Problem 18 - Calculus: Early Transcendentals

Question

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \( \displaystyle y = \int^1_{\sin x} \sqrt{1 + t^2} \,dt \)

Step-by-Step Solution

Answer

\(-\cos x \sqrt{1 + \sin^2 x}\)

Step 1: Understand the Problem

We need to find the derivative of a function defined by an integral. The problem involves using Part 1 of the Fundamental Theorem of Calculus.

Step 2: Apply the General Formula

According to the Fundamental Theorem of Calculus, if \( F(x) = \int_{a(x)}^{b(x)} f(t) \, dt \), then \( \frac{dF}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \). Here, our function \( y(x) \) is \( \int_{\sin x}^1 \sqrt{1 + t^2} \ dt \).

Step 3: Identify the Components

Identify \( a(x) \), \( b(x) \), and \( f(t) \) in the integral \( \int_{\sin x}^1 \sqrt{1 + t^2} \ dt \). Here, \( a(x) = \sin x \), \( b(x) = 1 \), and \( f(t) = \sqrt{1 + t^2} \).

Step 4: Calculate the Derivative of the Limits

The derivative of \( b(x) = 1 \) with respect to \( x \) is 0, i.e., \( b'(x) = 0 \). The derivative of \( a(x) = \sin x \) is \( \cos x \), i.e., \( a'(x) = \cos x \).

Step 5: Apply the Fundamental Theorem of Calculus

Using the theorem, compute the derivative: \( \frac{dy}{dx} = \sqrt{1 + (1)^2} \times 0 - \sqrt{1 + (\sin x)^2} \times \cos x \). This simplifies to \( -\sqrt{1 + \sin^2 x} \cdot \cos x \).

Key Concepts

Derivative of an Integral

In calculus, finding the derivative of a function that is defined by an integral involves using the Fundamental Theorem of Calculus. This theorem connects the processes of differentiation and integration, which are central concepts in calculus.

The problem at hand involves the function \( y = \int^1_{\sin x} \sqrt{1 + t^2} \, dt \). To find the derivative \( \frac{dy}{dx} \), we use Part 1 of the Fundamental Theorem of Calculus. It states that if you have an integral of the form \( F(x) = \int_{a(x)}^{b(x)} f(t) \, dt \), the derivative is given by:

• \( \frac{dF}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \)

This formula helps in calculating the derivative when the limits of integration are functions of \( x \). Here, by identifying the components \( a(x) = \sin x \), \( b(x) = 1 \), and \( f(t) = \sqrt{1 + t^2} \), we simplify this computation.

The derivative of the function thus results from substituting these components into the fundamental formula, yielding \( \frac{dy}{dx} = -\sqrt{1 + \sin^2 x} \cdot \cos x \).

Calculus

Calculus is a vital branch of mathematics that focuses on how things change. It provides tools for understanding the motion and behavior of quantities. Calculus is primarily concerned with two main operations: differentiation and integration.

• Differentiation involves finding the rate at which a quantity changes, commonly known as the derivative. It answers questions like, "How fast is something changing?"
• Integration is about accumulation. It calculates the total, considering all parts, often relating to area under a curve.

The Fundamental Theorem of Calculus forms a powerful bridge between these two concepts. This theorem simplifies many calculations by showing that differentiation and integration are inverse processes. This exercise and solution illustrate its usage to find derivatives of integrals within specific limits, which often appear in physics and engineering contexts.

Understanding this interrelation allows us to tackle complex problems and find precise solutions across different fields of study.

Integration Limits

Integration limits refer to the boundaries over which integration occurs. In definite integrals, these limits are crucial, representing the starting and ending points of the integral calculation. The limits can be constants or functions of another variable, as in our exercise with the function \( y = \int^1_{\sin x} \sqrt{1 + t^2} \, dt \).

In our exercise, the integration limits are:\

• Lower limit: \( a(x) = \sin x \)
• Upper limit: \( b(x) = 1 \)

These limits need careful consideration, especially when they are functions like \( \sin x \). They affect the computation of the derivative since the derivative of \( a(x) \) or \( b(x) \) needs to be determined.

Notably, when the limits are functions of \( x \), we apply Part 1 of the Fundamental Theorem of Calculus to address the change in the integral's value due to changes in \( x \). This explains why the calculation involves subtracting the effect of the lower limit's derivative when finding \( \frac{dy}{dx} \). It's crucial to ensure proper function evaluation as per the theorem for accurate results.