Problem 83 - Calculus: Early Transcendentals

Question

Find a function \( f \) and a number \( a \) such that \( \displaystyle 6 + \int^x_a \frac{f(t)}{t^2} \, dt = 2 \sqrt{x} \) for all \( x > 0 \)

Step-by-Step Solution

Answer

The function is \( f(x) = x^{3/2} \) and you can choose \( a = 1 \).

Step 1: Differentiate Both Sides

To solve this problem, begin by differentiating both sides of the given equation with respect to \( x \). The left side is \( 6 + \int_a^x \frac{f(t)}{t^2} \, dt \) and the right side is \( 2 \sqrt{x} \). Differentiating, we get:\[ \frac{d}{dx} \left( 6 + \int_a^x \frac{f(t)}{t^2} \, dt \right) = \frac{d}{dx}(2 \sqrt{x}) \]This simplifies to:\[ \frac{f(x)}{x^2} = \frac{d}{dx}(2 \sqrt{x}) \]

Step 2: Differentiate the Right Side

Differentiate the right side \( 2 \sqrt{x} \) with respect to \( x \).Using the power rule, recall that the derivative of \( x^{1/2} \) is \( \frac{1}{2} x^{-1/2} \).Thus, \( \frac{d}{dx}(2 \sqrt{x}) = 2 \times \frac{1}{2} x^{-1/2} = x^{-1/2} \).

Step 3: Equate the Derivatives

Now, equate the differentiated left side with the differentiated right side:\[ \frac{f(x)}{x^2} = x^{-1/2} \]This implies:\[ f(x) = x^{-1/2} \times x^2 = x^{3/2} \]

Step 4: Choose the Appropriate Value for \( a \)

Since the problem states that the equation holds for all \( x > 0 \), we can evaluate the constant term (6 - the integral at the lower limit \( a \)) to hold 2. Thus,Choose any convenient \( a \), say \( a = 1 \), and ensure that the equation balances. This does not change the form of \( f(x) \).

Key Concepts

Differentiation

Differentiation is a central operation in calculus that deals with finding the rate at which a function changes at any point. In the context of the given problem, we are interested in differentiating both sides of the equation to find a function that satisfies the integral relationship. This process allows us to extract information about the function inside the integral by analyzing how the entire expression changes with respect to the variable.

In our exercise, the equation contains an integral on the left side and a square root function on the right side. By differentiating, we leverage the property that differentiation and integration are inverse operations. Remember the formula for differentiation of simple powers and chain rules, as these are useful tools when handling composite variances like square roots. For example, the derivative of a power function like \(x^{n}\) is \(n \cdot x^{(n-1)}\).

By applying these principles, we were able to set the derivatives of each side equal to find the underlying function, which is crucial in solving the integral equation.

Definite Integrals

Definite integrals are used when we want to calculate the net area under a curve from one point (a) to another point (x) on the function. This type of integration results in a number, representing the accumulation of \(f(t)\) divided by \(t^2\) from \(t=a\) to \(t=x\).

In this exercise, the given equation involves a definite integral with a variable upper limit, which is a common scenario tackled in fundamental calculus exercises. The lower limit \(a\) is often a constant, influencing the base reference from which changes in the function are measured. By choosing an appropriate \(a\), we ensure the equation holds true under the specified conditions, simplifying the process of identifying unknown functions.

Thus, by understanding how definite integrals accumulate values over an interval and applying differentiation, we identify the behavior of function \(f(t)\) through the limit evaluation.

Integration Techniques

Integration techniques in calculus help us solve integrals that might not be straightforward. In our exercise, the integral is in the form of \(\int_a^x \frac{f(t)}{t^2} \, dt\). To tackle this, we use specific strategies, such as substitution or finding patterns through which the integration simplifies.

An important technique used in our solution is leveraging the derivative counterparts through differentiation, substituting values where necessary. This method bypasses the need for direct integration by equating derivatives, offering a path to identify \(f(t)\). The underlying aim is to simplify or transform the integral into a recognizable format where known rules apply.

Effective integration techniques require practice and familiarity with common forms, allowing for intuitive recognition and manipulation of complex integrals. Through these skills, finding solutions to integral equations becomes much more efficient.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the two main concepts of calculus—differentiation and integration. It holds that differentiation and integration are inverse processes. The theorem has two parts: the first establishes that an integral function can be reversed through differentiation, while the second states that the definite integral of a derivative returns the net change in the original function.

In this scenario, differentiating the integral's direct relation to the upper limit \(x\) helps determine not only function \(f(t)\) but also confirms the continuity and behavior of the expression inside the integral bounds. The practical application of this theorem comes into play when we differentiate both sides of the given equation, essentially reversing the integration to uncover the function \(f(x)\).

The fundamental theorem is invaluable for solving problems involving accumulations and rates, as it provides a structured way to interchange between these core calculus operations. This insight simplifies complex calculations and deepens understanding of how integration and differentiation shape functional relationships.