Problem 50 - University Calculus: Early Transcendentals

Question

Sketch the region of integration, reverse the order of integration, and evaluate the integral. $$\int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} d y d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(e^8 - 1 - \frac{3}{8}\).

Step 1: Determine the Region of Integration

The given limits are \(0 \leq x \leq 2\) and \(0 \leq y \leq 4 - x^2\). To visualize this, plot the parabola \(y = 4 - x^2\) which opens downwards and intersects the \(y\)-axis at 4. The region is bounded on the left by \(x = 0\), on the bottom by \(y = 0\), and on the right by \(x = 2\), below the curve.

Step 2: Reverse the Order of Integration

To reverse the order of integration, express \(x\) as a function of \(y\). From the curve, \(x = \sqrt{4-y}\). The bounds for \(y\) are \(0 \leq y \leq 4\). For each \(y\), \(x\) ranges from \(0\) to \(\sqrt{4-y}\). Thus, the new bounds are: \(0 \leq y \leq 4\) and \(0 \leq x \leq \sqrt{4-y}\). Replace the original integral with the new order of integration: \(\int_{0}^{4} \int_{0}^{\sqrt{4-y}} \frac{x e^{2 y}}{4-y} \, dx \, dy\).

Step 3: Evaluate Inner Integral

Evaluate the inner integral with respect to \(x\): \(\int_{0}^{\sqrt{4-y}} x e^{2y} \, dx\). Since \(e^{2y}\) is a constant relative to \(x\), integrate \(x\) to get \(\frac{1}{2} x^2 e^{2y}\) evaluated from \(x = 0\) to \(x = \sqrt{4-y}\). This becomes \(\frac{1}{2} (4-y) e^{2y}\).

Step 4: Simplify the Result

The result from Step 3 simplifies to \(2 e^{2y} - \frac{1}{2} y e^{2y}\). Be careful to account for all terms in the simplification process.

Step 5: Evaluate Outer Integral

Evaluate the outer integral \(\int_{0}^{4} \left(2 e^{2y} - \frac{1}{2} y e^{2y}\right) \, dy \). We split it into two integrals: \(2 \int_{0}^{4} e^{2y} \, dy\) and \(-\frac{1}{2} \int_{0}^{4} y e^{2y} \, dy\). Use integration by parts for the second term.

Step 6: Compute the Integral \(\int e^{2y} \, dy\)

Compute \(\int e^{2y} \, dy\) as \(\frac{1}{2} e^{2y}\). So, \(2 \int_{0}^{4} e^{2y} \, dy\) evaluates to \([e^{2y}]_{0}^{4} = e^8 - 1\).

Step 7: Compute the Integral \(\int y e^{2y} \, dy\) using Integration by Parts

Let \(u = y\) and \(dv = e^{2y} \, dy\). Then \(du = dy\) and \(v = \frac{1}{2} e^{2y}\). Integration by parts gives: \(\int y e^{2y} \, dy = \frac{1}{2} y e^{2y} - \int \frac{1}{2} e^{2y} \, dy\), which further simplifies to \(\frac{1}{2} y e^{2y} - \frac{1}{4} e^{2y}\). Evaluate from 0 to 4.

Step 8: Calculate \(\int_{0}^{4} \left(y e^{2y} - \frac{1}{2} e^{2y}\right) \, dy\)

Substitute the result from Step 7: Evaluate \(\left.\left(\frac{1}{4}y e^{2y} - \frac{1}{8} e^{2y}\right) \right|_0^4\) to get \(\frac{1}{4} (4e^8) - \frac{1}{8} (e^8 - 1)\), which simplifies to \(2e^8 - \frac{1}{8}(e^8 - 1)\).

Step 9: Combine All Parts to Get the Final Result

Add the results of Steps 6 and 8: \(([e^8 - 1] - [\frac{1}{4} e^8 - \frac{1}{8}])\). Simplify the expression to find the value of the original integral.

Key Concepts

Reversing Order of Integration

Reversing the order of integration is a crucial technique when dealing with double integrals. It can simplify the integration process, especially when the region of integration is bounded by curves that make one order of integration difficult to compute. To reverse the order:

• Analyze the given limits for both variables to understand the region of integration.
• Plot the region to visualize how horizontal and vertical strips cover it.
• Express one variable in terms of the other to determine the new limits.

In the exercise, the initial integration is set up with respect to the variable \(y\), bound by the parabola \(y = 4 - x^2\). By reversing the order, we express \(x\) as a function of \(y\): \(x = \sqrt{4-y}\). Then, we set new limits for \(y\) from 0 to 4. The new integral runs from \(x = 0\) to \(x = \sqrt{4-y}\), simplifying the integration for each horizontal strip across the region.

Integration by Parts

Integration by parts is a powerful technique derived from the product rule for differentiation. It's particularly useful for functions that are products of algebraic and exponential expressions, such as those found in the exercise.For the integration of terms like \(\int y e^{2y} \, dy\), break it down as follows:

• Choose \(u\) and \(dv\) such that the resulting integral is simpler than the original: let \(u = y\) and \(dv = e^{2y} dy\).
• Compute \(du = dy\) and \(v = \frac{1}{2} e^{2y}\), then apply the formula \(\int u \, dv = uv - \int v \, du\).
• This results in \(\frac{1}{2} y e^{2y} - \frac{1}{4} e^{2y}\), an expression easier to evaluate between limits.

This process helps in transforming the integral of a product into more manageable terms, revealing simpler components that can be integrated directly or further simplified.

Regions of Integration

Understanding regions of integration is fundamental to solving double integrals. The region of integration is determined by the given limits, reflecting how the variables interact within specified boundaries.For the problem at hand:

• Define the inequality bounds: \(0 \leq x \leq 2\) and \(0 \leq y \leq 4 - x^2\) describe a specific section of the plane.
• Sketch this as a region bounded by a downward-opening parabola intersecting the \(y\)-axis at 4, and the \(x\)-axis from 0 to 2, forming a curvy triangular area.
• This sketching allows you to clearly see how both \(x\) and \(y\) are constrained, providing insight into the best order of integration for simplification purposes.

Visualizing the region helps in making logical decisions when setting integration limits and assists in the transition between different orders of integration. It ensures you're working within the correct area for evaluating the integral accurately.