Problem 35 - Calculus Early Transcendentals

Question

Sketch each region and write an iterated integral of a continuous function \(f\) over the region. Use the order \(d x d y\). The region bounded by \(y=4-x, y=1,\) and \(x=0\)

Step-by-Step Solution

Answer

Based on the given step-by-step solution, write a short answer question. Question: Write the iterated integral of a continuous function f(x, y) over the region bounded by the equations y = 4-x, y = 1, and x = 0 using the order dxdy. Answer: \(\int_{1}^{4} \int_{0}^{3-y} f(x,y) \,dx \,dy\)

Step 1: Sketch the Region

First, let's sketch the region bounded by the three equations \(y=4-x\), \(y=1\), and \(x=0\). Begin by plotting the three lines on a coordinate plane. \(y = 4-x\) is a line with a slope of \(-1\) and a \(y\)-intercept of \(4\). \(y = 1\) is a horizontal line crossing the \(y\)-axis at \(1\). Finally, \(x = 0\) is the \(y\)-axis itself. The intersection of these lines forms a triangular region.

Step 2: Determine Intersection Points

Next, let's find the points where the lines intersect, as they will determine the limits of integration. \(y = 4-x\) and \(y = 1\) intersect when \(4-x=1\). Solving for \(x\) gives \(x = 3\). To find the \(y\)-coordinate of that point, substitute \(x=3\) into \(y=4-x\), which results in \(y=1\). Therefore, the point of intersection is \((3, 1)\). Two other intersection points are \((0, 1)\) and \((0, 4)\), which can be easily found as they lie on \(x=0\).

Step 3: Determine Limits of Integration

Now that we have all three intersection points, we can determine the limits of integration. As we are using the order \(dxdy\), we first need to find the limits for \(x\). Within the triangular region, \(x\) will vary from 0 (the line \(x=0\)) to \(3-y\) (the line \(y=4-x\)). Next, we need to find the limits for \(y\). The \(y\) values run from the minimum value of 1 (the line \(y=1\)) to the maximum value of 4 (the line \(x=0, y=4\)).

Step 4: Write the Iterated Integral

With the limits of integration determined, we can now write the iterated integral of a continuous function \(f\) over the region using the limits found in Step 3. The iterated integral will be: \(\int_{1}^{4} \int_{0}^{3-y} f(x,y) \,dx \,dy\) This is the required iterated integral of a continuous function \(f\) over the region using the order \(dxdy\).

Key Concepts

Iterated integral

To understand iterated integrals, let's start with what they are used for. An iterated integral allows you to integrate over a region in space by breaking the multi-variable integral into a series of single-variable integrals. It involves integrating a function in more than one step.
This is particularly useful when dealing with regions defined by complex boundaries. In the exercise, the iterated integral is expressed as \[ \int_{1}^{4} \int_{0}^{3-y} f(x,y) \,dx \,dy \]Here, the integral over \(x\) occurs first, followed by the integral over \(y\).

• This integral will help in finding the accumulated value of a function \(f\) over a defined region.
• The order of integration, here \(dxdy\), is essential to set up correctly as it influences the integration limits.
• Visualizing the region and understanding its boundaries helps in determining the correct limits and the order to solve the integral.

Integration limits

Setting integration limits is a key part of solving iterated integrals. They define the bounds within which you perform your integration and are crucial for solving problems accurately.
In our exercise, we overlapped two boundaries defined by equations \(y=4-x\), \(y=1\), and \(x=0\) to find the range of \(x\) and \(y\).

• For the x-direction, the region moves from \(x=0\) (the line \(x=0\)) to \(x=3-y\) (along the line \(y=4-x\)).
• For the y-direction, boundaries move from \(y=1\) to \(y=4\), defining the total vertical extent of the region.
• Identifying intersection points (like \(0,1\), \(0,4\), and \(3,1\)) is crucial for setting accurate limits.

Positioning these limits correctly ensures that your integral evaluates over the intended area, covering the entire triangular region formed by these lines.

Function evaluation

Once you have set up your iterated integral with the correct integration limits, the next step is evaluating the function \(f(x, y)\) over the region. In this case, \(f(x, y)\) is an arbitrary continuous function.
Evaluating this function implies carrying out the double integration process:

• First, integrate \(f(x, y)\) with respect to \(x\) while keeping \(y\) constant, using limits from \(x=0\) to \(x=3-y\).
• After performing this inner integral, proceed to integrate the result with respect to \(y\), using limits from \(y=1\) to \(y=4\).

This double integration process sums up the function's values across each dimension within the given bounds. The process aids in computing areas, volumes or other quantities over the specified region. Understanding how this function evaluation works in iterated integrals is crucial for applying the technique to real-world problems.