Problem 79 - Thomas Calculus

Question

In Exercises 79 and \(80,\) assume that \(f\) is continuous and \(u(x)\) is twice- differentiable. Calculate \(\frac{d}{d x} \int_{a}^{u(x)} f(t) d t\) and check your answer using a CAS.

Step-by-Step Solution

Answer

The derivative is \( f(u(x)) \cdot u'(x) \).

Step 1: Understanding the Problem

We're asked to differentiate the integral \( \int_{a}^{u(x)} f(t) \, dt \) with respect to \( x \). It involves the Fundamental Theorem of Calculus and the chain rule since \( u(x) \) is a function of \( x \).

Step 2: Applying the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(t) \) is an antiderivative of \( f(t) \), then: \( \frac{d}{dt} \int_{a}^{t} f(s) \, ds = f(t) \). We extend this to our integral to get: \( \frac{d}{dx} \int_{a}^{u(x)} f(t) \, dt = f(u(x)) \).

Step 3: Using the Chain Rule

Since \( u(x) \) is a function of \( x \), we also need to apply the chain rule. Thus, \( \frac{d}{dx} \int_{a}^{u(x)} f(t) \, dt = f(u(x)) \cdot \frac{d}{dx}[u(x)] = f(u(x)) \cdot u'(x) \).

Step 4: Combining the Results

From the previous steps, the derivative of the integral \( \int_{a}^{u(x)} f(t) \, dt \) with respect to \( x \) is given by: \( \frac{d}{dx} \int_{a}^{u(x)} f(t) \, dt = f(u(x)) \cdot u'(x) \).

Step 5: Verification using a CAS

Use a computer algebra system (CAS) to verify the result. Input the expression \( \int_{a}^{u(x)} f(t) \, dt \) and differentiate it. Check if the CAS provides \( f(u(x)) \cdot u'(x) \) as the derivative.

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus when dealing with functions composed of other functions. It helps us differentiate composite functions by relating the derivatives of the outer and inner functions. In our problem, the chain rule applies when differentiating the integral \( \int_{a}^{u(x)} f(t) \, dt \) because the upper limit of the integral, \( u(x) \), is itself a function of \( x \). Here's why the chain rule is important:

• Without the chain rule, we couldn't properly differentiate integrals with limits that are not constants.
• It ensures that we account for how changes in \( x \) affect \( u(x) \), and consequently, the value of the integral.

In our specific problem, after applying the chain rule, the differentiation becomes:\[ \frac{d}{dx} \int_{a}^{u(x)} f(t) \, dt = f(u(x)) \cdot u'(x). \]This relationship tells us the impact of \( x \) on both the integrand \( f(t) \) through \( u(x) \) and the direct derivative of \( u(x) \). The derivative \( u'(x) \) scales \( f(u(x)) \) depending on how \( u(x) \) changes with \( x \). Thus, the chain rule is vital to correctly compute the derivative of integrals with variable limits.

Differentiation Under the Integral Sign

Differentiation under the integral sign allows us to differentiate an integral where the limits or the integrand depend on a variable. This is closely tied to both the chain rule and the Fundamental Theorem of Calculus.When we have an integral like \( \int_{a}^{u(x)} f(t) \, dt \) and need to differentiate it with respect to \( x \), we apply this technique to handle the complexity that arises from \( u(x) \) being a function of \( x \). Here's how it works in practice:

• First, recognize that \( f(t) \) does not depend on \( x \) directly, but \( t \) does through the upper limit \( u(x) \).
• The derivative of the integral is the product \( f(u(x)) \cdot u'(x) \), where \( f(u(x)) \) comes from applying the Fundamental Theorem of Calculus.
• \( u'(x) \) appears due to applying the chain rule, acknowledging how \( u(x) \) changes with \( x \).

This method enables us to take the derivative of complex integrals easily, merging two powerful calculus concepts: differentiation and integration. It is particularly useful in advanced calculus and applications involving physics, engineering, and economics.

Continuous Functions

Continuous functions are a vital concept in calculus, often forming the backbone of theorems and techniques. A function \( f(t) \) is continuous if there are no breaks, jumps, or holes in its graph over its domain.Here's why continuity is crucial in calculus, particularly in our exercise:

• Continuous functions can be integrated over any interval, a property required when using the Fundamental Theorem of Calculus.
• They ensure the existence of an antiderivative, which allows us to apply differential calculus rules reliably.

In our specific example, the exercise states \( f \) is continuous. This assurance allows us to confidently differentiate the integral\( \int_{a}^{u(x)} f(t) \, dt \)by guaranteeing that there are no disruptions in \( f(t) \) that would otherwise complicate the problem. When applying the Fundamental Theorem of Calculus, continuity is essential to smoothly transition from integration to differentiation. Thus, continuous functions play a pivotal role in bridging these fundamental calculus processes.