This problem tests your ability to interpret definite integrals as accumulation in applied contexts. Since r(t) represents the filling rate in liters per minute, the definite integral ∫₂⁶ r(t)dt accumulates these rates over time, giving the total volume of water added to the tank. The integral multiplies rate (liters/minute) by time (minutes), yielding units of liters—representing the total amount added from t=2 to t=6. Choice A incorrectly suggests an average rate, but that would require dividing the integral by (6-2). Always verify units: when integrating a rate with respect to time, you get the total accumulated quantity, not a rate.

This problem tests your understanding of definite integrals as accumulation in energy contexts. Since P(t) represents power usage in kilowatts (energy per time), the definite integral ∫₃⁷ P(t)dt accumulates this power over time, giving the total energy consumed. The integral multiplies power (kilowatts) by time (hours), yielding units of kilowatt-hours—the standard unit for energy consumption from hour 3 to hour 7. Choice E incorrectly suggests "total power," but power is already a rate; we accumulate energy, not power itself. Always check units: rate × time = total quantity, so kilowatts × hours = kilowatt-hours of energy.

This problem tests your ability to interpret definite integrals as accumulation in meteorological contexts. Since R(t) represents the rainfall rate in inches per hour, the definite integral ∫₀³ R(t)dt accumulates these rates over time, giving the total rainfall accumulated. The integral multiplies rate (inches/hour) by time (hours), yielding units of inches—representing the total depth of rain that fell from t=0 to t=3. Choice C incorrectly suggests an average rate, but the integral gives a total amount, not a rate (average would require dividing by 3). Always verify units: rate × time = accumulated quantity, so inches/hour × hours = inches of rainfall.

This problem tests your ability to interpret definite integrals as accumulation in hydrological contexts. Since L(t) represents the rate at which water leaves in cubic meters per day, the definite integral ∫₂⁹ L(t)dt accumulates these rates over time, giving the total volume of water that left the reservoir. The integral multiplies rate (cubic meters/day) by time (days), yielding units of cubic meters—representing the total water loss from day 2 to day 9. Choice A incorrectly suggests the volume remaining in the reservoir, but the integral measures what left, not what remains. Always focus on what the rate describes: L(t) is a leaving rate, so its integral is total volume that left.

This problem requires interpreting a definite integral of speed as distance traveled in an applied context. Since s(t) represents the runner's speed in meters per second, the definite integral $\int_4^{10} s(t), dt$ accumulates these speeds over time, giving the total distance traveled. The integral multiplies speed (meters/second) by time (seconds), yielding units of meters—representing how far the runner traveled from t=4 to t=10. Choice B incorrectly suggests average speed, but that would require dividing the integral by $(10-4)$. Remember: speed integrated over time gives distance traveled (always positive), while velocity integrated gives displacement (can be negative).

This problem requires interpreting a definite integral of a population rate as net change in an applied context. Since p(t) represents the rate of population change in fish per day, the definite integral ∫₁₀²⁰ p(t)dt accumulates these rates over time, giving the net change in population size. The integral multiplies rate of change (fish/day) by time (days), yielding units of fish—representing how much the population increased or decreased from day 10 to day 20. Choice A incorrectly suggests the actual population size at day 20, but the integral gives change, not absolute value. Remember: integrating a rate of change gives the net change, not the final value.

This problem requires interpreting a definite integral of a gain rate as total accumulation in a digital context. Since g(t) represents the rate of gaining visitors in visitors per hour, the definite integral ∫₁₈²⁴ g(t)dt accumulates these rates over time, giving the total number of new visitors gained. The integral multiplies rate (visitors/hour) by time (hours), yielding units of visitors—representing how many visitors were added from t=18 to t=24. Choice A incorrectly suggests the total number on the site at t=24, but the integral only counts new arrivals during the interval. Remember: integrating a rate of increase gives the total increase, not the final total.

This question tests the skill of interpreting accumulation functions using definite integrals in applied contexts. The definite integral ∫ from 0 to 4 of I(t) dt represents the total charge that flows through the circuit over the interval from t=0 to t=4 seconds. Since I(t) is current in amperes (coulombs per second), integrating over time yields the net charge transferred, in coulombs. This measures the accumulated electric charge during that period. A tempting distractor is choice B, the average current, but that would involve dividing the integral by the time interval of 4 seconds to get amperes. To verify such interpretations, perform a units check: multiplying current units (coulombs/s) by time (s) gives coulombs, confirming total charge rather than a rate.

This question tests the skill of interpreting accumulation functions using definite integrals in applied contexts. The definite integral ∫ from 0 to 12 of v(t) dt represents the net displacement of the car over the interval from t=0 to t=12 seconds. Since v(t) is velocity in meters per second, integrating over time accumulates the net change in position, accounting for direction, in meters. This measures how far the car has moved from its starting point, positive or negative depending on the velocity's sign. A tempting distractor is choice D, the total distance traveled, but that would require integrating the absolute value |v(t)| to account for path length regardless of direction. To verify such interpretations, perform a units check: multiplying velocity units (m/s) by time (s) gives meters, matching displacement but not speed or acceleration units.

This question tests the skill of interpreting accumulation functions using definite integrals in applied contexts. The definite integral ∫ from 5 to 15 of P'(t) dt represents the net change in the rabbit population over the interval from t=5 to t=15 days. Since P'(t) is the rate of change in rabbits per day, integrating this rate yields the total net increase or decrease in population, in rabbits. This is equivalent to P(15) - P(5) by the Fundamental Theorem of Calculus. A tempting distractor is choice A, the population size at t=15, but that would require adding the initial population at t=5 to the integral. To verify such interpretations, perform a units check: multiplying rate units (rabbits/day) by time (days) gives rabbits, aligning with net change rather than a rate.