Let \(\mathrm{f}\) be a real-valued function defined on the interval \((0, \infty)\) by \(\mathrm{f}(\mathrm{x})=\ell n \mathrm{x}+\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} \mathrm{dt} .\) Then which of the following statement \((\mathrm{s})\) is (are) true ? A) \(\mathrm{f}^{\prime \prime}(\mathrm{x})\) exists for all \(\mathrm{x} \in(0, \infty)\) B) \(\mathrm{f}^{\prime}(\mathrm{x})\) exists for all \(\mathrm{x} \in(0, \infty)\) and \(\mathrm{f}^{\prime}\) is continuous on \((0, \infty)\), but not differentiable on \((0, \infty)\) C) there exists \(\alpha>1\) such that \(\left|f^{\prime}(x)\right|<|f(x)|\) for all \(x \in(\alpha, \infty)\) D) there exists \(\beta>0\) such that \(|\mathrm{f}(\mathrm{x})|+\left|\mathrm{f}^{\prime}(\mathrm{x})\right| \leq \beta\) for all \(\mathrm{x} \in(0, \infty)\)

Integration is a fundamental concept in calculus that involves finding the integral of a function. It can be thought of as the process of calculating the area under a curve or the accumulation of quantities. In the context of the exercise, integration is used to define the function f(x), which includes the integral of \( \sqrt{1 + \sin(t)} \) from 0 to x.

The integral of a real-valued function, like the one given, can be interpreted as the accumulated value starting from the lower limit of integration (0 in this case) to the upper limit (x). In practical terms, this integral can represent a wide array of physical quantities such as distance, area, volume, or other accumulative sums depending upon the problem's context.

An important aspect of integration is its close relationship with differentiation, called the Fundamental Theorem of Calculus, which allows us to evaluate definite integrals by finding an antiderivative. In this problem, the integral is connected to the differentiation part of the function when we take the derivative of f(x) in Step 1 of the solution.

Differentiation, the counterpart to integration, involves computing the derivative of a function. It measures how a function's value changes as its input changes, representing the rate of change or the slope of the function's graph at any given point.

For the real-valued function f(x) in our exercise, the process of differentiation is applied to find the first derivative f'(x), which conveys the instantaneous rate of change of f with respect to x. The derivative of \(\ln(x)\) is \(\frac{1}{x}\), and applying the Fundamental Theorem of Calculus, the derivative of the integral part is the integrand itself evaluated at x, which is \(\sqrt{1 + \sin(x)}\).

The exercise further asks us to find the second derivative, f''(x), which represents the rate of change of the rate of change, providing insights into the concavity of the function's graph and often associated with acceleration in physics.

Real-valued functions are functions that take a real number as input and provide a real number as output. They can be graphically represented on a two-dimensional plane, highlighting their behavior across different intervals. The function f(x) in our exercise is a real-valued function defined on the interval (0, ∞).

The components of f(x) include \(\ln(x)\) and an integral with a trigonometric function within its integrand. When combined, these components form the distinct real-valied function that must be analyzed for different properties like continuity, differentiability, and boundedness. Real-valued functions like f(x) are the bedrock of calculus and are used to model an enormous variety of real-world phenomena.

The concepts of continuity and differentiability are essential in understanding the behavior of functions. A function is continuous at a point if there are no breaks, jumps, or holes in its graph at that point, meaning it can be drawn without lifting the pen from the paper. Differentiability, on the other hand, implies that the function has a defined derivative at that point, indicating a smooth curve without any sharp turns.

In Step 3 of our exercise's solution, we discuss the continuity and differentiability of f'(x). Since f'(x) is formed by combining two continuous functions, it is continuous across its domain (0, ∞). Moreover, because these functions are also differentiable over the interval, we can conclude that f'(x) is differentiable, satisfying one of the conditions of the exercise.

The Fundamental Theorem of Calculus is a vital link between differentiation and integration. It consists of two parts: The first part guarantees that an indefinite integral of a function can act as an antiderivative. The second part allows us to evaluate definite integrals through antiderivatives. Specifically, the theorem tells us that if a function is continuous over an interval, then the integral over that interval can be calculated by finding the difference between the values of an antiderivative evaluated at the endpoints of the interval.

This theorem is directly applied in Step 1 of the solution, where the derivative of the integral part of f(x) is taken. By using the theorem, we evaluate the derivative as simply the integrand evaluated at x, which in this case is \(\sqrt{1 + \sin(x)}\). This principle is essential in solving many calculus problems, as it bridges the gap between the abstract concepts of integration and differentiation and provides a practical method for calculation.