Problem 68 - Calculus: Early Transcendentals

Question

If \( \displaystyle f(x) = \int^{\sin x}_0 \sqrt{1 + t^2} \, dt \) and \( \displaystyle g(y) = \int^y_3 f(x) \, dx \), find \( g''(\pi/6) \).

Step-by-Step Solution

Answer

\( g''(\pi/6) = \frac{\sqrt{5} \cdot \sqrt{3}}{4} \).

Step 1: Understand the Relationship

First, look at the function definitions. We have two functions: \( f(x) = \int_0^{\sin x} \sqrt{1 + t^2} \, dt \) and \( g(y) = \int_3^y f(x) \, dx \). This tells us that \( f(x) \) is the integral of a function in terms of a dummy variable \( t \), where the upper limit involves \( \sin x \), and \( g(y) \) is an integral of \( f(x) \) from a constant 3 to \( y \).

Step 2: Differentiate f(x) using the Leibniz Rule

To find \( f'(x) \), use the Leibniz Rule for differentiation under the integral sign. If \( F(x) = \int_{a(x)}^{b(x)} h(t) \, dt \), then \( F'(x) = h(b(x)) \cdot b'(x) - h(a(x)) \cdot a'(x) \). Here \( a(x) = 0 \) and \( b(x) = \sin x \). Thus, \( f'(x) = \sqrt{1 + (\sin x)^2} \cdot \cos x \) because \( \frac{d}{dx}(\sin x) = \cos x \).

Step 3: Differentiate g(y) once

The first derivative \( g'(y) \) involves differentiating the integral \( \int_3^y f(x) \, dx \), which is straightforward by the Fundamental Theorem of Calculus: \( g'(y) = f(y) \).

Step 4: Differentiate g(y) twice

Next, find \( g''(y) \) by differentiating \( g'(y) = f(y) \) with respect to \( y \). We know \( g'(y) = f(y) \), so differentiating once more gives \( g''(y) = f'(y) \).

Step 5: Compute f'(y) at \( y = \frac{\pi}{6} \)

Plug \( y = \frac{\pi}{6} \) into the expression for \( f'(x) \). From Step 2, \( f'(x) = \sqrt{1 + (\sin x)^2} \cdot \cos x \). Since \( \sin(\pi/6) = 1/2 \) and \( \cos(\pi/6) = \sqrt{3}/2 \), substitute these values to get: \( f'\left(\frac{\pi}{6}\right) = \sqrt{1 + (1/2)^2} \cdot \sqrt{3}/2 = \sqrt{1.25} \cdot \sqrt{3}/2 \). Compute it to find the answer.

Key Concepts

Integral Calculus

Integral Calculus is a fundamental part of calculus concerned with the concept of integration. In integration, we find the total accumulation of quantities, often represented as the area under a curve.
There are two main types of integrals:

• Definite Integral: Represents the area under a curve between two points on a graph. It's expressed as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
• Indefinite Integral: Represents a general form of antiderivatives, expressed as \( \int f(x) \, dx + C \), where \( C \) is the constant of integration.

In the exercise, we see integrals used in the functions \( f(x) = \int_0^{\sin x} \sqrt{1 + t^2} \, dt \) and \( g(y) = \int_3^y f(x) \, dx \).
Both require us to apply techniques of evaluating definite integrals. Understanding the boundaries and limits of these integrals helps solve problems related to area and accumulation of quantities.

Leibniz Rule

The Leibniz Rule, also known as the differentiation under the integral sign, is a powerful tool for solving integrals where limits are functions of the derivative variable. It offers a way to differentiate an integral with respect to a parameter, such as \( x \) in our case for \( f(x) \).
If you have the function \( F(x) = \int_{a(x)}^{b(x)} h(t) \, dt \), according to the Leibniz Rule, its derivative is given by:\[F'(x) = h(b(x)) \cdot b'(x) - h(a(x)) \cdot a'(x)\]This rule was essential to find \( f'(x) \) in the exercise:

• Here \( a(x) = 0 \) (a constant), so its derivative is zero.
• \( b(x) = \sin x \), making its derivative \( \cos x \).
• The function \( h(t) = \sqrt{1 + t^2} \) evaluated at \( b(x) \) yields \( \sqrt{1 + (\sin x)^2} \).

Plugging these into the formula, we derive \( f'(x) = \sqrt{1 + (\sin x)^2} \cdot \cos x \). This step is crucial for later finding the derivative \( f'(y) \) needed to obtain \( g''(\pi/6) \).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, providing a way to evaluate definite integrals. It is composed of two main parts:

• First part: If a function \( f \) is continuous on \([a, b]\) and \( F \) is its antiderivative, then:\[\int_a^b f(x) \, dx = F(b) - F(a)\]
• Second part: If \( F \) is defined as \( F(x) = \int_c^x f(t) \, dt \), then \( F'(x) = f(x) \).

This theorem is crucial for understanding how \( g'(y) = f(y) \) in the exercise, where \( g(y) = \int_3^y f(x) \, dx \). By differentiating \( g(y) \) with respect to \( y \), the result directly gives \( f(y) \). This straightforward differentiation using the fundamental theorem simplifies the problem significantly.

Differentiation

Differentiation is a fundamental concept in calculus that deals with the rate at which a function changes. It's the process of finding a derivative, which measures the function's sensitivity to change in its variable. Derivatives have practical applications across fields like physics, engineering, and economics.
In the context of the exercise, differentiation allows us to handle the transition from \( g'(y) = f(y) \) to \( g''(y) = f'(y) \). This is done by applying further differentiation to find the second derivative.
For the given problem,

• First, we used differentiation to derive \( f'(x) \) using the Leibniz Rule.
• Then, we found \( g''(y) \) by differentiating \( g'(y) = f(y) \), leading us to \( g''(y) = f'(y) \).

Finally, inserting \( y = \pi/6 \) into the equation \( f'(y) = \sqrt{1 + (\sin y)^2} \cdot \cos y \) helped compute \( g''(\pi/6) \). Differentiation here proved to be essential in moving from one derivative to another to achieve the solution.