Problem 78 - Thomas Calculus

Question

In Exercises \(75-78,\) let \(F(x)=\int_{a}^{u(x)} f(t) d t\) for the specified \(a, u,\) and \(f\) . Use a CAS to perform the following steps and answer the questions posed. a. Find the domain of \(F\) b. Calculate \(F^{\prime}(x)\) and determine its zeros. For what points in its domain is \(F\) increasing? decreasing? c. Calculate \(F^{\prime \prime}(x)\) and determine its zero. Identify the local extrema and the points of inflection of \(F\) . d. Using the information from parts (a)-(c), draw a rough hand-sketch of \(y=F(x)\) over its domain. Then graph \(F(x)\) on your CAS to support your sketch. $$ a=0, \quad u(x)=1-x^{2}, \quad f(x)=x^{2}-2 x-3 $$

Step-by-Step Solution

Answer

Domain: \([-1, 1]\); critical points and behavior from derivatives show where F increases or decreases; sketch verified by CAS.

Step 1: Setup the problem

First, we have the function \( F(x) = \int_{0}^{1-x^2} (t^2 - 2t - 3) \, dt \). We will use this expression to perform the upcoming operations.

Step 2: Find the domain of F

The domain of \( F(x) \) is determined by the valid values of \( u(x) = 1 - x^2 \). For \( u(x) \) to be valid, it must satisfy \( 1 - x^2 \geq 0 \), thus \( x^2 \leq 1 \), or \(-1 \leq x \leq 1 \). Therefore, the domain is \(-1, 1\).

Step 3: Calculate the derivative F'

Applying the Fundamental Theorem of Calculus, we find \( F'(x) = f(u(x)) \cdot u'(x) = ( (1-x^2)^2 - 2(1-x^2) - 3) \cdot (-2x) \). Simplifying, \( f(1-x^2) = (1-x^2)^2 - 2(1-x^2) - 3 \).

Step 4: Find F'(x) zeros and intervals of increase/decrease

Simplify \( F'(x) \) and solve \( F'(x) = 0 \) to find critical points. Evaluate the sign of \( F'(x) \) on intervals between critical points to determine where \( F \) is increasing or decreasing.

Step 5: Calculate the second derivative F''

Differentiate \( F'(x) \) to obtain \( F''(x) \). Simplify and solve \( F''(x) = 0 \) to find potential inflection points. Use these to determine concavity changes.

Step 6: Find local extrema and points of inflection

Combine results from Step 4 and Step 5. Use the second derivative test: if \( F''(x) > 0 \), a critical point is a local minimum; if \( F''(x) < 0 \), a local maximum. Points where \( F''(x) \) changes sign are points of inflection.

Step 7: Sketch the graph by hand and verify with CAS

Draw a rough sketch of \( y = F(x) \) using information about domain, critical points, intervals of increase/decrease, and concavity. Use a Computer Algebra System to graph \( y = F(x) \) and compare with your hand-drawn sketch to ensure accuracy.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects the concepts of differentiation and integration. Simply put, it states that differentiation and integration are inverse processes. This theorem comes in two main parts:

• The first part establishes that the integral of a function's derivative over an interval returns the function values at the boundaries of this interval.
• The second part provides a way to calculate the derivative of an integral function. Specifically, if you have a function defined as an integral, such as \( F(x) = \int_{a}^{u(x)} f(t) \, dt \), then its derivative \( F'(x) \) can be found using \( f(u(x)) \cdot u'(x) \).

In our exercise, we used the second part of the theorem to differentiate the integral function \( F(x) \) with respect to \( x \). This theorem simplifies the process by allowing us to focus primarily on the boundary value's behavior of the function inside the integral.

Derivative Analysis

Derivative analysis is essential in understanding the behavior of functions, like how they increase or decrease on different intervals. First, we find the first derivative \( F'(x) \), which tells us where the slopes of the tangent to the function are zero, known as critical points. These points often indicate potential maxima, minima, or inflection points.
In our step-by-step solution:

• We calculated \( F'(x) = ( (1-x^2)^2 - 2(1-x^2) - 3) \cdot (-2x) \).
• By solving \( F'(x) = 0 \), we found the critical points within the domain \(-1 \leq x \leq 1\).
• We evaluated \( F'(x) \) on intervals between these critical points to determine where the function is increasing or decreasing.

Understanding where a function increases or decreases helps in constructing a more accurate graph and predicting the function's behavior.

Integral Calculus

Integral calculus involves finding the function whose derivative is the given function, essentially reversing the process of differentiation. The definite integral represents the accumulated sum of areas under a curve over an interval. In this task:

• We set up \( F(x) = \int_{0}^{1-x^2} (t^2 - 2t - 3) \, dt \), leading to our exploration of area accumulation from \( t = 0 \) up to \( t = 1-x^2 \).
• The definite integral provides a new function \( F \) which stores information about the area beneath \( f(t) \) from its starting point to the upper limit defined by \( u(x) \).

By analyzing this integral function, we effectively translate the geometric interpretation of area under the curve into a functional understanding, allowing us to apply calculus techniques to study rates of change and accumulation.

Critical Points

Critical points of a function are where its derivative is zero or undefined, providing potential locations of relative maxima, minima, or inflection points, crucial for graphing and understanding the dynamics of a function. To find these:

• First, we solve \( F'(x) = 0 \) to locate critical points. These points mark where the function changes its increasing or decreasing trends.
• We also compute the second derivative \( F''(x) \) to conduct the second derivative test, helping to confirm if these critical points are indeed maxima or minima.
• Points where \( F''(x) = 0 \) indicate potential inflection points, where concavity changes.

Evaluating these gives a complete view of the function's overall structure by indicating changes in slope and curvature, essential for creating accurate sketches and analysis.