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. $$\int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} d y d x$$

Short answer

Based on the given step-by-step solution, the short answer to evaluate the double integral is: $$\int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} d y d x = \frac{1}{4}(e^8 + 1)$$

Step-by-step solution

(Step 1: Sketch the region of integration)

Let's analyze the given integral limits. The integral is given by: $$\int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} d y d x$$ The outer integral is with respect to \(x\), ranging from \(0\) to \(2\), and the inner integral is with respect to \(y\), ranging from \(0\) to \(4-x^2\). We can sketch the region of integration by finding the intersection points of the curves: \(y = 4-x^2\) and \(x = 2\). This way, we get the curve \(y = 4 - x^2\), \(x = 0\), \(x = 2\), and \(y = 0\). The region of integration is bounded by the curve \(y = 4-x^2\), \(x = 0\), \(x = 2\), and \(y = 0\). Notice that this region is a vertically-dissected area of a parabola.

(Step 2: Reverse the order of integration)

To reverse the order of integration, we'll first find the bounds for x in terms of y. We can rewrite the parabolic bounding curve as \(x^2 = 4 - y\) and then solve for \(x\): $$x = \pm \sqrt{4-y}$$ Since we look only at the positive side, we'll deal with the right half of the parabola, and \(x\) will go from \(0\) to \(\sqrt{4-y}\). For the \(y\)-bounds, we observe that it ranges from \(0\) to \(4\). Therefore, reversing the order of integration, we have: $$\int_{0}^{4} \int_{0}^{\sqrt{4-y}} \frac{x e^{2 y}}{4-y} d x d y$$

(Step 3: Evaluate the integral)

First, we'll find the inner integral with respect to \(x\). This can be done by integrating \(\frac{x e^{2 y}}{4-y}\) with respect to \(x\) and respecting the limits: $$\int_{0}^{\sqrt{4-y}} \frac{x e^{2 y}}{4-y} d x = \frac{1}{2}e^{2y}\int_{0}^{\sqrt{4-y}} x d x = \frac{1}{2}e^{2y} \left[\frac{1}{2}x^2\right]_0^{\sqrt{4-y}} = \frac{1}{4}e^{2y}(4-y)$$ Now, we evaluate the outer integral with respect to \(y\): $$\int_{0}^{4} \frac{1}{4}e^{2y}(4-y) dy = \frac{1}{4}\left[\int_{0}^{4} 4e^{2y} dy - \int_{0}^{4} ye^{2y} dy\right]$$ Evaluate each of the integrals separately: $$\int_{0}^{4} 4e^{2y} dy = 4\left[\frac{1}{2}e^{2y}\right]_0^4 = 2(e^{8}-1)$$ Now, we need to evaluate the second integral using integration by parts: $$\int_{0}^{4} ye^{2y} dy$$ Let \(u = y\) and \(dv = e^{2y} dy\). Thus, \(du = dy\) and \(v = \frac{1}{2}e^{2y}\). Integration by parts formula: $$uv\Big|_{0}^{4} - \int_{0}^{4} v du = \left(\frac{1}{2}ye^{2y} \right) \Big|_{0}^{4} - \int_{0}^{4}\frac{1}{2}e^{2y}dy$$ $$= 2e^8- \left[2\left(\frac{1}{2}e^{2y}\right)\right]_{0}^{4} = 2e^8 - (e^8 - 1)$$ Adding both integrals, we have: $$\frac{1}{4}(2(e^{8}-1) + (2e^8 -(e^8 - 1))) = \frac{1}{4}(e^8 + 1)$$ Therefore, the evaluated integral is: $$\int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} d y d x = \frac{1}{4}(e^8 + 1)$$

Key concepts

Reverse Order of Integration

Understanding the reverse order of integration can sometimes simplify an integral, especially when it cannot be easily solved in its original setup. In this context, the given double integral \(\int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} d y d x\) requires reversing the order to a more manageable form. To do this, we need to re-examine the integration limits in terms of the other variable. Originally, \(x\) varied between 0 and 2, while \(y\) was from 0 to \(4-x^2\). By changing the order, \(x\)'s limits become functions of \(y\), ranging from 0 to \(\sqrt{4-y}\), and \(y\) extends from 0 to 4. This reversal often provides a clearer understanding of the region we are dealing with in two dimensions—here, a part of a parabolic section. By re-evaluating the double integral with these new limits \(\int_{0}^{4} \int_{0}^{\sqrt{4-y}} \frac{x e^{2 y}}{4-y} d x d y\), solving becomes more intuitive. This approach can be quite helpful in tackling more complex integrals.

Integration Techniques

Integration techniques are essential for solving various types of integrals, whether single or multiple. In the present example, different techniques come into play when evaluating a double integral. Firstly, solving \(\int_{0}^{\sqrt{4-y}} \frac{x e^{2 y}}{4-y} d x\) involved basic antiderivative rules, leading to \(\frac{1}{2}x^2\) as the antiderivative of \(x\). This was calculated as

• \(\frac{1}{4}e^{2y}(4-y)\) which simplified the inner integral by substituting back the limits for \(x\).

Secondly, integration by parts was needed for the outer integral \(\int_{0}^{4} ye^{2y} dy\). This technique helps integrate functions that are products of simpler functions. Here, the formula \(\int u \, dv = uv - \int v \ du\) was key, with \(u = y\) and \(dv = e^{2y} \, dy\). Applying these steps made it possible to separate and solve complex compounds in an integral, enabling us to approach the final result patiently and accurately.

Definite Integrals

Definite integrals represent a fundamental concept in calculus, providing the accumulated value of a quantity, such as area under a curve, over a specific interval. Unlike indefinite integrals, which include a constant of integration \(C\), definite integrals result in a numerical value as seen through limits of integration. In the exercise, the definite integral \(\int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} d y d x\) represented a defined area under the given function. By reversing the order of integration, you simplify the evaluation and eventually calculate the numerical result. This solution led to the result \(\frac{1}{4}(e^8 + 1)\), illustrating how definite integrals provide a concrete value expressing area or volume in specific regions. Being aware of definite integral properties and techniques allows us to tackle complex mathematical problems systematically, directly applying concepts to find solutions effectively.