AP Calculus AB

AP Calculus AB Practice Test: Practice Test 6

Practice Test 6 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

What integral gives the volume if the base is bounded by $x=0$ and $x=\sqrt{9-y^2}$ for $-3\le y\le 3$ , with semicircular cross sections perpendicular to the $y$ -axis?

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Question 1

What integral gives the volume if the base is bounded by $x=0$ and $x=\sqrt{9-y^2}$ for $-3\le y\le 3$ , with semicircular cross sections perpendicular to the $y$ -axis?

$\displaystyle \int_{-3}^{3} \frac{\pi}{8}\,(9-y^2)\,dy$ (correct answer)
$\displaystyle \int_{-3}^{3} \frac{\pi}{8}\,\sqrt{9-y^2}\,dy$
$\displaystyle \int_{-3}^{3} \frac{\pi}{4}\,(9-y^2)\,dy$
$\displaystyle \int_{-3}^{3} \frac{1}{2}\,(9-y^2)\,dy$
$\displaystyle \int_{-3}^{3} \pi\,(9-y^2)\,dy$

Explanation: This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the y-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the horizontal distance between the curves x=0 and x=√(9-y²), which is √(9-y²). The area of a semicircle is (1/2) π r², where r = √(9-y²)/2. Therefore, the area simplifies to π (9-y²) / 8. The volume is the integral of this area from y=-3 to y=3. A common mistake is to use π √(9-y²) / 8, as in choice B, which forgets to square the radius properly. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 2

For $f(x)=\frac{|x|}{x}$ with $f(0)$ undefined, does $f'(0)$ exist, and why?

Yes; the derivative exists even if the function is undefined at $0$ .
No; $f$ is not continuous/defined at $0$ , so $f'(0)$ does not exist. (correct answer)
Yes; the one-sided derivatives match at $0$ .
No; $f$ has a corner at $0$ .
Yes; $f$ has a vertical tangent at $0$ .

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, stressing the need for function definition and continuity. The function f(x) = |x|/x is undefined at x=0, violating the basic requirement for differentiability. Without a function value at x=0, continuity cannot hold, and the derivative limit cannot be computed. Therefore, f'(0) does not exist. A tempting distractor might suggest yes if one-sided derivatives match, but the lack of definition overrides this. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 3

For $s(x)=\begin{cases}\sin x,&x\ne0\\1,&x=0\end{cases}$ , is $s$ continuous at $x=0$ , and why?

Yes; $\lim_{x\to0}s(x)=1=s(0)$ .
No; $\lim_{x\to0}s(x)=0\ne s(0)=1$ . (correct answer)
No; $s(0)$ is undefined.
No; $\lim_{x\to0}s(x)$ does not exist because $\sin x$ oscillates.
Yes; $\sin x$ is continuous, so redefining $s(0)$ cannot affect continuity.

Explanation: To determine continuity at x = 0, we verify: (1) s(0) exists, (2) lim[x→0] s(x) exists, and (3) they are equal. The function defines s(0) = 1, so condition (1) is met. For the limit, since s(x) = sin x for x ≠ 0, we have lim[x→0] s(x) = lim[x→0] sin x = sin(0) = 0. We have s(0) = 1 but lim[x→0] s(x) = 0, so 1 ≠ 0, violating condition (3). This is a classic example where redefining a continuous function at a single point creates a discontinuity—the natural value of sin(0) = 0 has been changed to 1. Remember the continuity checklist: (1) Function value exists? (2) Limit exists? (3) Are they equal?

Question 4

Define $F(x)=\int_{0}^{x} g(t)\,dt$ . If $g$ is positive on $(0,2)$ and negative on $(2,4)$ , what must be true about $F$ at $x=2$ ?

$F$ has a local maximum (correct answer)
$F$ has a local minimum
$F(2)=0$
$F$ is undefined
$F$ is concave down

Explanation: This problem tests identification of local extrema based on sign changes in the integrand. Since $F'(x) = g(x)$ by the Fundamental Theorem, critical points occur where $g(x) = 0$ . At $x = 2$ , the integrand changes from positive to negative, meaning $F'(x)$ changes from positive to negative. This indicates $F$ transitions from increasing to decreasing, creating a local maximum at $x = 2$ . Choice B would be incorrect since a minimum requires the opposite sign change. To find extrema in accumulation functions, locate where the integrand changes sign.

Question 5

An antiderivative of $p(x)$ is $P(x)=\frac{1}{x}-x^2$ . Evaluate $\int_{1}^{3} p(x)\,dx$ using FTC.

