Final answer:
To sketch the region of the given integral, create a trapezoid bounded by x=1 (left), y=3 (top), y=0 (bottom), and x=e^3 (right) on the x-y plane. The lower bound of x depends on y and is represented by the curve y=ln(x).
Step-by-step explanation:
To answer the student's question regarding the integral, we need to sketch the region of integration within the x-y plane for the given integral:
∫ from 0 to 3, ∫ from e^r to e^3 of (x / ln(x)) dx dy.
To start, let's consider the limits of integration. The outer integral has limits from 0 to 3, which defines the range for y. The inner integral has variable limits from e^r to e^3 for x, where y=r. This indicates the lower bound of the x-integration is not a constant but a variable dependent on y. To sketch the region, we will plot the curve y = ln(x) since x ranges from e^y to e^3. This curve will start at x=e (when y=1) and will go to x=e^3. As y increases from 0 to 3, the lower bound of x shifts from e^0 (which is 1) to e^3.
The region will be bounded on the left by the line x = 1 (since e^0 is 1), on the right by the line x = e^3, and on the top and bottom by y = 3 and y = 0, respectively. Thus, the region is a trapezoid on the x-y plane. To complete the sketch, one would draw this trapezoid and shade the region of integration. If the student provided graphs, the correct one would be the trapezoid with the aforementioned boundaries.