Problem 127 - Thomas Calculus

Question

find a function \(f\) satisfying each equation. \begin{equation}\int_{2}^{x} \sqrt{f(t)} d t=x \ln x\end{equation}

Step-by-Step Solution

Answer

The function is \(f(x) = (\ln x + 1)^2\).

Step 1: Express the derivative of the integral using the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, the derivative with respect to \(x\) of the integral \(\int_{2}^{x} \sqrt{f(t)} \ dt\) is equal to \(\sqrt{f(x)}\). So, we write:\[ \sqrt{f(x)} = \frac{d}{dx}\left(\int_{2}^{x} \sqrt{f(t)} \ dt\right) = \frac{d}{dx} (x \ln x) \]

Step 2: Differentiate the right-hand side

Next, differentiate \(x \ln x\) with respect to \(x\). We use the product rule. Let \(u = x\) and \(v = \ln x\), then:\[ \frac{d}{dx}(x \ln x) = \frac{d}{dx}(u \cdot v) = u'v + uv' = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \]

Step 3: Set the derivatives equal to solve for \(f(x)\)

Now, setting the derivatives equal, we have:\[ \sqrt{f(x)} = \ln x + 1 \]To solve for \(f(x)\), square both sides of the equation:\[ f(x) = (\ln x + 1)^2 \]

Step 4: Verify the solution

To verify \(f(x) = (\ln x + 1)^2\), differentiate the right-hand side \((x \ln x)\) and check if it matches \(\sqrt{f(x)}\). We have already done this and found that both sides are equal, so this function is our solution.

Key Concepts

Definite Integrals

Definite integrals provide a way to calculate the accumulation of quantities, where the result has definite limits. In this case, the integral is given as \( \int_{2}^{x} \sqrt{f(t)} \, dt \). This integral represents the area under the curve of the square root of some function \( f(t) \), between \( t = 2 \) and \( t = x \). This area changes as \( x \) changes, allowing us to understand how the accumulation shifts.

You can visualize definite integrals as the sum of infinitely small areas, between a starting point \( t = 2 \) and an endpoint \( t = x \), creating a precise value that represents the total area. Changes in these values are the bedrock for concepts like "net change" and "total growth," common in physics and engineering.

Differentiation

Differentiation focuses on determining how a function changes as its input changes. Here, we are interested in how the accumulation defined by the integral changes, using differentiation. By applying the Fundamental Theorem of Calculus, we find that the derivative of \( \int_{2}^{x} \sqrt{f(t)} \, dt \) with respect to \( x \) is \( \sqrt{f(x)} \).

This theorem connects the world of integrals with the world of derivatives, acting like a bridge. Differentiation helps us step back from the accumulation and observe the rate of change, giving us insights into how the underlying function behaves at any particular point. Through this process, we can unravel the particular shape and form of \( f(x) \) itself by looking at its instantaneous slopes.

Product Rule

The product rule is a powerful tool in calculus that helps find the derivative of the product of two functions. In this exercise, we need to differentiate \( x \ln x \), which requires applying the product rule. Here's how it works:

• Define \( u = x \) and \( v = \ln x \).
• Derive each: \( u' = 1 \) and \( v' = \frac{1}{x} \).
• The product rule states \( \frac{d}{dx}(uv) = u'v + uv' \).

So, the derivative is \( 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \). This is pivotal for evaluating the expressions and leads directly to finding \( f(x) \). The product rule simplifies finding derivatives involving products of functions, making calculus more manageable.

Finding Functions

The problem's goal is to identify the function \( f(x) \) that fits within our integral equation. To do this, we took the derivative of both sides and set them equal, resulting in \( \sqrt{f(x)} = \ln x + 1 \). To isolate \( f(x) \), we simply square each side of this equation:

\[ f(x) = (\ln x + 1)^2 \]

By following through on differentiation and algebraic manipulation, we unveil the precise form of \( f(x) \). This function explains the function's curvature that was hidden within the integral's square root relationship. Effectively, this moves us from calculus back to plain functions, bridging conceptual understanding and practical application.