Problem 39 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

a. Examining integration with respect to x and y Integrating with respect to x: \(\int_{0}^{1}(y - y^2) dx = \int_{0}^{1}(x - x^2) dx\) Integrating with respect to y: \(\int_{0}^{1}(\sqrt{x} - x) dy = \int_{0}^{1}(\sqrt{y} - y) dy\) We can see that both integrals are possible calculations for the area of the region bounded by the functions \(y=x\) and \(x=y^2\). Therefore, the statement is false. b. Calculating the integral: \(\int_{0}^{\pi / 2}(\cos x-\sin x) d x = [-\sin x - \cos x]_{0}^{\pi / 2} = (-1 - 0) - (0 - 1) = 1\) Since the integral evaluates to 1, the statement is true. c. Comparing the integrals: \(\int_{0}^{1}(x-x^{2}) d x = [\frac{1}{2}x^2 - \frac{1}{3}x^3]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\) \(\int_{0}^{1}(\sqrt{y}-y) dy = [\frac{2}{3}y^{\frac{3}{2}} - \frac{1}{2}y^2]_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}\) Both integrals evaluate to \(\frac{1}{6}\), so the statement is true.

Step 1: a. Examining the statement and finding the integration bounds for both x and y

First, we need to determine the intersection points of the functions \(y=x\) and \(x=y^2\). To do this, we substitute \(y=x\) into the second equation: \(x=(x)^2\). Solving for \(x\) gives us \(x=0\) and \(x=1\). Now we can find the corresponding \(y\) values: at \(x=0\), \(y=0^2=0\) and at \(x=1\), \(y=1^2=1\). Therefore, the region is bounded between \((0,0)\) and \((1,1)\), and its area can be found by integrating the difference of the two functions. Let's consider both options: integrating with respect to \(x\) and integrating with respect to \(y\).

Step 2: b. Examining the statement, finding the function difference, and calculating the integral

We are given two functions, \(y=\sin x\) and \(y=\cos x\). We need to find the region between these functions on the interval \([0,\pi/2]\). To do this, we need to determine which function has a higher value on the given interval. Since \(\cos x\) starts at 1 and goes to 0, and \(\sin x\) starts at 0 and goes to 1 on the interval \([0,\pi/2]\), the difference between the functions will be \(\cos x - \sin x\). The statement claims that the area of the region between these functions is given by the integral \(\int_{0}^{\pi / 2}(\cos x-\sin x) d x\). Let's calculate this integral.

Step 3: c. Examining the statement and comparing the integrals

We are given two integrals and asked whether they are equal: \(\int_{0}^{1}(x-x^{2}) d x\) and \(\int_{0}^{1}(\sqrt{y}-y) dy\). To determine this, let's evaluate both integrals and compare their values.