Question

Sketch the region of integration and evaluate the following integrals as they are written. $$\int_{0}^{\ln 2} \int_{e^{y}}^{2} \frac{y}{x} d x d y$$

Short answer

The result of evaluating the given double integral is: $$ (\ln 2)\frac{(\ln 2)^2}{2} - \frac{(\ln 2)^3}{3}$$

Step-by-step solution

Sketch the region of integration

Let's first plot the two lines \(x = e^y\) and \(x = 2\) in the \(xy\)-plane. Our region of integration is bounded by these two lines and the lines \(y = 0\) and \(y = \ln 2\). To sketch the region, follow these steps: 1. Plot the line \(x = e^y\), which is the inverse of the exponential function. This will be an increasing function of \(y\). 2. Sketch the vertical line \(x = 2\). 3. Mark the points where these lines intersect the \(y\)-axis: \((0, 0)\), \((2, 0)\), and \((2, \ln 2)\). 4. Connect these points, forming a region bounded by the curves and the lines \(y = 0\) and \(y = \ln 2\). This region is the area over which we'll integrate the function.

Evaluate the inner integral

We'll first evaluate the inner integral \(\int_{e^{y}}^{2} \frac{y}{x} d x\), treating \(y\) as constant. $$\int_{e^{y}}^{2} \frac{y}{x} d x = y\int_{e^{y}}^{2} \frac{1}{x} d x$$ Now, we can integrate with respect to \(x\). The antiderivative of \(\frac{1}{x}\) is \(\ln x\). Therefore, $$y\int_{e^{y}}^{2} \frac{1}{x} d x = y [\ln x]_{e^y}^{2} = y(\ln 2 - \ln (e^y)) = y(\ln 2 - y)$$

Evaluate the outer integral

Now, we can evaluate the outer integral with respect to \(y\): $$\int_{0}^{\ln 2} y (\ln 2 - y) dy$$ To integrate this, we can use the distributive property to expand and then integrate term by term: $$\int_{0}^{\ln 2} (\ln 2y - y^2) dy = (\ln 2) \int_{0}^{\ln 2} y dy - \int_{0}^{\ln 2} y^2 dy$$ The antiderivative of \(y\) is \(\frac{y^2}{2}\), and the antiderivative of \(y^2\) is \(\frac{y^3}{3}\). Applying this to the integrals above: $$ (\ln 2) \left[\frac{y^2}{2}\right]_{0}^{\ln 2} - \left[\frac{y^3}{3}\right]_{0}^{\ln 2} = (\ln 2) \left(\frac{(\ln 2)^2}{2} - 0\right) - \left(\frac{(\ln 2)^3}{3} - 0\right)$$

Simplify the result

Finally, we can simplify the expression: $$ (\ln 2)\frac{(\ln 2)^2}{2} - \frac{(\ln 2)^3}{3}$$ This is the final result and the evaluation of the given double integral.

Key concepts

Region of Integration

When dealing with double integrals, understanding the region of integration is crucial. The region of integration refers to the area in a plane over which you're integrating. For our specific example, we sketch the region given by the inequalities defined by our limits.

In this case, after sketching, the region is found between the functions defined by the lines \(x = e^y\) and \(x = 2\). These lines define our boundaries in the \(x\)-direction. Meanwhile, we integrate vertically between \(y = 0\) and \(y = \ln 2\).

The process of sketching involves identifying these bounds and drawing the associated lines and curves on a graph to visualize the area we are looking at. This helps us ensure our calculations correspond to the correct region.

• Line \(x = e^y\) increases with \(y\).
• The vertical line \(x = 2\) is our upper bound in the \(x\)-direction.
• Horizontal bounds \(y = 0\) and \(y = \ln 2\) set limits for \(y\).

Inner Integral

The inner integral involves integrating with respect to one variable while treating the other variables as constants. For the given function \(\int_{e^{y}}^{2} \frac{y}{x} dx\), we treat \(y\) as a constant and integrate over \(x\).

To simplify, notice \(y\) is indeed a constant in this operation, effectively pulling it outside the integral. Thus, we can rewrite it as \(y \int_{e^{y}}^{2} \frac{1}{x} dx\).

This setup focuses on evaluating the integratory path through the region by slicing it horizontally between the bounds \(x = e^y\) and \(x = 2\). Once solved, this yields a function purely in terms of \(y\), which we then use in our outer integral.

• Treat \(y\) as constant during integration.
• Simplifies to \(y [\ln x]_{e^y}^2\).
• Result: \(y (\ln 2 - y)\).

Outer Integral

With the result from the inner integral, our next step is focusing on the outer integral. Here, we integrate the expression resulting from the inner integral over the specified range for \(y\): \(\int_{0}^{\ln 2} y (\ln 2 - y) dy\).

This task involves using the distributive property for simplification. We break down the polynomial \(y (\ln 2 - y)\) into simpler components that can then be independently integrated.

This eventually resolves into the sum/difference of terms each integrated across the region's \(y\) direction, giving us the whole value of the double integral.

• Break the expression as \(\int_{0}^{\ln 2} (\ln 2y - y^2) dy\).
• Integrate each term separately.
• Apply limits to find exact contribution.

Exponential Function

Exponential functions, such as \(e^y\), appear frequently in integrations regarding real-world growth models and transformations. In our problem, the exponential function plays a crucial role in bounding the region of integration.

We encounter it as \(x = e^y\), impacting how we determine the region in the \(x\)-direction. The behavior of \(e^y\) is such that it grows rapidly as \(y\) increases, which affects where it intersects with other constraints.

This exponential relationship is what underpins one side of our integration limits, centered around its increasing nature as a natural logarithm ceiling for \(y\). Understanding how exponential functions affect integration is vital, making them a staple concept in calculus.

• Defines one of the bounds for \(x\).
• Important for calculating intersections and regions.
• Represent rapid growth commonly in calculations.