Problem 150 - Calculus Volume 1

Question

In the following exercises, use the Fundamental Theorem of Calculus, Part \(1,\) to find each derivative. $$ \frac{d}{d x} \int_{3}^{x} \sqrt{9-y^{2}} d y $$

Step-by-Step Solution

Answer

The derivative is \( \sqrt{9-x^2} \).

Step 1: Identify the Fundamental Theorem of Calculus, Part 1

The Fundamental Theorem of Calculus, Part 1 states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( \frac{dF}{dx} = f(x) \). This theorem allows us to find the derivative of an integral where the upper limit of integration is a variable.

Step 2: Apply the Theorem to the Given Function

In this problem, the integral is \( \int_{3}^{x} \sqrt{9-y^{2}} \, dy \). According to the theorem, the derivative of this integral with respect to \( x \) is simply the integrand evaluated at the upper limit \( x \).

Step 3: Differentiate the Integral

By applying the Fundamental Theorem of Calculus, Part 1, we find that \( \frac{d}{dx} \int_{3}^{x} \sqrt{9-y^2} \, dy = \sqrt{9-x^2} \). Here, we replace \( y \) with \( x \) in the integrand.

Key Concepts

Derivative of an Integral

The concept of finding the derivative of an integral might initially sound complex, but it's simplified significantly through the Fundamental Theorem of Calculus, Part 1. This theorem provides the groundwork that links differentiation and integration, two core operations in calculus.

• According to the theorem, if you have a function defined by an integral, where the upper limit is a variable, then the derivative of this function is simply the integrand evaluated at this variable upper limit.
• In simpler terms, for a function defined as \( F(x) = \int_{a}^{x} f(t) \, dt \), the derivative \( \frac{dF}{dx} \) is \( f(x) \).

This principle is highly powerful because it allows you to differentiate a seemingly complex integral expression quickly and efficiently. You don't need to integrate the function first to then differentiate it.
Instead, you apply this direct rule, which significantly reduces the complexity. The key is recognizing that the upper limit of integration is variable and using it directly in the integrand for differentiation.

Upper Limit of Integration

The upper limit of integration plays a crucial role when applying the Fundamental Theorem of Calculus. It is important to understand that the upper limit in an integral impacts how you apply the theorem.

• In the exercise provided, the upper limit is a variable \( x \), which modifies the approach to finding the derivative.
• When taking the derivative of the integral \( \int_{3}^{x} \sqrt{9 - y^{2}} \, dy \), the upper limit \( x \) tells us where to "plug in" this variable into the integrand.

The transition from \( y \) in the integrand to \( x \) demonstrates how the variable upper limit allows us to evaluate the function at this point.
This is a concept that, once understood, provides a shortcut to finding derivatives without the extensive need for manual integration. It allows the theorem's powerful simplicity to be fully leveraged in calculus problem-solving.

Calculus Problem-Solving

Calculus problem-solving often involves a strategic approach. Recognizing how core concepts like the Fundamental Theorem of Calculus apply can simplify the solution process significantly.

• In tackling integration and differentiation problems, identifying applicable theorems and rules can open up simpler pathways to the solution.
• The exercise demonstrated how understanding the relationship between an integral and its derivative can provide a direct route to solving for the derivative of an integral, by evaluating the integrand at the variable upper limit of integration.

By mastering these strategies, calculus problems that at first glance seem complex can often be broken down into their basic components.
This approach not only aids in problem-solving but also enhances the intuitive grasp of calculus, making it easier to tackle more challenging problems with confidence.