Problem 28 - Calculus Single Variable

Question

In each of Exercises \(21-28,\) calculate the derivative of \(F(x)\) with respect to \(x\). $$ F(x)=\int_{1}^{x} e^{t} \ln (t) d t $$

Step-by-Step Solution

Answer

The derivative is \( F'(x) = e^x \ln(x) \).

Step 1: Identify the function to differentiate

The given function is \( F(x) = \int_{1}^{x} e^{t} \ln(t) \, dt \), which represents an integral with a variable upper limit.

Step 2: Apply the Fundamental Theorem of Calculus Part 1

According to the Fundamental Theorem of Calculus Part 1, if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). In this case, \( f(t) = e^t \ln(t) \).

Step 3: Differentiate using the fundamental theorem

By applying the theorem, the derivative of \( F(x) \) with respect to \( x \) is simply the integrand evaluated at \( x \). Thus, \( F'(x) = e^x \ln(x) \).

Key Concepts

Integral with Variable Upper Limit

When we talk about an integral with a variable upper limit, we refer to a special type of integral where the upper boundary is a function, typically the variable of differentiation. In our exercise, the function given is \[ F(x) = \int_{1}^{x} e^t \ln(t) \, dt \] Here, the upper limit of the integral is the variable \( x \), which means as \( x \) changes, the upper boundary of the integral shifts as well. This setup is important because it sets the perfect stage for applying the Fundamental Theorem of Calculus, which connects differentiation and integration in a profound way. Practically, this type of integral expression allows us to explore how the accumulated area under the curve of \( f(t) = e^t \ln(t) \) from 1 to \( x \) affects our function \( F(x) \).

• The essence is in capturing how the accumulation of these tiny areas changes regarding the upper bound.
• Understanding this variable boundary helps in deriving how \( F(x) \) changes as \( x \) extends beyond its initial point "1".

This concept is fundamental as it opens up the interaction between integration and differentiation.

Differentiation of Integrals

Differentiation of integrals is a fascinating process brought to life by the Fundamental Theorem of Calculus. The theorem paves the way to directly differentiate integrals that have an upper limit which itself is a variable. To differentiate \( F(x) = \int_{1}^{x} e^{t} \ln(t) \, dt \), we utilize the Fundamental Theorem of Calculus Part 1, revealing:

• If you have an integral expressed as \( F(x) = \int_{a}^{x} f(t) \, dt \), the derivative, \( F'(x) \), is simply \( f(x) \) — the integrand at \( t = x \).
• This results from the property that differentiation effectively cancels out the integration, revealing the original function \( f(t) \) at \( x \).

The elegance lies in how the integral's rate of change, or derivative, is depicted by evaluating the function at \( x \). This makes differentiation of integrals a noble task of restoring the original function's behavior at its upper limit.

Calculation of Derivatives

Once you're familiar with integrals having a variable upper limit and the differentiation process of integrals, calculating derivatives becomes more intuitive. For our specific function: To find \( F'(x) \) for the function \( F(x) = \int_{1}^{x} e^t \ln(t) \, dt \), we proceed by:

• Recognizing the integrand: \( f(t) = e^t \ln(t) \).
• Utilizing the Fundamental Theorem of Calculus Part 1 to determine \( F'(x) = f(x) \).
• Simply evaluate \( f(t) \) at \( t = x \).

This results in the derivative: \[ F'(x) = e^x \ln(x) \]The derivative essentially tells you how \( F(x) \) changes at each point \( x \), unraveling the slope of the function \( F(x) \). With its simplicity and power, the calculation reaffirms how calculus beautifully ties integration and differentiation together, enabling precise analysis of change and accumulation.