Problem 76 - 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. $$ a=0, \quad u(x)=x^{2}, \quad f(x)=\sqrt{1-x^{2}} $$

Step-by-Step Solution

Answer

Domain of F: \([0, 1]\); Increasing on \(0, 1\); decreasing nowhere. Use CAS for precision.

Step 1: Determine the Domain of F

To find the domain of \(F(x) = \int_{0}^{x^2} \sqrt{1-t^2} \, dt\), we need to consider the definition of the integral's limits. The function \(\sqrt{1-t^2}\) is defined where \(1-t^2 \geq 0\), meaning \(-1 \leq t \leq 1\). The limits of integration are from 0 to \(x^2\), so we need \(-1 \leq x^2 \leq 1\). This constraint simplifies to \(0 \leq x \leq 1\). Hence, the domain of \(F\) is \([0, 1]\).

Step 2: Calculate the First Derivative F'(x)

Using the Fundamental Theorem of Calculus and the chain rule, we have \(F'(x) = \frac{d}{dx}\left(\int_0^{x^2} \sqrt{1-t^2} \, dt\right) = \sqrt{1-(x^2)^2} \cdot 2x = 2x\sqrt{1-x^4}\). To find the zeros of \(F'(x)\), set \(2x\sqrt{1-x^4} = 0\). The solution is \(x = 0\). To determine where \(F\) is increasing or decreasing, consider the sign of \(F'(x)\): \(F'(x) > 0\) for \(0 < x < 1\), meaning \(F\) is increasing there, and \(F'(x) = 0\) at \(x = 0\).

Step 3: Calculate the Second Derivative F''(x)

To find the second derivative, differentiate \(F'(x) = 2x\sqrt{1-x^4}\). Use the product rule and chain rule \((u = 2x, v = \sqrt{1-x^4})\): \(F''(x) = 2\sqrt{1-x^4} + 2x\left(\frac{d}{dx}\sqrt{1-x^4}\right) = 2\sqrt{1-x^4} - \left(\frac{4x^5}{\sqrt{1-x^4}}\right)\). Simplify to: \[ F''(x) = \frac{2(1-x^4) - 4x^6}{\sqrt{1-x^4}} \]. To find zeros of \(F''(x)\), solve \([2(1-x^4) - 4x^6] = 0\), which is complex but electronically computable; you should numerically approximate in a CAS tool to find extrema and possible inflection points.

Step 4: Sketch and Graph F(x)

Knowing the derivative behaviors, we sketch \(y = F(x)\) over \([0, 1]\). Since \(F(x)\) is increasing within its domain and has a potential inflection point determined from \(F''(x)\), draw \(y = F(x)\) starting at \(x = 0\) increasing upwards. After sketching, use a computer algebra system (CAS) like Desmos or GeoGebra to plot an accurate graph of \(F(x)\) and validate your hand sketch.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a crucial link between differentiation and integration. It tells us that if a function is continuous on an interval, then it can be integrated, and the integral can give back the original function through differentiation. This theorem is divided into two parts.

• The **first part** states that if you have a continuous function and you integrate it over an interval, the antiderivative of that function at the upper bound minus the antiderivative at the lower bound will give you the integral.
• The **second part** involves differentiation and states that if you take the derivative of an integral of a function, you get back that function. Essentially, it says that if you have a function that is the integral of another function, differentiating it basically "undoes" the integration process, leaving you with the original function.

For instance, in our problem, the function \( F(x) = \int_{0}^{x^2} \sqrt{1-t^2} dt \) uses this concept. By applying the Fundamental Theorem of Calculus, we can differentiate under the integral symbol and find \( F'(x) \).

Chain Rule

The Chain Rule is a key concept in calculus for understanding how to differentiate composite functions. It's a method for finding the derivative of a function that is composed of two or more functions. Simply put, if you have a function within a function, the Chain Rule helps you find the derivative.

• When using the Chain Rule, you multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function.
• In terms of formula, if you have a composition \( f(g(x)) \), the derivative \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \).

In the original problem, the Chain Rule is used in conjunction with the Fundamental Theorem of Calculus to find \( F'(x) \). We differentiate \( \int_{0}^{x^2} \sqrt{1-t^2} dt \) to get \( 2x\sqrt{1-x^4} \). The \( 2x \) comes from the derivative of the inner function \( x^2 \).

Derivative of Integrals

The derivative of an integral is a special application of the Fundamental Theorem of Calculus. It involves functions defined by definite integrals with variable limits. This method helps in evaluating how the accumulated value of a function changes as its integration bounds vary.
For example, if \( F(x) = \int_{a}^{u(x)} f(t) dt \), then by the Fundamental Theorem of Calculus Part 2, the derivative \( F'(x) = f(u(x)) \cdot u'(x) \). Here, \( f(u(x)) \) is evaluated at the upper limit function \( u(x) \), and \( u'(x) \) is the derivative of the upper limit function with respect to \( x \).
Applying this to our exercise, we have the integral from \( 0 \) to \( x^2 \) of \( \sqrt{1-t^2} \). Using this concept, we differentiate to find that \( F'(x) = 2x\sqrt{1-x^4} \), where \( \sqrt{1-x^4} \) is \( f(x^2) \) and \( 2x \) is the derivative of \( x^2 \).

Domain of Functions

Understanding the domain of a function is essential in calculus. The domain is the set of all possible input values (typically \( x \) values) for which the function is defined. Restrictions on domain often come from domains of square roots, logarithms, denominators of fractions, etc.

• For polynomial functions, the domain is usually all real numbers.
• For square root functions, the expression inside the square root must be non-negative.

In the given problem, the function \( \sqrt{1-t^2} \) is under a square root and hence must be non-negative. This restriction gives \(-1 \leq t \leq 1\). Since \( F(x) = \int_{0}^{x^2} \sqrt{1-t^2} dt \), we must ensure \( 0 \leq x^2 \leq 1 \), leading us to conclude that the domain of \( F(x) \) is \( [0, 1] \). This reflects the need to ensure that the integral's bounds are meaningful and valid within the function's calculations.