Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

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 \(\int_{0}^{\pi} \sin x d x=2\). (e) Find all relative extrema and inflection points of \(G\) on the interval \([0,4 \pi]\). (f) Plot a graph of \(y=G(x)\) over the interval \([0,4 \pi]\).

Step-by-Step Solution

Answer

(a) G(0)=0, G(2π)=0. (b) y=-cos x+1. (c) C=1, so y=-cos x+1. (d) Integral is 2. (e) Maxima: π,3π; Minima: 2π,4π. (f) Oscillates from [0,4π]

Step 1: Evaluate G(0) and G(2π)

Since \(G(x) = \int_{0}^{x} \sin t \, dt\), we start by evaluating this at \(x = 0 \) and \(x = 2\pi\). For \(G(0)\):\[G(0) = \int_{0}^{0} \sin t \, dt = 0.\]For \(G(2\pi)\):\[G(2\pi) = \int_{0}^{2\pi} \sin t \, dt.\]The integral of \(\sin t\) from \(0\) to \(2\pi\) evaluates to zero as the positive area from \(0\) to \(\pi\) is cancelled by the negative area from \(\pi\) to \(2\pi\). This gives:\[G(2\pi) = 0.\]

Step 2: Apply the First Fundamental Theorem of Calculus

According to the First Fundamental Theorem of Calculus, if \(y = G(x)\) where \(G(x) = \int_{0}^{x} \sin t \, dt \), then\the derivative of \(G(x)\) with respect to \(x\) (denoted \(\frac{dy}{dx}\)) is the integrand \(\sin x\):\[\frac{dy}{dx} = \sin x.\]

Step 3: Solve the Differential Equation dy/dx = sin x

To solve the differential equation \(\frac{dy}{dx} = \sin x\), we integrate both sides with respect to \(x\):\[y = \int \sin x \, dx.\]The integral of \(\sin x\) is:\[y = -\cos x + C,\]where \(C\) is the constant of integration.

Step 4: Find Particular Solution y = G(0) With x = 0

We know from Step 1 that \(G(0) = 0\). Therefore, using the solution to the differential equation from Step 3, we substitute \(x = 0\) and \(y = 0\):\[0 = -\cos(0) + C \Rightarrow C = 1.\]Thus, the particular solution is:\[y = -\cos x + 1.\]

Step 5: Show \(\int_{0}^{\pi} \sin x \, dx = 2\)

Calculate the definite integral from \(0\) to \(\pi\):\[\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = (-\cos(\pi)) - (-\cos(0)) = 1 + 1 = 2.\]

Step 6: Find Critical Points for Relative Extrema

To find relative extrema, consider the derivative \(G'(x) = \sin x\). Setting \(\sin x = 0\) yields critical points at \(x = k\pi\), where \(k\) is an integer.On the interval \([0, 4\pi]\), critical points are \(x = 0, \pi, 2\pi, 3\pi, 4\pi\). Evaluate the sign of \(\sin x\) around these points to determine the relative extrema:- Relative maxima at \(x = \pi, 3\pi\) - \(x = \pi\), \(y = -\cos(\pi) + 1 = 2\) - \(x = 3\pi\), \(y = -\cos(3\pi) + 1 = 2\)- Relative minima at \(x = 2\pi, 4\pi\) - \(x = 2\pi\), \(y = -\cos(2\pi) + 1 = 0\) - \(x = 4\pi\), \(y = -\cos(4\pi) + 1 = 0\).

Step 7: Find Inflection Points

Inflection points occur where the second derivative changes sign. Start with:\[G'(x) = \sin x.\]\[G''(x) = \cos x.\]Set \(G''(x) = 0\), giving \(x = \frac{\pi}{2} + k\pi\).On \([0, 4\pi]\), inflection points are \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\).

Step 8: Plot y = G(x) over [0, 4π]

The function \(y = -\cos x + 1\) oscillates between 0 and 2 over the interval \([0, 4\pi]\). Plotting this showsthe maxima at \(x = \pi, 3\pi\) and minima at \(x = 2\pi, 4\pi\). The curve changes concavity at the inflection points.

Key Concepts

Definite Integral

The concept of a definite integral is essential in calculus. It represents the signed area under a curve of a function, between two specific points on the x-axis. In the context of the exercise, the definite integral \(G(x) = \int_{0}^{x} \sin t \, dt\) is explored over different intervals.

In part (a) of the exercise, evaluating \(G(0)\) and \(G(2\pi)\) involves computing the integral of \(\sin t\) over these boundaries. Since the sine function is symmetrical over the interval \([0, 2\pi]\), the positive area from \(0\) to \(\pi\) cancels out with the negative area from \(\pi\) to \(2\pi\). Thus, the integral from \(0\) to \(2\pi\) evaluates to zero, making \(G(2\pi) = 0\).

Understanding definite integrals helps us compute areas and solve real-world problems involving accumulations, such as distance, area, and other quantities.

Differential Equation

A differential equation involves the relationship between a function and its derivatives. In this exercise, it is crucial to apply the First Fundamental Theorem of Calculus to understand it thoroughly. This theorem tells us that if \(y = G(x)\) where \(G(x) = \int_{0}^{x} \sin t \, dt\), the derivative \(\frac{dy}{dx}\) is \(\sin x\), which is the integrand.

This leads us to the differential equation \(\frac{dy}{dx} = \sin x\). Solving it involves finding \(y\) by integrating both sides with respect to \(x\). The integration results in \(y = -\cos x + C\), where \(C\) is the constant of integration. This representation is the general solution to the differential equation. By using initial conditions, we can find particular solutions for specific problems.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. Identifying these points is key to finding relative maxima and minima. In the given exercise, to find the critical points for the function \(y = G(x)\), we set the derivative \(G'(x) = \sin x\) to zero. This occurs at \(x = k\pi\), where \(k\) is an integer.

In the interval \([0, 4\pi]\), the critical points are located at \(x = 0, \pi, 2\pi, 3\pi, 4\pi\). At these points, we evaluate \(y = -\cos x + 1\):

• Relative maxima occur at \(x = \pi\) and \(x = 3\pi\) with a value of 2.
• Relative minima occur at \(x = 2\pi\) and \(x = 4\pi\) with a value of 0.

Analyzing critical points helps in understanding the behavior of the function over a certain interval, painting a clearer picture of its overall shape.

Inflection Points

Inflection points are where a function changes its concavity. To find these, it is necessary to examine the second derivative, which indicates concavity.

For \(y = G(x)\), the first derivative is \(G'(x) = \sin x\), and the second derivative is \(G''(x) = \cos x\). Setting \(G''(x) = 0\) finds points where concavity changes, giving \(x = \frac{\pi}{2} + k\pi\) for integer \(k\).

Within the interval \([0, 4\pi]\), the inflection points are at \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\). These points mark transitions between the curve being concave up and concave down, assisting in understanding the wave-like behavior of trigonometric functions like \(\sin x\). Recognizing these points ensures one appreciates how the graph of \(y = G(x)\) elegantly shifts as \(x\) increases.