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Magazine Chapter 6: Problem 57
For each exercise: a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. BIOMEDICAL: Drug Dosage A drug taken orally is absorbed into the bloodstream at the rate of \(t e^{-0.5 t}\) milligrams per hour, where \(t\) is the number of hours since the drug was taken. Find the total amount of the drug absorbed during the first 5 hours.
Short Answer
Expert verified
The total drug absorbed after 5 hours is approximately 1.59 mg.
Step by step solution
01
Understand the Problem
We are given a function that represents the rate of drug absorption, \(f(t) = t e^{-0.5 t}\), in terms of time \(t\). The function describes how fast the drug is absorbed into the bloodstream (in mg/hour). We need to find the total amount of drug absorbed over the first five hours (\(0 \leq t \leq 5\)).
02
Set Up the Integral
To find the total amount of drug absorbed, we must integrate the rate function \(f(t) = t e^{-0.5 t}\) over the interval from \(t = 0\) to \(t = 5\). This requires the integral \[ \int_{0}^{5} t e^{-0.5 t} \, dt. \]
03
Apply Integration by Parts
Integration by parts is suitable here because we have a product of a polynomial and an exponential function. Let \(u = t\) and \(dv = e^{-0.5t} \, dt\). Then, \(du = dt\) and, integrating \(dv\), we find \(v = -2e^{-0.5t}\). Using integration by parts, we get: \[ \int u \, dv = uv - \int v \, du. \] Substituting, \[ \int t e^{-0.5 t} \, dt = -2t e^{-0.5t} - \int -2 e^{-0.5t} \, dt. \]
04
Solve the Remaining Integral
We integrate the remaining integral \(-2 \int e^{-0.5t} \, dt\). This requires finding the antiderivative of \(e^{-0.5t}\), which is \(-2e^{-0.5t}\). Therefore, \[ -2 \int e^{-0.5t} \, dt = 4 e^{-0.5t}. \] Thus, the entire integral becomes \[-2t e^{-0.5t} + 4 e^{-0.5t} + C.\]
05
Evaluate the Definite Integral
Now, we evaluate the definite integral from \(t = 0\) to \(t = 5\):\[ \left[-2t e^{-0.5t} + 4 e^{-0.5t}\right]_{0}^{5}. \] First, evaluate at \(t = 5\):\[ -2(5) e^{-0.5\times5} + 4 e^{-0.5\times5}. \] Then at \(t = 0\):\[ -2(0)e^{-0.5\times0} + 4e^{-0.5\times0}. \]Compute these values and find the difference to get the total drug absorbed.
06
Calculate the Result
Calculate the values:For \(t = 5\), \(-10e^{-2.5} + 4e^{-2.5}\) simplifies to \(-6e^{-2.5}\).For \(t = 0\), this simplifies to \(4\).Thus, the integral evaluates as:\[ -6e^{-2.5} + 4. \]Simplifying this gives approximately \(1.59\) mg.
07
Verify with a Graphing Calculator
Use a graphing calculator to compute the integral \(\int_{0}^{5} t e^{-0.5t} \, dt\). Cross-check the numeric result, approximately 1.59 mg, with the calculator's output to ensure consistency and accuracy.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a technique used in calculus to integrate the product of two functions. It's based on the product rule for differentiation. This method can simplify integrals that are too complex to be solved directly. In our example, we have an integral of the form \( \int u \, dv \), where we choose \( u = t \) and \( dv = e^{-0.5t} \, dt \). Integration by parts uses the formula:
- \( \int u \, dv = uv - \int v \, du \)
Definite Integral
A definite integral calculates the accumulation of quantities, such as areas under curves. It provides the exact value of a function over a specified interval, from \( a \) to \( b \). To evaluate the definite integral, we first find the antiderivative of the function.
- In our exercise, we defined an integral from \( t = 0 \) to \( t = 5 \) to calculate the total amount of drug absorbed in the given timeframe.
Integral Calculus
Integral calculus focuses on finding quantities and functions from their rates of change. It is integral to many real-world applications, including calculating motion, areas, and volumes. In medical contexts, like the drug dosage problem, integral calculus helps determine the total amount of drug absorbed over time, critical for dosing and efficacy assessments.
- This problem exemplifies the use of integral calculus in practical scenarios by providing the formula \( \int_{0}^{5} t e^{-0.5 t} \, dt \) to solve for drug dosage effectiveness.
Drug Dosage Problem
In pharmacology, understanding how drugs are absorbed in the bloodstream over time is crucial. Our example outlines a drug dosage problem, necessitating a precise calculation to determine total drug absorption.
- The absorption rate is given by \( f(t) = t e^{-0.5 t} \), characterizing how quickly the drug enters the system.
- Finding how much is absorbed over five hours requires integrating this function; indicating integral calculus's role in solving real-life pharmacokinetic equations.
Exponential Functions
Exponential functions are prevalent in mathematics, representing growth and decay processes. In our example, the drug absorption rate involves the function \( e^{-0.5 t} \), signifying an exponential decay.
- This function describes how, over time, the rate of drug absorption decreases.
- Incorporating exponential decay in integration by parts allows us to evaluate how various factors influence cumulative measurements over time, such as drug efficiency.