$P(3)$
$P(1)-P(3)$
$P(3)-P(1)$ (correct answer)
$P(3)+P(1)$
$P(1)$

Explanation: This problem demonstrates the Fundamental Theorem of Calculus Part 2, which states that for an antiderivative P(x) of p(x), the definite integral ∫[a to b] p(x)dx equals P(b) - P(a). Given P(x) = 1/x - x², we need to evaluate ∫[1 to 3] p(x)dx = P(3) - P(1). Computing the values: P(3) = 1/3 - 3² = 1/3 - 9 = 1/3 - 27/3 = -26/3 and P(1) = 1/1 - 1² = 1 - 1 = 0. Therefore, ∫[1 to 3] p(x)dx = P(3) - P(1) = -26/3 - 0 = -26/3. The incorrect choice P(1) - P(3) = 0 - (-26/3) = 26/3 reverses the bounds and changes the sign of the result. When using FTC Part 2, maintain the order: antiderivative at upper bound minus antiderivative at lower bound.

Question 6

The derivative $f'(x)$ decreases on $(-3,0)$ and increases on $(0,5)$ . Where is $f$ concave up and down?

Concave up on $(-3,0)$ ; concave down on $(0,5)$
Concave up on $(-3,5)$ ; concave down on no interval
Concave up on $(0,5)$ ; concave down on $(-3,0)$ (correct answer)
Concave up on no interval; concave down on $(-3,5)$
Concave up on $(-3,0)$ and $(0,5)$ ; concave down on no interval

Explanation: Analyzing the concavity of a function over its domain is a key skill in AP Calculus AB. Concavity can be determined from the behavior of the first derivative: f is concave up where f'(x) is increasing and concave down where f'(x) is decreasing. Given that f'(x) decreases on (-3, 0), f is concave down there, and increases on (0, 5), so f is concave up there. This corresponds to f''(x) < 0 on (-3, 0) and f''(x) > 0 on (0, 5). A tempting distractor is choice A, which reverses these intervals, but it fails because it confuses the relationship between the monotonicity of f' and concavity. Always use the sign of f''(x) or the monotonicity of f'(x) to determine concavity: positive for up, negative for down.

Question 7

A physics integral is $\int \frac{x^2+8x+20}{x+4}\,dx$ . Which is an antiderivative?

$\frac{x^2}{2}+4x+4\ln|x+4|+C$ (correct answer)
$\frac{x^2}{2}+4x+\ln|x+4|+C$
$\frac{x^2}{2}+4x+4\ln|x-4|+C$
$\frac{x^2+8x+20}{x+4}+C$
$\frac{x^2}{2}+4x-\frac{4}{x+4}+C$

Explanation: This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice B fails because it uses a coefficient of 1 instead of 4 for the logarithmic term, resulting from an incorrect remainder in the division. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.

Question 8

A function satisfies $f'(x)=4-4x$ for $0\le x\le 2$ . What is $f(2)-f(0)$ ?

$0$ (correct answer)
$4$
$-4$
$8$
$-8$

Explanation: 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.

Question 9

A bank account balance changes at $-15$ dollars/week due to fees; what does this mean?

The balance is $-15$ dollars.
The balance decreases by $15$ dollars each week. (correct answer)
Weeks decrease by $15$ per dollar.
The balance decreases by $15$ weeks each dollar.
The balance increases by $15$ dollars each week.

Explanation: This question assesses your understanding of negative rates in financial contexts, requiring you to interpret what a negative dollar/week rate means for a bank balance. The rate -15 dollars/week indicates that the balance changes by -15 dollars each week, which means it decreases by $15 weekly - option B correctly states this. Students often confuse the rate of change with the actual balance (option A incorrectly thinks -15 is the balance itself) or misinterpret the negative sign (option E ignores it and says the balance increases). The negative sign in a rate always indicates the direction of change, not a negative value for the quantity itself. In financial contexts, negative rates typically represent losses, fees, or expenses. When you see a negative rate, always interpret it as a decrease in the quantity, stating it as a positive decrease rather than a negative increase.

Question 10

A rectangle with perimeter $60$ has length $x$ and width $30-x$ ; which $x$ maximizes area $A=x(30-x)$ ?

