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Chapter 6: Problem 39

Determine whether the following statements are true and give an explanation or counterexample. a. The area of the region bounded by \(y=x\) and \(x=y^{2}\) can be found only by integrating with respect to \(x\) b. The area of the region between \(y=\sin x\) and \(y=\cos x\) on the interval \([0, \pi / 2]\) is \(\int_{0}^{\pi / 2}(\cos x-\sin x) d x\) c. \(\int_{0}^{1}\left(x-x^{2}\right) d x=\int_{0}^{1}(\sqrt{y}-y) d y\)

Short Answer

Expert verified
Question: Determine if the following statements are true and provide explanations or counterexamples for each case: a) The area of the region bounded by \(y=x\) and \(x=y^{2}\) can be found only by integrating with respect to \(x\). Answer: False. We can also calculate the area integrating with respect to \(y\), as all the involved curves can be expressed as \(x=f(y)\). b) The area of the region between \(y=\sin x\) and \(y=\cos x\) on the interval \([0, \pi / 2]\) is equal to \(\int_{0}^{\pi / 2}(\cos x-\sin x) d x\). Answer: False. The correct integral expression should be \(\int_{0}^{\pi/4}(\cos x-\sin x) d x + \int_{\pi/4}^{\pi/2}(\sin x-\cos x) d x\). c) If \(\int_{0}^{1}\left(x-x^{2}\right) d x=\int_{0}^{1}(\sqrt{y}-y) d y\). Answer: False. After evaluating each integral, we found that the first integral results in \(\frac{1}{6}\), while the second integral results in \(\frac{2}{3}\).

Step by step solution

01

a. Determine if the area of the region bounded by \(y=x\) and \(x=y^{2}\) can be found only by integrating with respect to \(x\)

For this problem, we firstly find the intersection points between \(y=x\) and \(x=y^{2}\). Solve the equations and find the intersection points: \(x=y^{2}\Rightarrow y=\pm\sqrt{x}\) So we obtain two parametrizations for the region, \(y=\sqrt{x}\) and \(y=-\sqrt{x}\) with \(y=x\), i.e., we have two regions: 1. Bounded by \(y=\sqrt{x}\) and \(y=x\) 2. Bounded by \(y=-\sqrt{x}\) and \(y=x\) To determine if the area can be found only by integrating with respect to \(x\), we must first see if we can calculate the area with respect to \(y\). If we can, the statement is false. To calculate the area with respect to \(y\), we will have to find the equations of the curves in terms of \(x\). For \(y=x\), we have \(x=y\). For \(y=\sqrt{x}\) and \(y=-\sqrt{x}\), we have \(x=y^{2}\). Since we can express all curves in terms of \(x=f(y)\), it is possible to calculate the area with respect to \(y\). Thus, the statement is false.
02

b. Determine if the area between \(y=\sin x\) and \(y=\cos x\) on the interval \([0, \pi / 2]\) is \(\int_{0}^{\pi / 2}(\cos x-\sin x) d x\)

To verify this statement, we need to calculate the area between the two curves. First, note that the intersection point between the curves is \((\pi/4, 1/\sqrt{2})\). Therefore, the area can be found by the following integral, considering that \(\sin x < \cos x\) within the interval \([0, \pi/4]\): \(\int_{0}^{\pi/4}(\cos x-\sin x) d x + \int_{\pi/4}^{\pi/2}(\sin x-\cos x) d x\) If this expression is equal to the provided integral, the statement is true. Let's calculate both expressions: \(\int_{0}^{\pi / 2}(\cos x-\sin x) d x = \int_{0}^{\pi/4}(\cos x-\sin x) d x + \int_{\pi/4}^{\pi/2}(\cos x-\sin x) d x\) The integrand in the second integral should have been \((\sin x-\cos x)\) in order for these expressions to be equal. Thus, the statement is false.
03

c. Determine if \(\int_{0}^{1}\left(x-x^{2}\right) d x=\int_{0}^{1}(\sqrt{y}-y) d y\)

We have the following integrals: \(\int_{0}^{1}(x-x^{2}) d x\) and \(\int_{0}^{1}(\sqrt{y}-y) d y\) Change of variables can possibly make the two integrals equal, thus making the statement true. Suppose \(x=y^{2}\). Then \(dx = 2y dy\). The integral becomes: \(\int_{0}^{1}(y^{2}-y^{4})(2y dy)\) If we evaluate both integrals and they are equal, the statement is true. Let's evaluate: \(\int_{0}^{1}(x-x^{2}) d x = [\frac{1}{2}x^{2}-\frac{1}{3}x^{3}]_{0}^{1} = \frac{1}{2}-\frac{1}{3} = \frac{1}{6}\) \(\int_{0}^{1}(y^{2}-y^{4})(2y dy) = [y^{3}-\frac{1}{3}y^{6}]_{0}^{1} = 1-\frac{1}{3} = \frac{2}{3}\) The statement is false, as the two integrals yield different results.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating with Respect to x
When you integrate with respect to x, you're essentially summing up small slices, or differentials, of an area parallel to the y-axis. Imagine slicing a region vertically and then adding up all those slices from left to right, between two specific x-values. For instance, when calculating the area between the curves of y=x and x=y^2, we would write our integral with dx to indicate a sum of vertical slices from a starting x-value to an ending x-value.

