Problem 39 - Advanced Engineering Mathematics

Question

Evaluate the given iterated integral by reversing the order of integration. $$ \int_{0}^{1} \int_{x}^{1} \frac{1}{1+y^{4}} d y d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{4} \ln 2 \).

Step 1: Sketch the Region of Integration

To understand the region over which we integrate, consider the integral limits. For the given integral \( \int_{0}^{1} \int_{x}^{1} \frac{1}{1+y^{4}} \, dy \, dx \), the region described is \( 0 \leq x \leq 1 \) and \( x \leq y \leq 1 \). This represents a triangular region in the xy-plane.

Step 2: Determine New Limits for Integration

Reverse the order of integration by looking at the same region. The triangle has its base along the line \( y = x \), and it spans from \( x = 0 \) to \( x = 1 \). For a fixed \( y \), \( x \) will run from \( 0 \) to \( y \). Thus, the new limits are \( 0 \leq y \leq 1 \) and \( 0 \leq x \leq y \).

Step 3: Setup the New Iterated Integral

With the new limits determined in Step 2, the iterated integral changes to \( \int_{0}^{1} \int_{0}^{y} \frac{1}{1+y^{4}} \, dx \, dy \).

Step 4: Integrate Inner Integral

Compute the integral with respect to \( x \). Here, it's a constant function with respect to \( x \). Therefore, the integration is straightforward:\[ \int_{0}^{y} \frac{1}{1+y^{4}} \, dx = \frac{x}{1+y^{4}} \Big|_{0}^{y} = \frac{y}{1+y^{4}}. \]

Step 5: Integrate Outer Integral

Now integrate with respect to \( y \): \[ \int_{0}^{1} \frac{y}{1+y^{4}} \, dy. \]We use substitution: let \( u = 1+y^{4} \), then \( du = 4y^{3} \, dy \), or \( y \, dy = \frac{1}{4} \, du \).Change of limits: when \( y = 0, \ u = 1 \), and when \( y = 1, \ u = 2 \).Thus, it becomes: \[ \frac{1}{4} \int_{1}^{2} \frac{1}{u} \, du. \]

Step 6: Evaluate Resulting Integral

The integral \( \int_{1}^{2} \frac{1}{u} \, du \) is a standard logarithmic integral, so we have:\[ \frac{1}{4} (\ln|u| \big|_{1}^{2}) = \frac{1}{4} (\ln 2 - \ln 1) = \frac{1}{4} \ln 2. \]

Step 7: Conclusion

Thus, the value of the iterated integral after reversing the order of integration is \( \frac{1}{4} \ln 2 \).

Key Concepts

Order of Integration

In iterated integrals, the order of integration refers to the sequence in which we perform the integration operation with respect to different variables. Initially, we had the iterated integral

• \( \int_{0}^{1} \int_{x}^{1} \frac{1}{1+y^{4}} \, dy \, dx \)

where the order was from \(y\) first and then \(x\). Reversing the order means switching which variable we integrate first. In our example, this required us to change the integral to be with respect to \(x\) first, then \(y\), as

• \( \int_{0}^{1} \int_{0}^{y} \frac{1}{1+y^{4}} \, dx \, dy \).

To reverse this, we examined the region of integration on the \(xy\)-plane, adjusted our integration limits accordingly, and set up the new integral with the reversed order. This often simplifies the problem-solving process and can make certain aspects of integration more manageable.

Integral Limits

The integral limits in iterated integrals define the region over which we integrate. They determine the values and scope of the variables involved. For the original iterated integral,

• The limits \( 0 \leq x \leq 1 \) and \( x \leq y \leq 1 \) define a triangular region.

This region was explored graphically to see how we can reverse the order of integration. By reversing, we redefined the region as

• \( \int_{0}^{1} \int_{0}^{y} \cdots \),

where the new limits, \( 0 \leq y \leq 1 \) and \( 0 \leq x \leq y \), describe the same area but in a way that supports integrating with respect to \(x\) first. Recognizing and appropriately adjusting these limits is crucial in solving iterated integrals.

Substitution Method

The substitution method in integration involves changing the variable of integration to simplify the integral. It often makes complex integrals more straightforward to solve. In our example, during the outer integral,

• \( \int_{0}^{1} \frac{y}{1+y^{4}} \, dy \),

we applied substitution by setting \( u = 1+y^4 \). We then found the derivative with respect to \(y\), converting \(y \, dy\) into a form involving \(du\),

• giving \( y \, dy = \frac{1}{4} \, du \).

The limits changed from \(y\) values to \(u\) values as well: when \( y = 0 \), \( u = 1 \) and when \( y = 1 \), \( u = 2 \). This transformed the integral into a simpler form:

• \( \frac{1}{4} \int_{1}^{2} \frac{1}{u} \, du \).

Such a transformation facilitated finding the solution by switching to a standard logarithmic form.