Problem 64 - Calculus Early Transcendentals

Question

The following integrals can be evaluated only by reversing the order of integration. Sketch the region of integration, reverse the order of integration, and evaluate the integral. $$\int_{0}^{\pi} \int_{x}^{\pi} \sin y^{2} d y d x$$

Step-by-Step Solution

Answer

Based on the step by step solution provided, create a short answer: We are given the integral $$\int_{0}^{\pi} \int_{x}^{\pi} \sin y^{2} d y d x$$ and asked to reverse the order of integration and evaluate it. After sketching the region of integration and reversing the order, we get the new integral: $$\int_{0}^{\pi} \int_{0}^{y} \sin y^{2} d x d y$$ Evaluating the integral using integration by parts, we get the final form of the integral as: $$-\frac{1}{4} \int_{0}^{\pi} \sin y^{2} dy$$ The integral can be evaluated numerically, but it has no elementary function as its solution.

Step 1: Sketch the region of integration

First, let's examine the given integral. $$\int_{0}^{\pi} \int_{x}^{\pi} \sin y^{2} d y d x$$ The outer integral has limits \(0\) to \(\pi\), and the inner integral limits are from \(x\) to \(\pi\). Therefore, the region of integration is bounded by: 1. \(x = 0\) 2. \(x = \pi\) 3. \(y = x\) 4. \(y = \pi\) We can now sketch the region of integration. [asy] import graph; unitsize(1cm); real f(real x) {return x;} draw(graph(f,0,pi), Arrows); fill((0,0)--(pi,0)--(pi,pi)--(0,pi)--cycle, gray(0.8)); draw((pi,0)--(0,0)--(0,pi), linewidth(1.5)); draw((pi,0)--(pi,pi), linestyle(dashed), linewidth(1.5)); label("\(x=0\)", (0,pi), NE, fontsize(10)); label("\(x=\pi\)", (pi,0), SE, fontsize(10)); label("\(y=x\)", (pi,pi), W, fontsize(10)); label("\(y=\pi\)", (0,pi), NE, fontsize(10)); [/asy] The shaded area represents the region of integration, which we will use to reverse the order of integration.

Step 2: Reverse the order of integration

Now that we've sketched the region of integration, let's reverse the order of integration. The new limits for the outer integral are determined by the vertical lines \(x = 0\) and \(x = \pi\). The new limits for the inner integral are determined by the horizontal lines \(y = x\) and \(y = \pi\). After reversing the order, the integral becomes: $$\int_{0}^{\pi} \int_{0}^{y} \sin y^{2} d x d y$$

Step 3: Evaluate the inner integral

Now, let's evaluate the inner integral with respect to \(x\): $$\int_{0}^{y} \sin y^{2} d x$$ Since \(\sin y^{2}\) does not contain \(x\), it can be treated as a constant while integrating with respect to \(x\). The integral becomes: $$\left[\sin y^{2} x\right]_{0}^{y}$$ Now, substitute the limits of integration: $$\sin y^{2} y - \sin y^{2} (0) = \sin y^{2} y$$

Step 4: Evaluate the outer integral

Now, the new integral is: $$\int_{0}^{\pi} \sin y^{2} y d y$$ We can use integration by parts to evaluate this integral. Recall that the integration by parts formula is: $$\int u dv = uv - \int v du$$ Let's choose \(u = y\) and \(dv = \sin y^{2} dy\). Then \(du = dy\) and \(v = -\frac{1}{2} \cos y^{2}\). Applying integration by parts, we have: $$-\frac{1}{2} \int_{0}^{\pi} y \cos y^{2} dy = -\frac{1}{2} \left[ -\frac{1}{2} y \sin (y^2)\right]_{0}^{\pi} - \frac{1}{2} \int_{0}^{\pi} -\frac{1}{2} \sin y^{2} dy$$ Since \(\sin (\pi^2) = \sin(0) = 0\), the boundary term evaluates to zero: $$-\frac{1}{2} \left[ -\frac{1}{2} \sin (y^2) y\right]_{0}^{\pi} = 0$$ So, the integral becomes: $$-\frac{1}{4} \int_{0}^{\pi} \sin y^{2} dy$$ Unfortunately, there is no elementary function for the integral \(\int_{0}^{\pi} \sin y^{2} dy\), but it can be computed numerically. For this exercise, we'll keep it as an indefinite integral. The final answer for the given integral is: $$-\frac{1}{4} \int_{0}^{\pi} \sin y^{2} dy$$