Problem 10 - Thomas Calculus

Question

In Exercises \(1-10,\) sketch the region of integration and evaluate the integral. $$ \int_{1}^{4} \int_{0}^{\sqrt{x}} \frac{3}{2} e^{y / \sqrt{x}} d y d x $$

Step-by-Step Solution

Answer

The value of the integral is \(7(e - 1)\).

Step 1: Identify the Limits of Integration

We have the double integral \(\int_{1}^{4} \int_{0}^{\sqrt{x}} \frac{3}{2} e^{y / \sqrt{x}} d y d x\). The limits for \(y\) indicate that \(y\) ranges from 0 to \(\sqrt{x}\), and the limits for \(x\) indicate that \(x\) ranges from 1 to 4.

Step 2: Sketch the Region of Integration

The region of integration is defined by \(0 \leq y \leq \sqrt{x}\) and \(1 \leq x \leq 4\). This region can be visualized by plotting \(y = \sqrt{x}\) and analyzing the rectangle formed from \(x = 1\) to \(x = 4\) on the horizontal axis. The curve \(y = \sqrt{x}\) and the lines \(y = 0\) (the x-axis) bound the region vertically.

Step 3: Evaluate the Inner Integral

Evaluate the inner integral \(\int_{0}^{\sqrt{x}} \frac{3}{2} e^{y / \sqrt{x}} d y\). Let \(u = \frac{y}{\sqrt{x}}\), which gives \(du = \frac{1}{\sqrt{x}} dy\) or \(dy = \sqrt{x} \, du\). Changing the limits of \(y\) from 0 to \(\sqrt{x}\) results in \(u\) limits from 0 to 1. The integral becomes:\[\int_{0}^{1} \frac{3}{2} e^u (\sqrt{x} \, du) = \int_{0}^{1} \frac{3\sqrt{x}}{2} e^u \, du\]Evaluating, we get:\[\frac{3\sqrt{x}}{2} \int_{0}^{1} e^u \, du = \frac{3\sqrt{x}}{2} [e^u]_0^1 = \frac{3\sqrt{x}}{2} (e - 1)\].

Step 4: Evaluate the Outer Integral

With the result of the inner integral \(\frac{3\sqrt{x}}{2} (e - 1)\), substitute it into the outer integral:\[\int_{1}^{4} \frac{3\sqrt{x}}{2} (e - 1) \, dx\]Factor out the constant \(\frac{3(e - 1)}{2}\):\[\frac{3(e-1)}{2} \int_{1}^{4} \sqrt{x} \, dx\].Evaluate the integral of \(\sqrt{x}\):\[\int_{1}^{4} \sqrt{x} \, dx = \left[\frac{2}{3} x^{3/2}\right]_{1}^{4} = \frac{2}{3} [(4^{3/2}) - (1^{3/2})] = \frac{2}{3}[8 - 1] = \frac{14}{3}\].

Step 5: Combine the Results

Now, combine the result of the outer integral with the factored-out constant:\[\frac{3(e-1)}{2} \cdot \frac{14}{3} = \frac{14(e-1)}{2}\]. Simplify:\[\frac{14(e-1)}{2} = 7(e-1)\].

Step 6: Final Answer

The evaluated integral is \(7(e-1)\). This is the result of the requested double integral.

Key Concepts

Region of Integration

In the world of double integrals, the "region of integration" is all about determining the specific area over which the integration takes place. For our problem, the bounds have been set by the inequality \(0 \leq y \leq \sqrt{x}\) and \(1 \leq x \leq 4\). Here's how you can visualize this:

• The inequality \(1 \leq x \leq 4\) determines we are operating on the x-plane between these numbers. Essentially, imagine you're cutting a slice on this plane ranging from 1 to 4.
• On the vertical axis, \(y\) ranges from 0 up to \(\sqrt{x}\), which forms a curve, instead of a straight line. This is because as \(x\) increases, \(\sqrt{x}\) also increases, creating a shape similar to the top half of a sideways parabola.

Together, these constraints form a sort of curved sliver or a wedge-like region on the plane. Getting a clear grasp on this region is crucial, as it helps set the stage for accurately evaluating the integral.

Integral Calculus

Integral calculus, the foundation of our journey here, deals with accumulation and sums over continuous spaces. This is accomplished using integrals, which are the opposites of derivatives. Here's how they work in this context:

• We calculate the integral of a function to determine the area under its curve – this visualizes doing a sum across an infinite number of tiny strips beneath the curve.
• In our case, we're looking to find the accumulated total not just in one direction, but across a plane. This is where double integrals come into play.

With double integrals, we break down the original area defined by our region into smaller and simpler components. We then calculate an integral for one dimension and integrate again over the second dimension, thus adding up these areas to solve the problem. The result gives insights into how broad concepts of change and accumulation can be observed beyond single dimensions.

Change of Variables

In the process of solving our double integral, a major technique used is the change of variables. This approach is akin to adjusting the lens of your camera to bring the subject into clearer view. Here's why it's valuable:

• By substituting \(u = \frac{y}{\sqrt{x}}\), we simplify the integration process. This means we're changing variable representations to make the math easier.
• This substitution adjusts the problem into a more manageable form, transforming complicated integral expressions into simpler ones that are easier to evaluate.

With the change of variables, not only does the maths become more approachable, but it also allows us to deftly shift the bounds of integration as needed. Thus, this technique serves as an essential tool in integral calculus for tackling complex integrations.

Limits of Integration

Defining the boundaries of our integrals is where limits of integration play their part. They are the edges that define the slice of space over which we’re integrating. Here’s how they come into play:

• Limits for \(y\) are from 0 to \(\sqrt{x}\). So, for each fixed \(x\), \(y\) starts at the base of the area (like the x-axis) and runs up to the curve \(\sqrt{x}\).
• For \(x\), these limits are between 1 and 4, signaling that integration occurs horizontally across this span before proceeding vertically.

These bounds are key to solving the integral accurately because they ensure every part of the region is accounted for without overlap or omission. By understanding the limits of integration, students can confidently tackle problems requiring keen attention to how integrals sum up areas.