Problem 25 - University Calculus: Early Transcendentals

Question

Integrate \(f\) over the given region. Quadrilateral \(f(x, y)=x / y\) over the region in the first quadrant bounded by the lines \(y=x, y=2 x, x=1,\) and \(x=2\)

Step-by-Step Solution

Answer

The integral over the region is \(\frac{3}{2} \ln 2\).

Step 1: Visualize the Region

To understand the region, consider the lines: \(y=x\) and \(y=2x\), which are sloped lines in the first quadrant; \(x=1\) and \(x=2\), which are vertical lines. The region bounded by these lines is a quadrilateral in the first quadrant.

Step 2: Determine the Limits of Integration

Using the visualization, see that the limits for \(x\) will be from \(x=1\) to \(x=2\). For a given \(x\) in this interval, \(y\) will range from \(y=x\) (lower line) to \(y=2x\) (upper line). The integration will be performed over this bounded region.

Step 3: Set Up the Double Integral

The function to integrate is \(f(x, y) = \frac{x}{y}\). The integral is set up as: \[\int_{x=1}^{2} \int_{y=x}^{2x} \frac{x}{y} \, dy \, dx\]. This is because for each \(x\) between 1 and 2, \(y\) ranges from \(x\) to \(2x\).

Step 4: Integrate with Respect to y

Perform the inner integral first: \(\int_{y=x}^{2x} \frac{x}{y} \, dy\). The antiderivative of \(\frac{x}{y}\) with respect to \(y\) is \(x \ln|y|\). Substitute the limits to get:\[x \ln|2x| - x \ln|x| = x \ln \left(\frac{2x}{x}\right) = x \ln 2\].

Step 5: Integrate with Respect to x

Now integrate the result of the first integral over \(x\) from 1 to 2: \[\int_{x=1}^{2} x \ln 2 \, dx\]. This simplifies to \(\ln 2 \int_{x=1}^{2} x \, dx\). The antiderivative of \(x\) is \(\frac{x^2}{2}\). Evaluating from 1 to 2 gives:\[\ln 2 \left(\frac{2^2}{2} - \frac{1^2}{2}\right) = \ln 2 \left(2 - \frac{1}{2}\right)\].This equals \(\ln 2 \cdot \frac{3}{2}\).

Step 6: Simplify the Final Result

The final result from the previous step is \(\frac{3}{2} \ln 2\). This represents the integral of the function \(f(x, y)=\frac{x}{y}\) over the specified quadrilateral region.

Key Concepts

Integration Techniques

A double integral is a powerful tool in calculus that allows us to calculate the volume under a surface in a specific region. When we approach integrating a function like \(f(x, y) = \frac{x}{y}\), we need to break the process into manageable steps. We start by performing an inner integration followed by an outer integration.

• Inner integration: Involves integrating with respect to one variable, while treating the other variable as a constant. For \(\int_{y=x}^{2x} \frac{x}{y} \, dy\), we integrate \(\frac{x}{y}\) with respect to \(y\).
• Outer integration: After resolving the inner integral, we perform integration with respect to the other variable, which in our case is \(x\).

The antiderivative skills are crucial here. For example, knowing that the antiderivative of \(\frac{x}{y}\) with respect to \(y\) is \(x \ln|y|\) allows us to integrate effectively. Remember, double integrals can be quite intricate, but by breaking them into parts, they become more approachable.

Region of Integration

Understanding the region of integration in a double integral problem is akin to setting a stage where the integration dance takes place. The specified region gives boundaries to our calculations. Visualizing or sketching this helps immensely in setting up the integral correctly

For this exercise, we're dealing with a quadrilateral defined by the lines \(y = x\), \(y = 2x\), \(x = 1\), and \(x = 2\). Once visualized, these lines enclose an area in the first quadrant of the Cartesian plane:

• Diagonal lines: These create slopes where \(y = x\) is steeper than \(y = 2x\).
• Vertical lines: These are straightforward as they establish fixed \(x\)-coordinates of 1 and 2.

The defined region in a coordinate plane is crucial, as the limits of integration for \(x\) and \(y\) are derived from these boundaries. Without correctly identifying this region, our integration could consider extraneous or incorrect parts of the plane.

Limits of Integration

Setting the limits of integration is like drawing the borders for our integration process. In the mentioned exercise, the first task is to identify the correct range for \(x\) and \(y\) based on the region we are working with.

• \(x\)-limits: These denote constant bounds for the outer integral. In our case, \(x\) ranges from 1 to 2, effectively creating a boundary from where to start and end the calculation along the \(x\)-axis.
• \(y\)-limits: These are functions of \(x\) since they vary with \(x\). For any fixed \(x\), \(y\) varies from \(x\) that represents the lower bound, reaching up to \(2x\) as the upper bound in the inner integral.

So for our specific function and region, choosing limits like \(\int_{1}^{2} \int_{x}^{2x} \frac{x}{y} \, dy \, dx\) properly reflects the constraints of our region. These constraints ensure that every part of the region is accounted for in the integration process. Adjusting these limits according to the lines bounding the region is key to obtaining an accurate result.