Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(G^{\prime}(x)\). $$ G(x)=\int_{x}^{\pi / 4}(s-2) \cot 2 s d s ; 0

Step-by-Step Solution

Answer

The derivative is \( G'(x) = (x-2) \cot(2x) \).

Step 1: Recognize the Function Type

The function \( G(x) = \int_{x}^{\pi/4} (s-2) \cot(2s) \, ds \) is a definite integral with variable limits of integration. This is an application for the Fundamental Theorem of Calculus Part 1.

Step 2: Fundamental Theorem of Calculus Application

The Fundamental Theorem of Calculus states that if \( F(t) \) is the antiderivative of \( f(t) \), then \( \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x))b'(x) - f(a(x))a'(x) \). Apply this theorem to \( G(x) \).

Step 3: Identify the Functions

Here, \( f(s) = (s-2)\cot(2s) \), \( a(x) = x \) and \( b(x) = \pi/4 \). Since \( b(x) = \pi/4 \) is constant, \( b'(x) = 0 \). For \( a(x) = x \), \( a'(x) = 1 \).

Step 4: Differentiate Using the Theorem

Using the derivative formula: \( G'(x) = (s-2)\cot(2s) \bigg|_{s=x} a'(x) = (x-2)\cot(2x) \).

Step 5: Final Expression for the Derivative

The derivative of \( G(x) \), considering all the steps is: \( G'(x) = (x-2) \cot(2x) \).

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus, representing the integral of a function over a specific interval. It calculates the net area under a curve, allowing us to evaluate the accumulated quantity over a range. In the given exercise, the definite integral is \[G(x) = \int_{x}^{\pi / 4}(s-2) \cot 2s \, ds\]This integral has specific limits of integration, from \(x\) to \(\pi/4\). These limits indicate that we are interested in the accumulated area between these two points on the curve described by the function \((s-2)\cot(2s)\). Understanding definite integrals is crucial because they form the basis for real-world applications like calculating distances, areas, and overall changes in quantities. Definite integrals also set the stage for approaching more complex calculus problems.

Variable Limits of Integration

In calculus, variable limits of integration occur when one or both limits of an integral are expressed as functions of a variable, rather than being fixed numbers. This can make integration more dynamic, as changes in the variable can affect the interval over which the function is integrated.In this exercise, the lower limit of the integral is \(a(x) = x\), a function of \(x\), while the upper limit \(b(x) = \pi/4\) remains constant. This scenario is typical when applying the Fundamental Theorem of Calculus, which helps us differentiate integrals with variable limits. By understanding this theorem, we can efficiently find the derivative of a function like \(G(x)\) under such conditions.The concept of variable limits is particularly useful in physics and engineering, where conditions change with respect to certain variables, requiring recalculation of phenomena, such as force or velocity, over new boundaries.

Calculus Differentiation

Differentiation is a process in calculus that involves finding the rate at which a function is changing at any point. It is essential for understanding how dynamic systems behave and is used extensively in fields that require detailed analytical insights, such as physics, economics, and engineering.In the context of the exercise, differentiation using the Fundamental Theorem of Calculus allows us to find \(G'(x)\). The theorem states that for a function defined as an integral with variable limits, the derivative can be determined by evaluating the integrand at the boundary defined by the variable limits. Hence, we derive\[G'(x) = (x-2) \cot(2x)\]This expression tells us how \(G(x)\) changes with \(x\), considering the integration limit and the behavior of the function \((s-2)\cot(2s)\). Differentiation transforms a static mathematical situation into a dynamic one, showing the instantaneous rate of change, which is key for modeling and predictions.