This problem uses accumulated-change reasoning to find the train's net displacement from velocity. Since velocity $v(t) = 10 - \tfrac{t}{2}$ is the rate of change of position, integrating from 2 to 8 gives the net displacement: $$\int_2^8 (10 - \tfrac{t}{2}) , dt$$. The velocity changes from positive to negative at $t = 20$, but since we're only considering $t \in[2,8]$, the velocity remains positive throughout our interval. Choice C ($v(8) - v(2)$) would incorrectly give the change in velocity rather than position. When given a velocity function, integrate it over the specified interval to find net position change.

This problem applies accumulated-change reasoning to find population change from growth rate. The rate P'(t)=200e^(-0.1t) represents how fast the population changes per year, so the integral ∫₀¹⁰ 200e^(-0.1t)dt gives the total population change from t=0 to t=10. The integral accumulates all instantaneous growth rates, and since the exponential is always positive, the population increases throughout. Choice D incorrectly adds the initial population P(0), but we want only the change, not the final population. To find total change from a rate of change, integrate the rate function over the given interval.

This problem requires using accumulated-change reasoning to find the net change in position from a velocity function. The definite integral of the velocity function over the time interval $[0, \pi]$ represents the net change in position, as velocity is the rate of change of position with respect to time. By the Fundamental Theorem of Calculus, $\int_0^\pi v(t) , dt$ equals the position at $t=\pi$ minus the position at $t=0$. This integral incorporates the direction of motion through the sign of $v(t)$, providing the net displacement. A tempting distractor is choice C, which uses the absolute value and computes total distance rather than net change. In general, to find the total change in a quantity from its rate of change, integrate the rate function over the interval.

This problem requires using accumulated-change reasoning to find the displacement from a velocity function. The definite integral of the velocity function over the time interval [0,9] represents the net change in position, as velocity is the rate of change of position with respect to time. By the Fundamental Theorem of Calculus, ∫ from 0 to 9 of v(t) dt equals the position at t=9 minus the position at t=0. This captures the net effect with v(t) negative initially and positive later. A tempting distractor is choice A, which uses absolute value and computes total distance rather than net displacement. In general, to find the total change in a quantity from its rate of change, integrate the rate function over the interval.

This problem requires using accumulated-change reasoning to find the net change in position from a velocity function. The definite integral of the velocity function over the time interval [0,6] represents the net change in position, as velocity is the rate of change of position with respect to time. By the Fundamental Theorem of Calculus, ∫ from 0 to 6 of v(t) dt equals the position at t=6 minus the position at t=0. This accounts for the continuous decrease in velocity, which remains positive over the interval. A tempting distractor is choice C, which uses absolute value unnecessarily since v(t) doesn't change sign here. In general, to find the total change in a quantity from its rate of change, integrate the rate function over the interval.

This problem requires using accumulated-change reasoning to find the displacement from a velocity function. The definite integral of the velocity function over the time interval [0,2π] represents the net change in position, as velocity is the rate of change of position with respect to time. By the Fundamental Theorem of Calculus, ∫ from 0 to 2π of v(t) dt equals the position at t=2π minus the position at t=0. This integral captures the net effect despite v(t) changing signs during the interval. A tempting distractor is choice A, which uses absolute value and calculates total distance instead of net displacement. In general, to find the total change in a quantity from its rate of change, integrate the rate function over the interval.

This problem requires accumulated-change reasoning to find net displacement from velocity. The integral ∫₀^(2π) (2+cos t) dt gives the cyclist's net displacement because velocity is the rate of change of position. Since 2+cos t is always positive (ranging from 1 to 3), the net displacement equals the total distance traveled. Choice C (v(2π)-v(0)) would give the change in velocity rather than position, confusing the rate with the accumulated quantity. When given a velocity function, integrate it over the time interval to find the net change in position.

This question uses accumulated-change reasoning to find net displacement from velocity. The integral ∫₁⁵(t-3)dt gives the net change in position because velocity is the derivative of position. Note that v(t) = t-3 is negative for t < 3 and positive for t > 3, so the particle changes direction at t = 3. Choice B with absolute value would give total distance traveled, not net displacement, which are different when velocity changes sign. To find net position change from velocity, integrate the velocity function (without absolute value) over the time interval.

This problem uses accumulated-change reasoning to find total volume added from a filling rate. Since r(t) = 5e^(-t/2) gives the rate of volume change in L/min, integrating it from 0 to 4 yields the total volume added: ∫₀⁴ 5e^(-t/2) dt. The integral accumulates all the instantaneous volume additions over the 4-minute period. Choice A (r(4)-r(0)) incorrectly finds the change in rate rather than the total volume accumulated. To find total accumulated quantity from a rate of change, integrate the rate function over the time interval.

This question applies accumulated-change reasoning to find total volume drained from a drain rate. Since r(t) = 4-ln(t+1) gives the rate of volume change in L/min, integrating from 0 to 2 yields the total volume drained: ∫₀²(4-ln(t+1))dt. The integral accumulates all the instantaneous volume changes over the 2-minute period. Choice A (r(2)-r(0)) would give the change in drain rate, not the total volume drained. To find total accumulated quantity from a rate of change, integrate the rate function over the time interval.