Question

When converted to an iterated integral, the following double integrals are easier to evaluate in one order than the other. Find the best order and evaluate the integral. $$\iint_{R} \frac{x}{(1+x y)^{2}} d A ; R=\\{(x, y): 0 \leq x \leq 4,1 \leq y \leq 2\\}$$

Short answer

Question: Evaluate the double integral of the function \(\frac{x}{(1+xy)^2}\) over the region R, where R is defined by \(0 \leq x \leq 4, 1 \leq y \leq 2\). Answer: \(\frac{1}{4}\ln(\frac{20}{9})\)

Step-by-step solution

Integrate with respect to x

First, we will integrate the function with respect to x. To do this, we will use substitution. Let \(u = 1 + xy\), so \(\frac{du}{dx} = y\). Rearranging for dx, we have \(dx = \frac{du}{y}\). The new limits of integration for x would have to be adjusted according to the substitution. When \(x = 0\), \(u = 1 + 0\cdot y = 1\). When \(x = 4\), \(u = 1 + 4y\). Substituting back into the integral, we have: $$\int_{1}^{2} \int_{1}^{1 + 4y} \frac{x}{u^{2}} \frac{du}{y} dy$$ Now we need to replace x in the integral. From our substitution, \(u = 1 + xy\) or \(x = \frac{u - 1}{y}\). Plugging this in, we have: $$\int_{1}^{2} \int_{1}^{1 + 4y} \frac{u - 1}{u^{2} \cdot y} du\, dy$$ Now integrate with respect to u: $$\int_{1}^{2} -\frac{1}{y} \int_{1}^{1 + 4y} \frac{u - 1}{u^2} du\, dy$$

Integrate with respect to u

Integrating the function with respect to u, we get: $$\int_{1}^{2} -\frac{1}{y} \left[ -\frac{1}{u}\right]_{1}^{1 + 4y} dy$$ Substitute back the limits of integration: $$\int_{1}^{2} -\frac{1}{y} \left(-\frac{1}{1 + 4y} + \frac{1}{1}\right) dy$$ We are integrating a relatively simpler function with respect to y now.

Integrate with respect to y

Lastly, integrate the function with respect to y: $$-\int_{1}^{2} \left(\frac{1}{y} - \frac{1}{(1 + 4y)y}\right) dy$$ Split the integral into two and then integrate: $$-\left[\int_{1}^{2} \frac{1}{y} dy - \int_{1}^{2} \frac{1}{(1 + 4y)y} dy\right]$$ $$=(-\ln |y|-\frac{1}{4}\ln |1+4y| )|_{1}^{2}$$

Evaluate the integral

Now, evaluate the integral at its limits of integration: $$=-\left[\ln(2)-\frac{1}{4}\ln(9) - \ln(1)+\frac{1}{4}\ln(5)\right]=\frac{1}{4}\ln(\frac{20}{9})$$ Hence, the value of the double integral is \(\frac{1}{4}\ln(\frac{20}{9})\).

Key concepts

Double Integrals

Double integrals are a powerful tool in calculus used to compute the volume under a surface or the accumulated value of a function over a given region. They involve integrating a function of two variables, typically denoted as \( x \) and \( y \), over a rectangular or more complex region. Double integrals take the form \( \iint_R f(x, y) \, dA \), where \( R \) is the region of integration and \( dA \) is the differential area element.

• Double integrals can be solved by integrating in a specific order, either \( x \)-first or \( y \)-first.
• The order of integration can significantly affect the difficulty of the problem, prompting choosing the "easier" direction first.

To solve double integrals, we typically transform them into iterated integrals, where the integration is performed step-by-step over each variable, holding one constant while integrating the other.

Substitution Method

The substitution method in integration is akin to the chain rule in differentiation. It transforms a complex integral into a simpler one by changing variables. In double integrals, substitution can help handle complicated bounds or integrands. For instance, if the integrand or limits involve complex expressions, substitution like \( u = 1 + xy \) can simplify calculations. Here's how it helps:

• It redefines the integration variable to make the integral more manageable.
• The limits of integration need adjustment according to the new variable.

By substituting, our original double integral becomes more straightforward with easier-to-handle functions and limits. Remember to back-substitute your original variables after integration when needed.

Integration by Parts

Integration by parts is a method used to integrate products of functions and is based on the product rule of differentiation. For the functions \( u \) and \( v \), it is expressed by the formula:\[ \int u \, dv = uv - \int v \, du \]This technique simplifies the integration by reducing complex products into manageable chunks. In the context of double integrals, integration by parts can help in the inner integral if the integrand is a product of easily differentiable functions. However, it's more commonly used in single integrals or when explicitly needed for step simplification, but it is good to know how it functions generally in calculus, as it could be handy depending on the integral's complexity.

Limits of Integration

Determining limits of integration is crucial for setting up an iterated integral correctly. These limits specify the "boundaries" of the region over which we integrate. In rectangular regions, these can be constants, whereas in more complex-shaped regions, they might be functions themselves.In our exercise, the region \( R \) is given by \( 0 \leq x \leq 4 \) and \( 1 \leq y \leq 2 \). Adjusting these bounds according to any substitution made (like \( u = 1 + xy \)) ensures that the integral covers the desired area correctly.

• The outer integral's limits (like for \( y \)) are evaluated first, setting the stage for the inner bounds.
• When using substitutions, remember to adjust these limits to match your new variables.

Precise determination of these limits is essential to ensure the accuracy of the integration results.