The skill here is calculating the average value of a function over an interval, applied to finding the average speed from a speed function. The formula (1/(b-a)) ∫$a^b$ f(t) dt gives the average speed, equivalent to total distance divided by time if speed is non-negative. For s(t) = $t^2$ on [1,3], we use (1/2) ∫$1^3$ $t^2$ dt, evaluating to 13/3 m/s. This reflects the increasing nature of $t^2$, so the average is between s(1) = 1 and s(3) = 9, specifically weighted by the integral. A tempting distractor is choice E, 26/3, which is the integral without dividing by the interval length of 2. To find the average value, always: identify the interval [a,b], compute ∫_$a^b$ f(t) dt accurately, divide by (b-a), and simplify the result.

The skill here involves calculating the average value of a function over an interval, representing the constant power that would deliver the same total energy as the varying output. For the heater's power P(t) = 6 - 2t kW from t = 0 to t = 2, the average is (1/(2 - 0)) ∫_$0^2$ P(t) dt, with the integral giving total energy in kWh if time is hours, but here it's the accumulation for averaging. Conceptually, this captures the linear decrease in power, finding the equivalent steady level. To apply it, integrate 6t - t² from 0 to 2, getting 8, then divide by 2 to yield 4 kW. A tempting distractor like 8 could result from computing the integral but forgetting to divide by the interval length, representing total instead of average. For any average value problem, checklist: identify the interval [a, b], compute the definite integral of the function, divide by (b - a), and interpret as the mean level over that span.

The skill here involves understanding the average value of a function over an interval, which geometrically is the height of a rectangle with width (b - a) and area equal to the integral. For f(x) = 2x - 1 from x = -2 to x = 4, the average value is a constant c such that ∫_${-2}^4$ f(x) dx = c(4 - (-2)), directly from the definition. Conceptually, this c represents the mean output level over the linear function's span. While computing gives c = 1, the representation is about equating the integral to c times the length. A tempting distractor like the total signed area might confuse the integral itself with the average, but the average scales it by the interval. For any average value problem, checklist: identify the interval [a, b], compute the definite integral of the function, divide by (b - a), and interpret as the mean level over that span.

This problem asks for the average rate of change given $P'(t) = 6t - 2$ on $[0,3]$. The average value of the derivative $P'(t)$ equals the average rate of change of $P(t)$, calculated as $\frac{1}{3-0} \int_0^3 (6t-2) , dt = \frac{1}{3} \int_0^3 (6t-2) , dt$. Integrating: $\int(6t-2) , dt = 3t^2 - 2t$, evaluated from 0 to 3 gives $(27-6) - 0 = 21$. The average is $21/3 = 7$. Choice B (the integral alone) represents the total change $P(3) - P(0)$, not the average rate. When finding average rates: use the average value formula on the derivative, integrate carefully, and divide by the interval length to get the average rate per unit time.

This problem asks for the average inflow rate given r(t) = 6 - 2t on [0,2]. The average value formula gives us (1/(2-0))∫0 to 2dt = (1/2)∫0 to 2dt. Evaluating the integral: [6t - t²] from 0 to 2 = 12 - 4 - 0 = 8. Therefore, the average value is 8/2 = 4. Choice E shows the correct formula setup, while choice B incorrectly uses the function value at the midpoint. Average value represents the constant rate that would produce the same total flow.

This problem asks for the average value of a velocity function over a time interval. The average value formula for a function f(t) on [a,b] is (1/(b-a))∫[a to b] f(t)dt, which gives the mean height of the function over that interval. For velocity v(t) = 6-2t on [0,2], we calculate (1/(2-0))∫0 to 2dt = (1/2)∫0 to 2dt. Choice C represents the total displacement (integral alone), not the average velocity. Choice A gives the velocity at just one moment, missing the averaging process entirely. Remember the average value checklist: determine interval endpoints, apply the formula with correct interval length in denominator, and integrate the given function.

This problem asks for the average rate of population change over a time interval. The average value formula states that for f(t) on [a,b], the average is (1/(b-a))∫[a to b] f(t)dt, representing the constant rate giving the same total change. For rate p(t) = 8-4t on [0,1], we calculate (1/(1-0))∫0 to 1dt = (1/1)∫0 to 1dt = ∫0 to 1dt. Choice B omits the division by interval length, giving total population change instead of average rate. Choice D evaluates the rate at the midpoint, which may not equal the average for non-constant rates. To find average values: determine interval length (here 1), integrate the rate function, then divide by 1.

This conceptual problem asks what the average value represents for m(x) = 2x - 1 on [1,4]. The average value is the constant height c such that the rectangle with base [1,4] and height c has the same area as the region under m(x). This means c(4-1) = ∫1 to 4dx, or equivalently, ∫[1 to 4]c dx = ∫[1 to 4]m(x)dx. Choice A gives total area but not average height, while choices C and D give specific function values that don't generally equal the average. Average value = constant function with same integral over the interval.

This problem requires finding the average value of a trigonometric function. The average value of f(x) on [a,b] equals (1/(b-a))∫[a to b] f(x)dx, which finds the horizontal line with the same area underneath. For g(x) = cos(x) on [0,π/2], the interval length is π/2-0 = π/2, so we need (1/(π/2))∫[0 to π/2] cos x dx = (2/π)∫[0 to π/2] cos x dx. Choice B gives just the integral (total area under cosine), not the average height. Choice C evaluates cosine at one point rather than averaging over the entire interval. Remember: calculate interval length, integrate the function, then divide by interval length—here multiply by 2/π.

This problem requires finding the average value of a rate function over a time interval. The average value of a function f(t) on interval [a,b] is given by (1/(b-a))∫[a to b] f(t)dt, which represents the total accumulated quantity divided by the time elapsed. For the inflow rate r(t) = 2+t on [0,4], we need (1/(4-0))∫0 to 4dt = (1/4)∫0 to 4dt. Choice B gives the total water that flowed in (integral without the 1/4 factor), not the average rate. Choice C evaluates the rate at one instant rather than averaging over the entire interval. To find average values: identify the interval length, set up the integral of the function, then divide the integral by the interval length.