Problem 44 - Calculus Single Variable

Question

In Exercises \(43-46,\) compute \(F^{\prime}\) and \(F^{\prime \prime} .\) Determine the intervals on which \(F\) is increasing, decreasing, concave up, and concave down. $$ F(x)=\int_{0}^{x} t \exp (-t) d t $$

Step-by-Step Solution

Answer

First derivative: \(F'(x) = x\exp(-x)\); second derivative: \(F''(x) = (1-x)\exp(-x)\). Increasing: \((0,\infty)\); decreasing: \((-\infty,0)\); concave up: \((-\infty,1)\); concave down: \((1,\infty)\).

Step 1: Compute the First Derivative of F

To find the first derivative of \( F(x) = \int_{0}^{x} t \exp(-t) \ dt \), we apply the Fundamental Theorem of Calculus Part 1. This states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). Here, \( f(t) = t \exp(-t) \), thus \( F'(x) = x \exp(-x) \).

Step 2: Compute the Second Derivative of F

The second derivative of \( F \) is the derivative of \( F'(x) = x \exp(-x) \). We'll use the product rule: if \( u = x \) and \( v = \exp(-x) \), then \( u' = 1 \) and \( v' = -\exp(-x) \). Thus, \[ F''(x) = u'v + uv' = 1 \cdot \exp(-x) + x \cdot (-\exp(-x)) = \exp(-x) - x \exp(-x) = (1-x) \exp(-x) \].

Step 3: Determine Intervals of Increase and Decrease

The function \( F(x) \) is increasing where \( F'(x) > 0 \) and decreasing where \( F'(x) < 0 \). Since \( F'(x) = x \exp(-x) \), solve the inequality \( x > 0 \). Thus, \( F(x) \) is increasing on \( (0, \infty) \) and decreases on \( (-\infty, 0) \).

Step 4: Determine Intervals of Concavity

\( F(x) \) is concave up where \( F''(x) > 0 \) and concave down where \( F''(x) < 0 \). Given that \( F''(x) = (1-x) \exp(-x) \), solve \( 1-x > 0 \) to find \( x < 1 \). Therefore, \( F(x) \) is concave up on \( (-\infty, 1) \) and concave down on \( (1, \infty) \).

Step 5: Summarize Results

The first derivative is \( F'(x) = x \exp(-x) \) and the second derivative is \( F''(x) = (1-x) \exp(-x) \). Function \( F(x) \) is increasing on \( (0, \infty) \) and decreasing on \( (-\infty, 0) \). It is concave up on \( (-\infty, 1) \) and concave down on \( (1, \infty) \).

Key Concepts

First Derivative

The first derivative of a function provides crucial information about the rate of change of the function at any point. In simple terms, it tells us how the function is moving or changing. For the function given by the integral \( F(x) = \int_{0}^{x} t \exp(-t) \ dt \), the Fundamental Theorem of Calculus Part 1 helps us find the first derivative. This theorem states that, if you have a function of \( F(x) = \int_{a}^{x} f(t) \ dt \), then the derivative of this function, \( F'(x) \), is simply \( f(x) \).

We are given \( f(t) = t \exp(-t) \). Applying the theorem, the first derivative becomes \( F'(x) = x \exp(-x) \).

Here's a quick look at what this means:

• \( F'(x) \) is positive where the function \( F(x) \) is increasing.
• \( F'(x) \) is negative where the function is decreasing.

Knowing \( F'(x) \) allows us to determine intervals of increase and decrease over the domain of the function.

Second Derivative

The second derivative of a function tells us about its concavity, or the way it bends. It also helps in determining the function's points of inflection, which are points where the function changes from being concave up to concave down, or vice versa.
Finding the second derivative means differentiating the first derivative. When we differentiate \( F'(x) = x \exp(-x) \), we use the product rule.

For background, the product rule is a method used to differentiate functions that are products of two functions: \( u(x) \) and \( v(x) \). Here's how it works:

• If \( u = x \) and \( v = \exp(-x) \), then \( u' = 1 \) and \( v' = -\exp(-x) \).
• The product rule states \( F''(x) = u'v + uv' \).

Applying this, the second derivative becomes \( F''(x) = (1-x) \exp(-x) \).

This gives insight into the concavity behavior of \( F(x) \).

Product Rule

The product rule is a fundamental tool in calculus used to differentiate functions that are products of two separate functions. It becomes especially handy when the function is a complex multiplication of terms.
According to the product rule:

• You identify the two functions within the product. In our case, \( u = x \) and \( v = \exp(-x) \).
• The derivatives are \( u' = 1 \) and \( v' = -\exp(-x) \).
• The overall derivative becomes the sum of the product of the first derivative of the first function and the second function, plus the original first function times the derivative of the second function.

So, the derivative can be represented as \( F''(x) = 1\cdot\exp(-x) + x(-\exp(-x)) = (1-x)\exp(-x) \).
Apply this in complex problems to manage different parts piece by piece.

Concavity

Concavity refers to the direction in which a curve bends. A function is concave up where it holds a cup shape (like a smile), meaning the second derivative is positive. Conversely, if the second derivative is negative, the function is concave down, resembling a frown or a dome shape.

Understanding concavity helps in visualizing the graph of functions and in optimization problems to identify maxima and minima.
For our function \( F(x) \), the second derivative is \( F''(x) = (1-x) \exp(-x) \).

• To find where \( F(x) \) is concave up, solve \( F''(x) > 0 \).
• This occurs when \( 1-x > 0 \) or simply \( x < 1 \).
• For concave down regions, solve \( F''(x) < 0 \), meaning \( x > 1 \).

Determining concavity helps sketch the function more accurately.