Problem 63 - Applied Calculus

Question

Fluid Absorption Runners and other athletes know that their ability to absorb water when exercising varies over time as their electrolyte levels change. Under certain circumstances, the absorption rate of water through the gastrointestinal tract will be \(f(x)=2 x e^{-\frac{1}{2} x}\) liters per hour, where \(t\) is the number of hours of exercise. Find the number of liters of water absorbed during the first 5 hours of exercise.

Step-by-Step Solution

Answer

Approximately 2.582 liters of water are absorbed in the first 5 hours.

Step 1: Understand the Function

Given the function, \(f(x) = 2x e^{-\frac{1}{2}x}\), which represents the rate of water absorption in liters per hour. \(x\) is in hours and we aim to find the total amount absorbed over 5 hours.

Step 2: Set Up the Integral

To find the total amount of water absorbed over a period, integrate the rate function, \(f(x)\), from \(x = 0\) to \(x = 5\). This can be represented as the integral \(\int_{0}^{5} 2x e^{-\frac{1}{2}x} \, dx\).

Step 3: Use Integration by Parts

Use integration by parts to solve \(\int 2x e^{-\frac{1}{2}x} \, dx\). Let \(u = 2x\) and \(dv = e^{-\frac{1}{2}x} \, dx\). Then, \(du = 2 \, dx\) and \(v = -2e^{-\frac{1}{2}x}\).

Step 4: Apply Integration by Parts Formula

Apply the formula \(\int u \, dv = uv - \int v \, du\). This results in \(-4xe^{-\frac{1}{2}x} + 4\int e^{-\frac{1}{2}x} \, dx\).

Step 5: Complete the Integral

Solve \(4\int e^{-\frac{1}{2}x} \, dx\), resulting in \(-8e^{-\frac{1}{2}x}\). So the final integrated function becomes \(-4xe^{-\frac{1}{2}x} - 8e^{-\frac{1}{2}x}\) plus constant.

Step 6: Evaluate the Definite Integral

Evaluate from 0 to 5: \([-4xe^{-\frac{1}{2}x} - 8e^{-\frac{1}{2}x}]_{0}^{5}\). Compute at \(x=5\): \(-4(5)e^{-\frac{5}{2}} - 8e^{-\frac{5}{2}}\) and at \(x=0\): \(-8e^{0}\).

Step 7: Compute and Simplify

Calculate values and simplify: \([-20e^{-\frac{5}{2}} - 8e^{-\frac{5}{2}} + 8]\). Estimate exponential terms for an approximate numerical result.

Key Concepts

Integration

Integration is a key concept in calculus, especially useful for finding areas under curves or total quantities over time. In this exercise, we use integration to determine the total amount of water absorbed by the athletes over a time period.

The absorption rate is given by a function, and integration helps us accumulate this rate over the desired hours of exercise. To start, we need to understand how an integral helps accumulate tiny changes (in this case, water absorption rates) to find a total quantity.

When we set up the integral in this scenario, \[\int_{0}^{5} 2x e^{-rac{1}{2}x} \, dx\]we interpret it as the sum of continuously changing values of water absorption from time 0 to 5 hours. This integral gives us the total liters absorbed by the gastrointestinal tract during that period.

Integration by Parts

Integration by parts is an essential technique for solving integrals that involve a product of functions. It's particularly useful when one part of the product is easily differentiable and the other is easily integrable.

In our problem, we use integration by parts because the integral \[\int 2x e^{-rac{1}{2}x} \, dx\] involves a product of the polynomial \(2x\) and an exponential function \(e^{-rac{1}{2}x}\).

The integration by parts formula, \[\int u \, dv = uv - \int v \, du\] helps transform the initial complex integral into more manageable parts. Here, we choose \(u = 2x\) so it differentiates to \(du = 2 \, dx\) and \(dv = e^{-rac{1}{2}x} \, dx\) which integrates to \(v = -2e^{-rac{1}{2}x}\). This choice simplifies our work, allowing us to break down the calculation step by step. After applying the formula, we resolve the integral into parts that are simpler to evaluate.

Definite Integrals

Definite integrals are used when you want to find the net value over a set interval rather than just finding an antiderivative. For definite integrals, the limits of integration specify where to start and stop measuring the accumulation, in this case from 0 to 5 hours.

The aim here is to evaluate the expression \([-4xe^{-rac{1}{2}x} - 8e^{-rac{1}{2}x}]_{0}^{5}\) which gives us the exact total quantity absorbed over those 5 hours.

This involves evaluating the integrated function at the upper limit (5 hours) and subtracting the value when evaluated at the lower limit (0 hours). This approach calculates the precise total liters absorbed within the specified time frame for the exercise scenario described.

Utilizing definite integrals allows us to precisely quantify the area under the absorption rate curve, offering an exact measure of water uptake essential for athletes monitoring their hydration status during exercise.