Questions: A colony of bacteria has a total weight which varies according to (w(t)=15.5 e^0.5 t) (weight in Newtons, time in hours). Find the average weight of the colony over the interval [1, 5]. (Round your answer to two decimal places.) Find a value (c) in ([1,5]) such that (w(c)) attains this average. (Round your answer to two decimal places.) c=3.32

Solution Steps

To find the average weight of the bacteria colony over the interval [1, 5], we need to compute the integral of the weight function \( w(t) = 15.5 e^{0.5 t} \) over this interval and then divide by the length of the interval.

Next, to find the value \( c \) in [1, 5] such that \( w(c) \) equals the average weight, we solve for \( c \) by setting \( w(c) \) equal to the computed average weight and solving for \( c \).

1. Compute the integral of \( w(t) = 15.5 e^{0.5 t} \) from 1 to 5.
2. Divide the result by the length of the interval (5 - 1) to get the average weight.
3. Solve \( 15.5 e^{0.5 c} = \text{average weight} \) for \( c \).

To find the average weight of the bacteria colony over the interval \([1, 5]\), we first compute the integral of the weight function \( w(t) = 15.5 e^{0.5 t} \) over this interval: \[ \int_{1}^{5} 15.5 e^{0.5 t} \, dt \]

Next, we divide the result of the integral by the length of the interval, which is \(5 - 1 = 4\): \[ \text{Average weight} = \frac{1}{4} \int_{1}^{5} 15.5 e^{0.5 t} \, dt = 81.64 \]

We need to find a value \( c \) in \([1, 5]\) such that \( w(c) \) equals the average weight. This means solving the equation: \[ 15.5 e^{0.5 c} = 81.64 \] Solving for \( c \), we get: \[ c = 3.32 \]

Final Answer