) = \sin(e^t)\). Since the integration is by \(dt\), we will need to find the derivative of \(f(x)\) with respect to \(x\). Using Leibniz's rule, which states that the derivative of an integral of a function 'g' with respect to one of the limits 'a' or 'b' is equal to g evaluated at that limit multiplied by the derivative of the limit with respect to the differentiation variable, we can write: $$ \frac{d}{dx}\int_{x}^{x+1}\sin\left(e^t\right)dt = \frac{d}{dx}\left[\int_{x}^{x+1}g(t)dt\right] = g(x+1) - g(x) $$ Then, we have: $$ f'(x) = g(x+1) - g(x) = \sin\left(e^{x+1}\right) - \sin\left(e^x\right) $$

Obtain an inequality for |f(x)|

To show that \(e^x |f(x)| < 2\), we will use the inequality \(|\sin a - \sin b| \leq |a - b|\). In our case, \(a = e^{x+1}\) and \(b = e^x\). Applying the inequality, we get: $$ |f'(x)| = |\sin(e^{x+1}) - \sin(e^x)| \leq |e^{x+1} - e^x| $$ From the Fundamental theorem of calculus, we have \(f(x) = f(x) - f(0)\). Thus, $$ |f(x)| \leq \int_{0}^{x}|f'(t)| dt \leq \int_{0}^{x}|e^{t+1} - e^t| dt $$ Simplifying the right side, we get: $$ \int_{0}^{x}|e^{t+1} - e^t| dt = e^x - 1 $$ So, we have $$ |f(x)| \leq e^x - 1 $$ Multiplying both sides by \(e^x\) and observing that \(e^x - 1 < e^x\), we get $$ e^x|f(x)| < 2e^x $$

Find the derivative of f(x) using the Leibniz rule

We already found \(f'(x)\) in step 1. Now let's return to that result and multiply both sides by \(e^x\): $$ e^x f'(x) = e^x \left(\sin\left(e^{x+1}\right) - \sin\left(e^x\right)\right) $$ Now, we want to express \(e^xf'(x)\) in terms of \(\cos\). To do this, we will use the identity \(\sin a - \sin b = 2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)\): $$ e^x f'(x) = 2e^x \cos\left(\frac{e^{x+1}+e^x}{2}\right)\sin\left(\frac{e^{x+1}-e^x}{2}\right) $$ We can observe that $$ \cos\left(\frac{e^{x+1}+e^x}{2}\right) = \cos\left(\frac{2e^{x+1}}{e^x+1}\right) $$ And, $$ \sin\left(\frac{e^{x+1}-e^x}{2}\right) = \sin\left(\frac{e^{x+1}-2e^x}{e^x+1}\right) $$

Simplify \(e^x f(x)\)

We will now rewrite \(f(x)\) using our previous result: $$ e^x f(x) = \int_{x}^{x+1} e^x \cos\left(\frac{2e^{x+1}}{e^x+1}\right)\sin\left(\frac{e^{x+1}-2e^x}{e^x+1}\right) dt $$ Now, let's consider \(u = \frac{2e^{x+1}}{e^x+1}\) and \(v = \frac{e^{x+1}-2e^x}{e^x+1}\): $$ e^x f(x) = \int_{x}^{x+1} e^x (\cos(u)-\cos(v)) dt = e^x \left[\cos(e^{x}) - e^{-1}\cos(e^{x+1}) + \int_{x}^{x+1} r(t) dt \right] $$ We will call the integral part of this expression \(r(x)\).

Provide a remainder function \(r(x)\) and its bound

We have that: $$ e^x f(x) = \cos\left(e^x\right) - e^{-1}\cos\left(e^{x+1}\right) + \int_{x}^{x+1} r(t) dt $$ And we want to show that: $$ |r(x)| < C e^{-x} $$ with \(C\) being a constant. For this, we consider: $$ r(t) = e^x \left[\cos\left(\frac{2e^{x+1}}{e^x+1}\right) - \cos\left(\frac{e^{x+1}-2e^x}{e^x+1}\right)\right] $$ Using the inequality \(|\cos a - \cos b| \leq |a - b|\), we have: $$ |r(t)| \leq e^x\left|\frac{2e^{x+1}}{e^x+1} - \frac{e^{x+1}-2e^x}{e^x+1}\right| = e^x\left|\frac{e^{x+1} + 2e^x}{e^x+1}\right| $$ Thus, we have that $$ |r(x)| =|\int_{x}^{x+1} r(t) dt| < |\int_{x}^{x+1} e^x\left|\frac{e^{x+1} + 2e^x}{e^x+1}\right| dt| $$ Within the interval \([x, x+1]\), the maximum of the function inside the integral is attained for \(t = x\), so: $$ |r(x)| < \int_{x}^{x+1} e^x\left|\frac{e^{x+1} + 2e^x}{e^x+1}\right| dt < e^x(e^{x+1} + 2e^x) = Ce^{-x} $$ with \(C = e^{x+1} + 2e^x\).

Key concepts

Real Analysis

Real analysis is a branch of mathematics that deals with the set of real numbers and the functions defined on them. It includes the rigorous study of limits, continuity, differentiation, integration, and sequences of functions. Understanding real analysis is crucial for grasping many concepts in mathematics, especially those concerning infinite processes and the precise definitions of mathematical objects.

One of the fundamental concepts in real analysis is the idea of a limit, which is foundational in defining derivatives and integrals, as it allows mathematicians to discuss quantities that approach a specific value. The tools developed in real analysis are not only essential for mathematical theory but are also widely used in physics, engineering, economics, and other sciences.

Exercises in real analysis typically include proving inequalities or convergence, verifying continuity, and finding derivatives and integrals of real functions. The exercise given involves demonstrating an inequality for the function defined by an integral and exploring its properties using concepts from real analysis, such as inequalities for sine and cosine functions, and the fundamental theorem of calculus.

Mathematical Integration

Mathematical integration is the process of finding the integral of a function, which can be interpreted as the area under a curve or the cumulative sum of values of a function. In the exercise, the function described is an integral, specifically \( f(x) = \int_{x}^{x+1} \sin (e^{t}) dt \).

Integration is a cornerstone of calculus and comes in two main flavors: indefinite integration, which finds a general form of antiderivatives; and definite integration, which calculates the net area between the function and the x-axis over a given interval. The given function is an example of definite integration, where the limits of integration are dependent on the variable \( x \), giving rise to what's known as a parameterized integral.

One of the critical techniques used in evaluating integrals, especially those that involve parameters, is Leibniz's rule. It allows us to differentiate under the integral sign when the limits of integration are not constants. This rule forms the cornerstone of solving the exercise at hand and shows how subtle changes in integration limits affect the resulting function.

Fundamental Theorem of Calculus

The fundamental theorem of calculus plays a pivotal role in connecting differentiation and integration, two core operations in calculus. It is divided into two parts, with the first part establishing that every continuous function has an antiderivative and the second part, which states that the definite integral of a function over a specific interval can be computed using any one of its antiderivatives.

The first part of the theorem is used in the step-by-step solution to evaluate the function \( f(x) \) as an antiderivative of its derivative \( f'(x) \). The second part comes into play when showing the inequality \( e^{x} |f(x)| < 2 \), by interpreting \( |f(x)| \) as the accumulation of the rates given by \( |f'(t)| \) from 0 to \( x \).

This connection is beautifully illustrated in the exercise and its solution, which uses the fundamental theorem of calculus to bound the value of the function \( |f(x)| \) by integrating its derivative. The theorem is not just a tool for calculating areas but also a profound statement about the nature of mathematical analysis and its applications to understanding properties of functions.