Problem 45 - 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_{1}^{x} t \ln (t) d t \quad 0

Step-by-Step Solution

Answer

\( F(x) \) increases on \( x > 1 \), decreases on \( 0 < x < 1 \), is concave up on \( x > \frac{1}{e} \), and concave down on \( 0 < x < \frac{1}{e} \).

Step 1: Understanding the Problem

We need to compute the first derivative \( F'(x) \) and the second derivative \( F''(x) \) of the given integral function \( F(x)=\int_{1}^{x} t \ln(t) \, dt \). Then, determine the intervals where \( F \) is increasing or decreasing, as well as where \( F \) is concave up or concave down.

Step 2: Compute the First Derivative Using the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(x) = \int_{a}^{x} f(t) \ dt \), then \( F'(x) = f(x) \). Here, \( f(t) = t \ln(t) \), so \( F'(x) = x \ln(x) \).

Step 3: Compute the Second Derivative

To find the second derivative \( F''(x) \), we differentiate \( F'(x) = x \ln(x) \) with respect to \( x \). Apply the product rule: \( d(uv)/dx = u'v + uv' \) where \( u = x \) and \( v = \ln(x) \). Thus, \( F''(x) = (1) \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1 \).

Step 4: Determine Intervals of Increase and Decrease

\( F(x) \) increases where \( F'(x) > 0 \) and decreases where \( F'(x) < 0 \). Since \( F'(x) = x \ln(x) \), analyze \( x \ln(x) \). \( x \ln(x) > 0 \) for \( x > 1 \) and \( x \ln(x) < 0 \) for \( 0 < x < 1 \). Thus, \( F(x) \) is increasing for \( x > 1 \) and decreasing for \( 0 < x < 1 \).

Step 5: Determine Intervals of Concavity

\( F(x) \) is concave up where \( F''(x) > 0 \) and concave down where \( F''(x) < 0 \). From Step 3, \( F''(x) = \ln(x) + 1 \). \( F''(x) > 0 \) when \( \ln(x) > -1 \) which is when \( x > \frac{1}{e} \), and \( F''(x) < 0 \) when \( x < \frac{1}{e} \). Thus, \( F(x) \) is concave up for \( x > \frac{1}{e} \) and concave down for \( 0 < x < \frac{1}{e} \).

Key Concepts

Derivative

The concept of derivatives is central to calculus. It is essentially the rate at which a function is changing at any given point. In simpler terms, it tells us how steep or flat the function's graph is at a specific point, like the speedometer of a car showing speed at an instant.

In the context of the function \( F(x) = \int_{1}^{x} t \ln(t) \, dt \), we find the first derivative \( F'(x) \) using the Fundamental Theorem of Calculus. According to the theorem, if we have an integral of the form \( F(x) = \int_{a}^{x} f(t) \, dt \), then the derivative \( F'(x) \) is simply \( f(x) \). Thus, here, \( F'(x) = x \ln(x) \).

Understanding \( F'(x) \) tells us where the function \( F(x) \) is increasing or decreasing. \( F(x) \) is increasing where \( F'(x) > 0 \) and decreasing where \( F'(x) < 0 \). For \( x \ln(x) \), it is positive when \( x > 1 \), meaning \( F(x) \) is increasing, and negative when \( 0 < x < 1 \), so \( F(x) \) is decreasing.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differential and integral calculus. It connects the concept of an integral, which measures area, with that of a derivative, which measures slope.

This theorem states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). The function \( F(x) \) is called an antiderivative or an indefinite integral of \( f(x) \).

For the given problem, \( F(x) = \int_{1}^{x} t \ln(t) \, dt \), using the theorem, we immediately find that \( F'(x) = t \ln(t) \), evaluated at \( x \), or simply \( F'(x) = x \ln(x) \).

This direct replacement allows us to find a derivative of complex expressions that might otherwise require elaborate calculations. Understanding the Fundamental Theorem helps simplify solving integrals, and it's truly revolutionary in calculus for simplifying evaluations of definite integrals.

Concavity

Concavity helps us understand the shape and curvature of a function's graph. It tells us whether the graph is curving upwards or downwards:

• A function is concave up when it curves like a cup or a smile. Mathematically, this occurs when the second derivative \( F''(x) \) is greater than 0.
• A function is concave down when it curves like a frown. In math terms, this happens when the second derivative \( F''(x) \) is less than 0.

The second derivative test is crucial in determining concavity. In the problem, we calculated the second derivative as \( F''(x) = \ln(x) + 1 \). Using this, we can predict how \( F(x) \) behaves:

• \( F''(x) > 0 \): \( F(x) \) is concave up for \( x > \frac{1}{e} \), because a positive second derivative indicates a curve upwards.
• \( F''(x) < 0 \): \( F(x) \) is concave down for \( 0 < x < \frac{1}{e} \), suggesting a curve downwards.

Understanding these intervals of concavity is vital for analyzing and graphing functions, helping predict the peaks, troughs, and overall shape of the graph.