This is a powerful technique for finding areas of regions bounded by curves that are functions of x, that is, where y is given in terms of x, like y=f(x). However, it's crucial to note that if the function can also be written as x in terms of y, then integrating with respect to y is also possible. This flexibility allows you to choose the more convenient variable to integrate with based on the given functions and the nature of the region.
Integrating with Respect to y
Conversely, integrating with respect to y involves summing horizontal slices of an area. To picture this, visualize cutting a region horizontally and tallying these slices from bottom to top, over a range of y-values. This method is used effectively when the given functions are better expressed as x being a function of y, or x=g(y).

An example is the region bound between y=x (which becomes x=y when solved for x) and x=y^2. Finding the area can be done by setting up an integral with dy as the differential, summing the horizontal slices along the y-axis. This is especially useful when vertical slicing would require multiple integrals due to intersections or cusps, whereas horizontal slicing would involve a single, straightforward integral.
Definite Integral
The definite integral is a fundamental concept that captures the net area under a curve between two boundaries. You can identify it by the presence of lower and upper limits on the integral sign. It's not just about finding a function that gives the area under a curve鈥攊t's also about evaluating this function at specific values.

For example, the definite integral from 0 to \( \pi/2 \) of \(\cos x-\sin x\) dx doesn鈥檛 calculate the entire area between the sine and cosine curves, but rather the area from x=0 to x=\( \pi/2 \). Paying attention to when the curves intersect, and the regions they enclose is crucial to setting up the correct limits of integration. These boundaries (limits of integration) ensure that we're calculating the area of the specific section of interest.
Change of Variables in Integrals
The method of change of variables in integrals, often referred to as u-substitution, is a strategic tool to simplify complex integrals. It involves replacing a part of the integrand with a new variable and its differential. This technique allows us to transform a difficult integral into an easier one that we can solve with basic rules.

For instance, to verify if \(\int_{0}^{1}(x-x^{2})dx \) is equal to \(\int_{0}^{1}(\sqrt{y}-y) dy\), we attempt a change of variables. If we let \(x=y^{2}\), and thus \(dx=2ydy\), our integrals might look the same. However, after computing we find different results, indicating the equality does not hold. The method of substitution does not always lead to equivalent expressions, underscoring the need to appropriately apply it and check the outcomes meticulously.

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Most popular questions from this chapter

In Chapter \(8,\) we will encounter the harmonic \(\operatorname{sum} 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} .\) Use a left Riemann sum to approximate \(\int_{1}^{n+1} \frac{d x}{x}(\) with unit spacing between the grid points) to show that \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}>\ln (n+1) .\) Use this fact to conclude that \(\lim _{n \rightarrow \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\) does not exist.

Starting at the same time and place, Abe and Bob race, running at velocities \(u(t)=4 /(t+1) \mathrm{mi} / \mathrm{hr}\) and \(v(t)=4 e^{-t / 2} \mathrm{mi} / \mathrm{hr},\) respectively, for \(t \geq 0\) a. Who is ahead after \(t=5\) hr? After \(t=10\) hr? b. Find and graph the position functions of both runners. Which runner can run only a finite distance in an unlimited amount of time?

Find the critical points of the function \(f(x)=\sinh ^{2} x \cosh x.\)

Some species have growth rates that oscillate with an (approximately) constant period \(P\). Consider the growth rate function $$N^{\prime}(t)=r+A \sin \frac{2 \pi t}{P}$$ where \(A\) and \(r\) are constants with units of individuals/yr, and \(t\) is measured in years. A species becomes extinct if its population ever reaches 0 after \(t=0\) a. Suppose \(P=10, A=20,\) and \(r=0 .\) If the initial population is \(N(0)=10,\) does the population ever become extinct? Explain. b. Suppose \(P=10, A=20,\) and \(r=0 .\) If the initial population is \(N(0)=100,\) does the population ever become extinct? Explain. c. Suppose \(P=10, A=50,\) and \(r=5 .\) If the initial population is \(N(0)=10,\) does the population ever become extinct? Explain. d. Suppose \(P=10, A=50,\) and \(r=-5 .\) Find the initial population \(N(0)\) needed to ensure that the population never becomes extinct.

Let \(R\) be the region bounded by the curve \(y=\sqrt{x+a}(\text { with } a>0),\) the \(y\) -axis, and the \(x\) -axis. Let \(S\) be the solid generated by rotating \(R\) about the \(y\) -axis. Let \(T\) be the inscribed cone that has the same circular base as \(S\) and height \(\sqrt{a} .\) Show that volume \((S) /\) volume \((T)=\frac{8}{5}\).
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