Problem 40 - Calculus

Question

Let \(G(x)=\int_{0}^{x} \sin t d t\) (a) Find \(G(0)\) and \(G(2 \pi)\). (b) Let \(y=G(x)\). Apply the First Fundamental Theorem of Calculus to obtain \(d y / d x=G^{\prime}(x)=\sin x\). Solve the differential equation \(d y / d x=\sin x\). (c) Find the solution to this differential equation that satisfies \(y=G(0)\) when \(x=0\) (d) Show that \(\int_{0}^{\pi} \sin x d x=2\).

Step-by-Step Solution

Answer

(a) \(G(0) = 0\), \(G(2\pi) = 0\). (b) \(y = -\cos x + C\). (c) \(y = -\cos x + 1\). (d) \(\int_{0}^{\pi} \sin x \, dx = 2\).

Step 1: Calculating G(0)

To find \(G(0)\), substitute \(x = 0\) in the integral: \[ G(0) = \int_{0}^{0} \sin t \, dt. \] Since the limits of integration are the same, the integral evaluates to 0. Therefore, \(G(0) = 0\).

Step 2: Calculating G(2π)

To calculate \(G(2\pi)\), substitute \(x = 2\pi\) in the integral: \[ G(2\pi) = \int_{0}^{2\pi} \sin t \, dt. \] Evaluate the integral: \[ G(2\pi) = [-\cos t]_{0}^{2\pi} = (-\cos(2\pi)) - (-\cos(0)) = -1 + 1 = 0. \] Thus, \(G(2\pi) = 0\).

Step 3: Finding dy/dx Using Fundamental Theorem of Calculus

By the First Fundamental Theorem of Calculus, \(\frac{dy}{dx} = G'(x) = \sin x\). Thus, the derivative of \(y = G(x)\) with respect to \(x\) is \(\sin x\).

Step 4: Solving the Differential Equation

Given the differential equation \(\frac{dy}{dx} = \sin x\), integrate both sides to solve for \(y\): \[ y = \int \sin x \, dx = -\cos x + C, \] where \(C\) is the integration constant.

Step 5: Finding the Particular Solution

Since \(y = G(0)\) when \(x = 0\), and we know \(G(0) = 0\), substitute into \(y = -\cos x + C\): \[ 0 = -\cos(0) + C = -1 + C, \] solving for \(C\), we get \(C = 1\). Hence, the particular solution is \(y = -\cos x + 1\).

Step 6: Evaluate Another Integral

To show \(\int_{0}^{\pi} \sin x \, dx = 2\), evaluate the integral: \[ \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = (-\cos(\pi)) - (-\cos(0)) = 1 - (-1) = 2. \] Therefore, the evaluated integral confirms the result.

Key Concepts

Integration

Integration is a mathematical process that helps us find the total accumulation of a quantity. Think of it like adding up a series of increments to get a whole. In the context of the exercise, we are working with definite integrals, which give us the total change of a function over an interval. In our case, the integral \( G(x) = \int_{0}^{x} \sin t \, dt \) computes the area under the sine wave from 0 to \( x \).
This integral is fundamental for understanding the accumulation of change given by the sine function. In step-based exercises, integration is used to solve both parts \( (a) \) and \( (d) \). For example:

• Step 1 finds \( G(0) \) by integrating with limits from 0 to 0, yielding an accumulated area of 0.
• Step 6 evaluates \( \int_{0}^{\pi} \sin x \, dx = 2 \), thus solidifying the theoretical aspect of integration by visualization of area under a curve from 0 to \( \pi \).

Through integration, we move from understanding local behavior (what a function is doing at a point) to global behavior (how the function affects an entire interval). This makes integration a powerful tool in calculus.

Differential Equations

Differential equations are equations involving derivatives, and they play a crucial role in mathematical modeling. They can define systems that change over time. In this exercise:
We start with the differential equation \( \frac{dy}{dx} = \sin x \). This equation tells us how the function \( y \) changes with respect to \( x \).
The First Fundamental Theorem of Calculus states that if you have a function like our \( G(x) \), its derivative is equal to the inside function of the integral, in this case, \( \sin x \).
Solving this differential equation involves integrating \( \sin x \), giving \( y = -\cos x + C \). Here are a few key points:

• This solution includes a constant \( C \) because indefinite integration can add a constant.
• The particular solution is found using initial conditions, like \( G(0) = 0 \), to solve for \( C \).
• In our solution, substituting gives \( C = 1 \), resulting in \( y = -\cos x + 1 \).

Understanding differential equations provides insight into how continuous functions behave, allowing predictions and descriptions of physical phenomena across various disciplines.

Trigonometric Functions

Trigonometric functions are key in modeling periodic phenomena, such as waves and oscillations. They form the basis for understanding complex relationships in both mathematics and physics. In this exercise, both the sine and cosine functions are prominently featured.
The integral \( \int \sin x \, dx \) leads to an antiderivative \( -\cos x \). This connection highlights how derivatives and integrals work together, and how trigonometric identities simplify calculus problems.
Consider the following elements when dealing with trigonometric functions:

• The sine function is continuous and periodic, repeating every \( 2\pi \).
• This repetition is used in evaluating the integral \( \int_{0}^{2\pi} \sin x \, dx = 0 \), as the area over one complete sine wave cycle adds up to zero.
• The cosine function is the derivative of \( \sin x \), demonstrating the cyclic nature of these functions' derivatives and antiderivatives.

Trigonometric functions' intrinsic patterns and behaviors underpin much of calculus, unraveling the dynamics of rotating wheels, oscillating springs, and even the flow of electrical currents.