This problem requires accumulation reasoning to find displacement from a piecewise constant velocity. From t = 0 to t = 1, the particle moves 2(1 - 0) = 2 units forward, and from t = 1 to t = 3, it moves -2(3 - 1) = -4 units (backward). The net displacement is 2 + (-4) = -2 units, indicating the particle ends up 2 units behind its starting position. A common error is to add the distances traveled (2 + 4 = 6) without considering direction, but displacement is a signed quantity. For piecewise constant velocities, compute the displacement for each interval and sum the signed results.

This problem uses accumulation reasoning to find the change in bank balance from a constant negative rate. The change B(7) - B(0) equals the integral of B'(t) = -5 from t = 0 to t = 7. With a constant rate, this equals -5(7 - 0) = -35 dollars. The negative result indicates the balance decreased by 35 dollars. A potential mistake is to compute 5 × 7 = 35 and select answer A, forgetting that the negative rate means the balance is decreasing. When working with constant rates of change, the net change equals rate times time interval, keeping track of the sign.

This problem applies accumulation reasoning to find function change from the derivative of natural logarithm. The rate f'(x) = 1/(x+1) over [0,1] integrates to ∫₀¹ 1/(x+1) dx = [ln(x+1)]₀¹ = ln(2) - ln(1) = ln(2) - 0 = ln(2). The natural logarithm of (x+1) is the antiderivative of 1/(x+1). Students might choose 1 by evaluating ln(2) ≈ 0.693 and rounding, but the exact answer is ln(2). When the derivative involves 1/(x+a), the antiderivative is ln|x+a| plus a constant.

This problem uses accumulation reasoning to find function change from a decreasing linear rate. The rate f'(x) = 1 - x starts positive at x = 0 and becomes negative for x > 1 over [0,3]. We integrate ∫₀³ (1-x) dx = [x - x²/2]₀³ = (3 - 9/2) - 0 = 3 - 4.5 = -1.5 = -3/2. The rate crosses zero at x = 1, with more negative contribution than positive. Students might choose 3 by evaluating the positive part only, but the full interval includes significant negative accumulation. When linear rates cross zero, compute the full integral to capture both positive and negative contributions.

This problem requires accumulation reasoning to find displacement from cosine velocity over a half period. The velocity $v(t) = \cos t$ decreases from 1 to -1 over $[0,\pi]$, so we integrate $∫_0^π \cos t , dt = [\sin t]_0^π = \sin(π) - \sin(0) = 0 - 0 = 0$. The cosine function creates equal positive and negative areas over this half period, resulting in zero net displacement. Students might choose 2 by confusing this with the sine integral over $[0,\pi]$, but cosine over $[0,\pi]$ integrates to zero. When cosine is integrated over a half period starting from zero, the result is always zero due to symmetry.

This problem applies accumulation reasoning with an initial condition to find a function value. Given f'(x) = -1 is constant over [-3,1] and f(-3) = 5, the change f(1) - f(-3) = -1×(1-(-3)) = -1×4 = -4. Therefore, f(1) = f(-3) + (-4) = 5 - 4 = 1. The negative constant rate produces linear decrease from the initial value. Students might choose -1 by confusing the rate with the final value, but f(1) equals the initial value plus the accumulated change. When rates of change are negative constants, subtract the accumulated change magnitude from the initial value.

This problem uses accumulation reasoning to find function change from a linear derivative with specific symmetry. The rate f'(x) = 4 - 4x = 4(1-x) decreases linearly from 4 to -4 over [0,2], crossing zero at x = 1. We integrate ∫₀² (4-4x) dx = [4x - 2x²]₀² = (8 - 8) - 0 = 0. The equal positive and negative areas on either side of x = 1 cancel exactly. Students might choose 4 by evaluating the initial rate, or -4 by evaluating the final rate, but the symmetric accumulation yields zero net change. When linear rates cross zero at the midpoint of the interval, the positive and negative contributions cancel completely.

This problem applies accumulation reasoning to find volume change from a constant draining rate. The tank drains at r(t) = -3 L/min over the 4-minute interval [2,6], so we accumulate this negative rate over time. The net change equals -3 L/min × (6-2) min = -3 × 4 = -12 liters, indicating the tank loses 12 liters. A student might choose -3 by confusing the rate with the total change, ignoring the 4-minute duration. When dealing with constant negative rates (outflow/draining), multiply the rate by the time interval to find total volume lost.

This problem applies accumulation reasoning to find temperature change from a reciprocal rate function. The rate $T'(t) = \frac{1}{t^2}$ decreases rapidly over $[1,2]$, so we integrate $\int_1^2 \frac{1}{t^2} , dt = \int_1^2 t^{-2} , dt$. This equals $[-t^{-1}]_1^2 = -\frac{1}{2} - (-\frac{1}{1}) = -\frac{1}{2} + 1 = \frac{1}{2}$. The negative reciprocal antiderivative captures the decreasing contribution as t increases. Students might choose 1 by evaluating $1/1^2$ at the left endpoint, but this gives the initial rate, not the accumulated change. When integrating power functions with negative exponents, apply the power rule with careful attention to signs.

This problem uses accumulation reasoning to find function change from a quadratic derivative that crosses zero. The rate $f'(x) = 2x - 4 = 2(x-2)$ is negative for $x < 2$ and positive for $x > 2$ over $[0,4]$. We integrate $\int_0^4 (2x-4) , dx = [x^2 - 4x]_0^4 = (16 - 16) - 0 = 0$. The symmetric quadratic creates equal negative and positive contributions that cancel exactly. Students might choose -8 by evaluating only the negative part or 8 by considering only the positive part, but the full interval yields zero net change. When quadratic rates are symmetric about their zero, the accumulated change can be zero despite significant variation.