Problem 92 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

POPULATION The population density is \(f(x, y)=1,000 y^{2} e^{-0.01 x}\) people per square mile at each point \((x, y)\) within the region \(R\) bounded by the parabola \(x=y^{2}\) and the vertical line \(x=4\). Find the total population in the region \(R\).

Step-by-Step Solution

Answer

Evaluate the integrals step-by-step using proper methods.

Step 1: Understand the problem

We need to find the total population in the region bounded by the parabola given by the equation \(x = y^2\) and the vertical line \(x = 4\). The population density function is given as \(f(x, y) = 1,000 y^2 e^{-0.01 x}\)

Step 2: Set up the double integral

To find the total population, we will integrate the population density function over the region \(R\). This can be done by setting up a double integral. The limits for \(y\) will be from \(-2\) to \(2\) (since \(x = y^2\) reaches \(4\) when \(y = \pm 2\)), and the limits for \(x\) will be from \(y^2\) to \(4\). The integral to evaluate is: total population = \(\int_{-2}^{2}\int_{y^2}^{4} 1,000 y^2 e^{-0.01x} \, dx \, dy\)

Step 3: Integrate with respect to \(x\)

Evaluate the inner integral with respect to \(x\) first. Let \(\int_{y^2}^{4} 1,000 y^2 e^{-0.01x} \, dx\)Rewrite the integrand as follows: \(\int_{y^2}^{4} 1,000 y^2 e^{-0.01x} \, dx\)= \(1,000 y^2 \int_{y^2}^{4} e^{-0.01x} \, dx\)Now integrate: The antiderivative of \(e^{-0.01x}\) is \(-100 e^{-0.01x}\)So we get: \(1,000 y^2 [-100 e^{-0.01x}]_{y^2}^{4}\)= \(1,000 y^2 [-100 e^{-0.01(4)} + 100 e^{-0.01(y^2)}]\)= \(1,000 y^2 [-100 e^{-0.04} + 100 e^{-0.01 y^2}]\)= \(1,000 y^2 \cdot 100 (e^{-0.01 y^2} - e^{-0.04})\)

Step 4: Simplify the result

Simplify the result obtained from the inner integral to get: \(100,000 y^2 (e^{-0.01 y^2} - e^{-0.04})\)

Step 5: Integrate with respect to \(y\)

continue with the following steps - substitution methods and evaluating common integrals

Key Concepts

Population Density Function

In this exercise, you are provided with a population density function defined as \(f(x, y) = 1,000 y^2 e^{-0.01 x} \). Such a function tells us the number of people per square mile at any point \(x, y\) within a given region. To fully understand the density function, break it down into parts: the term \(1,000 y^2\) represents how the density varies with \(y\), and the term \(e^{-0.01 x}\) shows how the density changes with \(x\). Notice that the function uses exponential decay, indicating the population density decreases as \(x\) increases.

Integration Bounds

The region of integration is bounded by the curves \(x = y^2\) and the vertical line \(x = 4\). To determine the bounds for our double integral, consider the following:

• For \(x\), it ranges from \(y^2\) to 4. This is because for a fixed \(y\), \(x\) starts at the parabola \(x=y^2\) and extends to the line \(x=4\).
• For \(y\), it ranges from -2 to 2. This range is derived from solving \(x=y^2\) when \(x=4\) which gives \(y=\pm2\).

These bounds set the limits for our integral, creating an area where we can calculate the population density.

Antiderivative Calculation

To find the total population, we need to evaluate a double integral. This involves calculating the antiderivative of the inner function. Starting with the inner integral, \[\int_{y^2}^{4} 1,000 y^2 e^{-0.01x} \, dx\requirementintegrating with respect to \(x\). Factor out constants: \[1,000 y^2 \int_{y^2}^{4} e^{-0.01x} \, dx.\] The integral of \(e^{-0.01 x}\) is \(-100 e^{-0.01x}\),\] and evaluating from \(y^2\) to 4 gives \[(1,000 y^2)[-100 e^{-0.01 (4)} + 100 e^{-0.01 (y^2)}]\] Simplifying, we get: \[100,000 y^2 (e^{-0.01y^2} - e^{-0.04})\] as the result of the inner integral.

Region Bounded by Curves

The region of interest is bounded by the parabola \(x = y^2\) and the vertical line \(x = 4\). Geometrically, this forms a region that resembles a symmetric area on either side of the \(y\)-axis. By visualizing the area, we see that the bounds for the double integral are derived from these curves, making it crucial:

• To recognize where each curve intersects or defines a limit for \(x\) or \(y\).
• To ensure these bounds are correctly used when setting up the integral.

This spatial understanding aids in properly solving integral problems involving areas bounded by curves.