Problem 38
Question
Net area from graphs The accompanying figure shows four regions bounded by the graph of \(y=x \sin x: R_{1}, R_{2}, R_{3},\) and \(R_{4},\) whose areas are \(1, \pi-1, \pi+1,\) and \(2 \pi-1,\) respectively. (We verify these results later in the text.) Use this information to evaluate the following integrals. $$\int_{0}^{3 \pi / 2} x \sin x d x$$
Step-by-Step Solution
Verified Answer
Answer: The value of the integral is \(4 - \pi\).
1Step 1: Understand the meaning of the given areas
We are provided with the areas \(R_1\), \(R_2\), \(R_3\), and \(R_4\) which can be expressed by the following integrals:
$$R_1 = \int_{0}^{\pi} x\sin{x} dx$$
$$R_2 = \int_{\pi}^{2\pi} x\sin{x} dx$$
$$R_3 = \int_{2\pi}^{9\pi/4} x\sin{x} dx$$
$$R_4 = \int_{9\pi/4}^{3\pi/2} x\sin{x} dx$$
2Step 2: Determine the signs of the given areas
We have to determine the respective signs of these areas based on their positions on the graph. The function \(y = x\sin{x}\) alternates between positive and negative values as the sine function changes its sign. We can observe the following about the function and the areas:
- \(R_1\): The function is positive from 0 to \(\pi\), so the integral representing \(R_1\) is positive.
- \(R_2\): The function is negative from \(\pi\) to \(2\pi\), so the integral representing \(R_2\) is negative.
- \(R_3\): The function is positive from \(2\pi\) to \(9\pi/4\), so the integral representing \(R_3\) is positive.
- \(R_4\): The function is negative from \(9\pi/4\) to \(3\pi/2\), so the integral representing \(R_4\) is negative.
3Step 3: Calculate the required integral
Now, we can determine the required integral, which is the sum of the four areas with their respective signs:
$$\int_{0}^{3\pi/2} x\sin{x} dx = R_1 - R_2 + R_3 - R_4$$
$$\int_{0}^{3\pi/2} x\sin{x} dx = 1 - (\pi - 1) + (\pi + 1) - (2\pi - 1)$$
4Step 4: Simplify the expression
Simplifying the expression, we get:
$$\int_{0}^{3\pi/2} x\sin{x} dx = 1 - \pi + 1 + \pi + 1 - 2\pi + 1$$
$$\int_{0}^{3\pi/2} x\sin{x} dx = 4 - \pi$$
Therefore, the integral \(\int_{0}^{3\pi/2} x\sin{x} dx\) is equal to \(4 - \pi\).
Key Concepts
Net AreaTrigonometric FunctionsGraphical AnalysisIntegration by Parts
Net Area
When dealing with definite integrals, it's crucial to understand the concept of the net area under a curve. The net area takes into account both the magnitude and the direction (whether above or below the x-axis) of the area.
The integral \( \int_{a}^{b} f(x)\,dx \) measures the net area, which infers substantive meaning about the function’s behavior over \( \left[a, b\right] \). In this problem, we have specific regions, \( R_{1}, R_{2}, R_{3},\) and \( R_{4},\) each representing integrals over different intervals.
The key takeaway is considering whether each component area adds positively to the sum (above the x-axis) or negatively (below the x-axis). This alternation is critical for properly interpreting the integral's result.
The integral \( \int_{a}^{b} f(x)\,dx \) measures the net area, which infers substantive meaning about the function’s behavior over \( \left[a, b\right] \). In this problem, we have specific regions, \( R_{1}, R_{2}, R_{3},\) and \( R_{4},\) each representing integrals over different intervals.
The key takeaway is considering whether each component area adds positively to the sum (above the x-axis) or negatively (below the x-axis). This alternation is critical for properly interpreting the integral's result.
Trigonometric Functions
The function in consideration, \( y = x \sin x,\) uses the basic trigonometric function \( \sin x,\) which is one of the essential trigonometric functions you often encounter. Trigonometric functions like sine oscillate through cycles as their angles increase.
The sine function changes sign every half-period, resulting in alternating positive and negative intervals when multiplied by \( x,\) a linearly increasing term. This principle is why the areas \( R_{1} \) through \( R_{4} \) show alternating signs.
Recognizing how and where these changes occur on the graph is a fundamental skill when analyzing trigonometric functions graphical evaluations.
The sine function changes sign every half-period, resulting in alternating positive and negative intervals when multiplied by \( x,\) a linearly increasing term. This principle is why the areas \( R_{1} \) through \( R_{4} \) show alternating signs.
Recognizing how and where these changes occur on the graph is a fundamental skill when analyzing trigonometric functions graphical evaluations.
Graphical Analysis
Graphical analysis involves interpreting the placement and behavior of functions on a set of axes. When visualizing \( y = x \sin x,\) we can see how the mixture of a linear and a trigonometric function produces a series of undulating waves.
Identifying where this function is above or below the x-axis helps us predict the areas \( R_{1}, R_{2}, R_{3}, \) and \( R_{4}.\) Note that each region shows whether the integral takes a positive or negative value based on its positioning relative to the x-axis.
Using graphical tools markedly aids in understanding complex integrals and predicting the result behavior, as it allows for visualizing positive versus negative integration intervals.
Identifying where this function is above or below the x-axis helps us predict the areas \( R_{1}, R_{2}, R_{3}, \) and \( R_{4}.\) Note that each region shows whether the integral takes a positive or negative value based on its positioning relative to the x-axis.
Using graphical tools markedly aids in understanding complex integrals and predicting the result behavior, as it allows for visualizing positive versus negative integration intervals.
Integration by Parts
Integration by parts is an essential technique in solving definite integrals, especially when dealing with equations like \( \int x \sin x\, dx.\) It is particularly effective when the function is a product of an algebraic and a trigonometric function.
The formula for integration by parts is given by:\[\int u\, dv = uv - \int v\, du\]
For this function, you typically select the algebraic part as \( u = x, \) and the differential of the trigonometric part as \( dv = \sin x\, dx.\)
Applying integration by parts in this context simplifies solving complex integrals, allowing you to break down a tough integral into more manageable components. This technique transforms complex integration into simpler sums and differences that are easier to compute.
The formula for integration by parts is given by:\[\int u\, dv = uv - \int v\, du\]
For this function, you typically select the algebraic part as \( u = x, \) and the differential of the trigonometric part as \( dv = \sin x\, dx.\)
Applying integration by parts in this context simplifies solving complex integrals, allowing you to break down a tough integral into more manageable components. This technique transforms complex integration into simpler sums and differences that are easier to compute.
Other exercises in this chapter
Problem 38
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