$x=30$
$x=0$
$x=15$ (correct answer)
$x=10$
$x=20$

Explanation: This problem involves solving optimization problems by maximizing the area $A(x) = x(30 - x)$ for a rectangle with perimeter 60. To find the maximum, compute the derivative $A'(x) = 30 - 2x$ and set it to zero, yielding $x = 15$ as the critical point. Evaluating $A$ at $x = 15$ gives the maximum area of 225. The problem implies a closed interval, but the critical point determines the solution without needing endpoints. A tempting distractor is $x = 30$ , where area is zero, but this is a minimum at the boundary. A transferable optimization-solving strategy is to identify critical points by setting the derivative to zero, evaluate the function at those points and any endpoints if applicable, and compare values to determine the extreme.

Question 11

Does $y=x^2+1$ satisfy $\dfrac{dy}{dx}=2x$ at every $x$ ?

Yes, because $y'=2x$ and the right side is $2x$ . (correct answer)
No, because $y'=x^2$ but the right side is $2x$ .
No, because $y'=2x+1$ but the right side is $2x$ .
Yes, because $y'=x^2+1$ equals $2x$ .
No, because $y'=2$ but the right side is $2x$ .

Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = x² + 1, we find y' = 2x using the power rule. The differential equation dy/dx = 2x requires that y' equals 2x. Substituting our derivative: y' = 2x, which exactly matches the right side 2x. Since both sides are identical, the function satisfies the equation. Choice B incorrectly states y' = x², confusing the original function with its derivative. Always compute derivatives carefully and verify both sides of the equation are equal.

Question 12

A particle’s average velocity is $\dfrac{(x+1)^2-4}{x-1}$ ; find $\lim\limits_{x\to 1}\dfrac{(x+1)^2-4}{x-1}$ .

$0$
$4$ (correct answer)
$2$
$1$
$-4$

Explanation: This limit presents an indeterminate form 0/0 that requires algebraic simplification. First, expand the numerator: (x+1)² - 4 = x² + 2x + 1 - 4 = x² + 2x - 3. Factor this quadratic as (x + 3)(x - 1). The expression becomes (x + 3)(x - 1)/(x - 1), which simplifies to (x + 3) after canceling the common factor. Substituting x = 1 yields 1 + 3 = 4. A frequent mistake is expanding incorrectly or not recognizing the factored form. The transferable approach is to expand, factor, and cancel common terms that cause the indeterminate form.

Question 13

For $0<x<5$ , maximize $f(x)=x(5-x)^2$ ; which $x$ gives the maximum?

$x=\frac{5}{3}$ (correct answer)
$x=\frac{5}{2}$
$x=5$
$x=0$
$x=\frac{10}{3}$

Explanation: This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize $f(x) = x(5 - x)^2$ , compute the derivative $f' = (5 - x)(5 - 3x)$ and set it to zero, yielding $x = \frac{5}{3}$ . Evaluating at $x = \frac{5}{3}$ gives the maximum, while endpoints yield zero. The second derivative confirms a maximum. A tempting distractor is $x = \frac{5}{2}$ , but it yields a lower value as it is not the critical point. To solve optimization problems, express the quantity as a function of one variable, find critical points by setting the derivative to zero, and evaluate at critical points and endpoints to find the maximum or minimum.

Question 14

A product’s demand price is $p(x)=50-2x$ dollars; revenue is $R(x)=xp(x)$ for $0<x<25$ . Which $x$ maximizes revenue?

$x=12.5$ (correct answer)
$x=25$
$x=0$
$x=10$
$x=15$

Explanation: This problem involves solving optimization problems by finding the value of x that maximizes revenue R(x) = x(50 - 2x) for 0 < x < 25. To find the maximum, compute the derivative R'(x) = 50 - 4x and set it to zero, yielding the critical point x = 12.5. Evaluating R at x = 12.5 and considering the open interval shows this is the maximum. Since the interval is open, endpoints are approached but not included, confirming the maximum at the critical point. A tempting distractor is x = 25, where revenue is zero, but this is the minimum revenue point near the boundary. A transferable optimization-solving strategy is to identify critical points by setting the derivative to zero, evaluate the function at those points and any endpoints if applicable, and compare values to determine the extreme.

Question 15

A differentiable one-to-one function $p$ satisfies $p(0)=7$ and $p'(0)=5$ ; compute $(p^{-1})'(7)$ .

$\dfrac{1}{5}$ (correct answer)
$5$
$\dfrac{1}{p'(7)}$
$\dfrac{1}{7}$
$\dfrac{1}{p'(0)}$

Explanation: This problem involves finding the derivative of an inverse function at a specific value. The inverse function derivative formula tells us $(p^{-1})'(b) = rac{1}{p'(a)}$ when $p(a) = b$ . We have $p(0) = 7$ and $p'(0) = 5$ , and need $(p^{-1})'(7)$ . Since $p(0) = 7$ , we get $p^{-1}(7) = 0$ , so we use $a = 0$ . Thus $(p^{-1})'(7) = rac{1}{p'(0)} = rac{1}{5}$ . Choice B ( $5$ ) represents $p'(0)$ without the reciprocal, a common error. To find an inverse derivative: locate the input that produces your output, then take the reciprocal of the derivative there.

