Problem 25 - Thomas Calculus

Question

In Exercises \(25-28,\) 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 evaluates to \(\frac{3\ln(2)}{2}\).

Step 1: Sketch the Region

Begin by sketching the region bounded by the lines \(y = x\), \(y = 2x\), \(x = 1\), and \(x = 2\) in the first quadrant. This will help you visualize the intersection points, which will be useful for setting up integrals.

Step 2: Determine Intersection Points

Identify the intersection points of the lines.- The lines \(y = x\) and \(x = 1\) intersect at \((1,1)\).- The lines \(y = x\) and \(x = 2\) intersect at \((2,2)\).- The lines \(y = 2x\) and \(x = 1\) intersect at \((1,2)\).- The lines \(y = 2x\) and \(x = 2\) intersect at \((2,4)\).

Step 3: Set Up the Integral Limits for y

Determine the range for \(y\) which varies between \(y = x\) and \(y = 2x\). For each fixed \(x\) between 1 and 2, \(y\) starts at \(x\) and goes to \(2x\).

Step 4: Define the Double Integral

The integral to be evaluated is:\[ \int_{1}^{2} \int_{x}^{2x} \frac{x}{y} \, dy \, dx \]This sets \(y\) limits from \(x\) to \(2x\) for a given \(x\), and \(x\) limits from 1 to 2.

Step 5: Integrate with Respect to y

Evaluate the inner integral with respect to \(y\):\[ \int_{x}^{2x} \frac{x}{y} \, dy \]The antiderivative of \(\frac{1}{y}\) is \(\ln|y|\), so:\[ x \left[ \ln|y| \right]_{y=x}^{y=2x} = x (\ln(2x) - \ln(x)) = x \ln(2) \]

Step 6: Integrate with Respect to x

Now evaluate the outer integral:\[ \int_{1}^{2} x \ln(2) \, dx \]This can be factored to \(\ln(2) \cdot \int_{1}^{2} x \, dx \), where:\[ \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]Thus, the final result is:\( \ln(2) \cdot \frac{3}{2} = \frac{3\ln(2)}{2} \)

Step 7: State the Final Answer

The solution to the integral of \(f(x, y) = \frac{x}{y}\) over the given region is \(\frac{3\ln(2)}{2}\).

Key Concepts

Integration Limits

When solving double integrals, determining the integration limits is crucial. The integration limits define the region over which the function will be integrated. In this exercise, the region of integration is bounded by specific lines in the first quadrant:

• \(y = x\)
• \(y = 2x\)
• \(x = 1\)
• \(x = 2\)

These boundaries guide us to establish the integration limits for the variable \(y\) and \(x\).
For any fixed \(x\) between 1 and 2, \(y\) changes from \(x\) (given by the line \(y = x\)) to \(2x\) (from the line \(y = 2x\)). Consequently, the integration limit for \(y\) is set between \(x\) and \(2x\). Meanwhile, the integration limit for \(x\) is constant from 1 to 2. Understanding these limits will help you define the complete domain for your double integral.

Antiderivative

In calculus, finding the antiderivative involves determining the original function that was differentiated to produce the given function. For integrals, it's necessary to compute the antiderivative of the integrand.
In our exercise, the initial function to integrate in the inner integral is \(\frac{x}{y}\). To evaluate this with respect to \(y\), identify the antiderivative of \(\frac{1}{y}\), which is \(\ln|y|\).
Therefore, the antiderivative for the inner integral with respect to \(y\) is expressed as:

• x \([ \ln|y| ]_{y=x}^{y=2x}\)

Upon simplification, this becomes

• x (\ln(2x) - \ln(x)) = x \ln(2)

Keep in mind, understanding antiderivatives forms the basis for resolving many types of integration problems, including those involving double integrals.

Region of Integration

The region of integration refers to the specific section of the plane over which the double integral is computed. It defines where both integrations occur.

• In this exercise, the region of interest is the quadrilateral determined by intersecting lines in the first quadrant.
• It is crucial to visualize this region to correctly establish the limits of integration and ensure that the integral accurately represents the area under study.

By sketching or analyzing the lines that describe this area—\(y = x\), \(y = 2x\), \(x = 1\), and \(x = 2\)—we can ascertain where each variable changes across its range. For \(x\), it ranges from 1 to 2; for \(y\), it varies depending on the line equations, moving between \(x\) and \(2x\).
Understanding the region of integration ensures clarity in defining the scope of your integral calculations.

Definite Integral

The concept of a definite integral involves determining the area under a curve within specified limits. In double integrals, this translates to finding the volume under a surface across a defined region. Let's dive into this further.

• First, evaluate the inner integral with respect to \(y\), producing a result dependent on \(x\): \(x \ln(2)\)
• Then, resolve the outer integral with respect to \(x\) from 1 to 2, which leads to solving \(\ln(2) \cdot \int_{1}^{2} x \, dx\).

The outer integral evaluates as follows:

• \(\int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \)

Thus, the overall computation is \( \frac{3}{2} \cdot \ln(2) = \frac{3\ln(2)}{2} \).
Understanding this process, you can see how definite integrals help quantify areas or volumes in multivariable calculus.