Problem 457 - Calculus Volume 1

Question

Find the derivative. $$\frac{d}{d x} \int_{1}^{\ln (x)}\left(4 t+e^{t}\right) d t$$

Step-by-Step Solution

Answer

The derivative is \(\frac{4\ln(x)}{x} + 1\).

Step 1: Recognize the Use of the Fundamental Theorem of Calculus

The problem involves finding the derivative of an integral with variable limits of integration. The Fundamental Theorem of Calculus states that if \( F(x) = \int_{a}^{g(x)} f(t)\, dt\), then the derivative \( \frac{d}{dx} F(x) = f(g(x)) \cdot g'(x)\). This is known as Leibniz's rule for differentiating under the integral sign.

Step 2: Identify Parts of the Integral

The integral given is \( \int_{1}^{\ln(x)} (4t + e^{t}) \, dt \). Here, the upper limit of integration \( g(x) = \ln(x) \) and the function inside the integral is \( f(t) = 4t + e^{t} \).

Step 3: Differentiate the Function with Respect to the Upper Limit

According to the theorem, we first evaluate the integral function at the upper limit \( \ln(x) \) by substituting it into \( f(t) \). This gives us \( 4(\ln(x)) + e^{\ln(x)} \). The fundamental theorem specifies substituting directly into the function for the upper limit.

Step 4: Compute the Derivative of the Upper Limit

Next, differentiate the upper limit with respect to \( x \). Therefore, \( \frac{d}{dx} \ln(x) = \frac{1}{x} \). This is the chain rule part of the fundamental theorem when applied to integrals with upper variable limits.

Step 5: Apply the Chain Rule and Fundamental Theorem

Multiply the result from Step 3 by the derivative from Step 4. Thus, \( \frac{d}{dx} \left[\int_{1}^{\ln(x)} (4t + e^{t}) \, dt\right] = \left(4\ln(x) + e^{\ln(x)} \right) \cdot \frac{1}{x} \).

Step 6: Simplify the Expression

Simplify the multiplication: \(4\ln(x)\cdot\frac{1}{x} + e^{\ln(x)}\cdot\frac{1}{x} = \frac{4\ln(x)}{x} + \frac{x}{x} = \frac{4\ln(x)}{x} + 1 \), since \( e^{\ln(x)} = x \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. A composite function is one where you have a function within another function, like \( g(x) \) inside \( f(g(x)) \). Here's a straightforward explanation:

• Suppose you have a function \( y = f(u) \) where \( u = g(x) \).
• To find the derivative \( \frac{dy}{dx} \), you would use the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

In our context, the upper limit of the integral is \( \ln(x) \). We treat this upper limit as \( g(x) \) which is part of the composite function. We find \( g'(x) \) or the derivative of \( \ln(x) \), resulting in \( \frac{1}{x} \).
Thus, the chain rule helps us incorporate the derivative of the variable upper limit into the derivative of the integral expression as a whole.

Integrals with Variable Limits

Integrals with variable limits play a crucial role in advanced calculus. What makes them intriguing is their ability to combine integration and differentiation. Let's clarify this concept:

• An integral with a variable limit is expressed generally as \( \int_{a}^{g(x)} f(t) \, dt \).
• The function \( g(x) \) is a variable limit, meaning it depends on \( x \).
• This requires us to use the Fundamental Theorem of Calculus when finding the derivative.

By understanding this, we realize that the integral's upper limit is not just a number but a dynamic function. This means the rate of change, or the derivative, also involves evaluating the function inside the integral at this dynamic limit.
In the exercise, \( \int_{1}^{\ln(x)}(4t+e^t)dt \) uses the upper limit \( \ln(x) \), demonstrating the principle of variable limits.

Derivatives of Integrals

When dealing with derivatives of integrals, especially with variable limits, we turn to a specific application of the Fundamental Theorem of Calculus. Here's how it uniquely applies:

• First, evaluate the integrand at the upper limit, turning \( f(t) \) into \( f(g(x)) \).
• The derivative is then found by multiplying this by the derivative of the upper limit: \( g'(x) \).

This means we effectively "plug in" the upper limit into the integrand before taking the derivative, as we did with \( f(t) = 4t + e^t \) where \( g(x) = \ln(x) \).
In practical terms, the derivative \( \frac{d}{dx}\left[\int_{a}^{g(x)} f(t) \, dt\right] \) equals \( f(g(x)) \cdot g'(x) \). We applied this in the problem by evaluating \( 4\ln(x) + e^{\ln(x)} \) and multiplying by \( \frac{1}{x} \), thus embracing the nature of variable limits in derivative calculations.