Problem 41 - Calculus Single Variable

Question

A patient with a fever has temperature at time \(t\) (measured in hours) given by $$ T(t)=99.6-t+0.8 t^{2} \text { degrees } $$ where temperature is measured in degrees Fahrenheit. Find the patient's average temperature as \(t\) ranges from 0 to \(3 .\)

Step-by-Step Solution

Answer

The patient's average temperature over the interval is 100.5 degrees Fahrenheit.

Step 1: Understand the Formula for Average Value

The average temperature is found using the formula for the average value of a function over an interval \([a, b]\). This formula is: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(t) \, dt \] where \(f(t) = T(t) = 99.6 - t + 0.8t^2\), and \(a = 0\) and \(b = 3\).

Step 2: Set Up the Integral for the Average Temperature

Substitute the function \(T(t)\) and the limits of integration into the average value formula: \[ \text{Average temperature} = \frac{1}{3-0} \int_0^3 (99.6 - t + 0.8t^2) \, dt \] This becomes \[ \frac{1}{3} \int_0^3 (99.6 - t + 0.8t^2) \, dt \].

Step 3: Calculate the Integral

Begin by integrating each term of the function \(99.6 - t + 0.8t^2\) separately: \( \int (99.6) \, dt = 99.6t \), \( \int (-t) \, dt = -\frac{t^2}{2} \), \( \int (0.8t^2) \, dt = 0.8 \cdot \frac{t^3}{3} = \frac{0.8t^3}{3} \). Combine these to get the integrated function: \[ \left[ 99.6t - \frac{t^2}{2} + \frac{0.8t^3}{3} \right]_0^3 \].

Step 4: Evaluate the Integrated Function at Bounds

Substitute the upper limit, \(t = 3\), into the integrated function:\( 99.6(3) - \frac{3^2}{2} + \frac{0.8(3)^3}{3} \).Simplify each term: \( = 298.8 - 4.5 + 7.2 = 301.5 \).Now substitute the lower limit, \(t = 0\): \( = 0 \) (since all terms in the integrated function become zero).Calculate the definite integral, which is \(301.5 - 0 = 301.5\).

Step 5: Divide by the Time Interval Length

Now, divide the value of the definite integral by 3 (the length of the time interval, from 0 to 3):\( \text{Average temperature} = \frac{301.5}{3} = 100.5 \). This is the average temperature.

Key Concepts

Understanding Integral Calculus

Integral calculus is a branch of mathematical analysis dealing with the accumulation of quantities. It focuses on integration, which is a way to sum up small quantities over an interval. This is similar to finding the area under a curve on a graph. In practical terms, integral calculus helps us solve problems related to finding areas, volumes, and other accumulative properties.
For students tackling problems in integral calculus, the main task often involves setting up and evaluating definite integrals. This allows you to measure these accumulative values over a specific range and apply them to real-world problems like the one we are discussing here.

Definite Integral and Its Application

A definite integral is a specific concept in integral calculus that calculates the integral of a function between two fixed limits. It's represented as \[ \ \int_{a}^{b} f(x) \, dx \\] Where \(a\) and \(b\) are the lower and upper bounds of integration, respectively. For our exercise, the function is the patient's temperature over a specified time interval from 0 to 3 hours.
By evaluating the definite integral of the function \(T(t) = 99.6 - t + 0.8t^2\), we find the total temperature accumulation over that interval.
This step is essential before calculating the average value of the function as it provides the necessary cumulative value that the function represents over its domain.

Calculating the Average Temperature

The average temperature in this context is simply the mean temperature of the patient over the specified time. The formula to find the average value of a function between \(a\) and \(b\) is: \[ \ \text{Average value} = \frac{1}{b-a} \int_a^b f(t) \, dt \\] Using this formula, you substitute the integral result and the interval length. The process is as follows:

• Calculate \(\int_0^3 (99.6 - t + 0.8t^2) \, dt\), which results in a value of 301.5.
• Find the interval length: \(b-a = 3 - 0 = 3\).
• Divide the integral result by the interval length: \(\frac{301.5}{3} = 100.5\).

The calculation gives an average temperature of 100.5 degrees Fahrenheit, providing insight into the patient's overall temperature trend during the observation period.