Question

a pump allows water to flow into a tank at the rate of $r(t) = 3t$ liters per minute, where $t$ is the time in minutes since the pump was turned on. which of the following defines a function that measures the accumulation of water in the tank during the time period from $t = 1$ to $t = x$ as the variable $x$ moves along the $t$-axis as shown in the figure above?\
\
a $f(x) = 3x$\
\
b $f(x) = 1 + 3x$\
\
c $f(x) = \int_{0}^{x} 3t dt$\
\
d $f(x) = \int_{1}^{x} 3t dt$

Explanation:

Step1: Recall the concept of accumulation function

The accumulation of a quantity with a rate function \( r(t) \) over an interval \([a, x]\) is given by the definite integral of the rate function from \( a \) to \( x \), i.e., \( f(x)=\int_{a}^{x}r(t)dt \).

Step2: Identify the interval and rate function

Here, the rate function is \( r(t) = 3t \) liters per minute, and the interval is from \( t = 1 \) to \( t = x \). So, the accumulation function \( f(x) \) should be the integral of \( 3t \) with respect to \( t \) from \( 1 \) to \( x \).

Step3: Analyze each option

• Option A: \( f(x)=3x \) is just the rate function evaluated at \( x \), not the accumulation. So, A is incorrect.
• Option B: \( f(x)=1 + 3x \) is a linear function, not related to the integral of the rate function. So, B is incorrect.
• Option C: The integral is from \( 0 \) to \( x \), but we need the interval from \( 1 \) to \( x \). So, C is incorrect.
• Option D: The integral \( \int_{1}^{x}3t\ dt \) represents the accumulation of water from \( t = 1 \) to \( t = x \) with rate \( r(t)=3t \). So, D is correct.

Answer:

D. \( f(x)=\int_{1}^{x}3t\ dt \)