Question 16

Let $n(x)=\sin\!\left(\ln(7-4x)\right)$ . What is $n'(x)$ ?

$\cos\!\left(\ln(7-4x)\right)\cdot\dfrac{-4}{7-4x}$ (correct answer)
$\cos\!\left(\ln(7-4x)\right)\cdot\dfrac{1}{7-4x}$
$\cos\!\left(\ln(7-4x)\right)$
$\sin\!\left(\ln(7-4x)\right)\cdot\dfrac{-4}{7-4x}$
$\dfrac{-4}{7-4x}$

Explanation: This is sin composed with ln(7-4x). The derivative of sin(u) is cos(u), giving cos(ln(7-4x)) times the derivative of ln(7-4x). For ln(7-4x), we get 1/(7-4x) times the derivative of 7-4x, which is -4. Therefore, n'(x) = cos(ln(7-4x)) · 1/(7-4x) · (-4) = cos(ln(7-4x)) · (-4)/(7-4x). The negative sign comes from the -4x term and is often overlooked. Pattern recognition: when the innermost function has a negative coefficient (like -4x), that negative will appear in the final derivative.

Question 17

Which of the following is the most appropriate method to find the limit $\lim_{x \to \infty} \frac{\sqrt{9x^2 + x}}{2x+1}$ ?

Multiply the numerator and denominator by the conjugate of the numerator, $\sqrt{9x^2+x}$ .
Divide the numerator and denominator by $x$ , remembering that for $x>0$ , $x = \sqrt{x^2}$ . (correct answer)
Apply the Squeeze Theorem, since the function contains a square root and is difficult to evaluate directly.
Use direct substitution, which is the first step for all limit problems and will give the answer here.

Explanation: For a limit at infinity involving a rational-like expression with a radical, the most effective method is to divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$ . To divide the numerator by $x$ , we use the fact that $x = \sqrt{x^2}$ for positive $x$ to bring the term inside the square root. This simplifies the expression and allows for evaluation of the limit.

Question 18

For $0<x<9$ , maximize $f(x)=x(9-x)^{1/2}$ ; which $x$ gives the maximum?

$x=3$
$x=6$ (correct answer)
$x=0$
$x=9$
$x=\frac{9}{2}$

Explanation: This problem involves solving optimization problems by maximizing f(x) = x √(9 - x) for 0 < x < 9. To find the maximum, compute the derivative f'(x) = √(9 - x) - x/(2√(9 - x)) and set it to zero, yielding x = 6 as the critical point. Evaluating f at x = 6 gives the maximum of 6√3 ≈ 10.39. Since the interval is open, endpoints are not evaluated, but give limits of 0. A tempting distractor is x = 3, where f=3√6 ≈ 7.35, less than the maximum. A transferable optimization-solving strategy is to identify critical points by setting the derivative to zero, evaluate the function at those points and any endpoints if applicable, and compare values to determine the extreme.

Question 19

Let $f$ be continuous on $[1,5]$ with $f(1)=-2$ and $f(5)=6$ . What must be true?

There exists $c\in(1,5)$ such that $f(c)=0$ . (correct answer)
There exists $c\in(1,5)$ such that $f(c)=8$ .
There exists $c\in(1,5)$ such that $f(c)=-3$ .
There exists $c\in(1,5)$ such that $f(c)=7$ .
No conclusion can be drawn because continuity is not enough.

Explanation: The Intermediate Value Theorem (IVT) states that if f is continuous on [a,b] and k is any value between f(a) and f(b), then there exists at least one c in (a,b) such that f(c) = k. Here, f is continuous on [1,5] with f(1) = -2 and f(5) = 6. Since 0 is between -2 and 6, IVT guarantees there exists c in (1,5) where f(c) = 0. A common error is thinking IVT only applies to positive values or that it guarantees values outside the range [-2,6]. Note that while 7 and -3 are outside this range, 8 is not guaranteed despite being close to 6. IVT Checklist: ✓ Function is continuous on closed interval, ✓ Target value is between endpoint values, ✓ Conclusion: value exists in open interval.

Question 20

In a heat-transfer model, find $\displaystyle \int \frac{1}{x^2+6x+13}\,dx$ .

$\dfrac{1}{2}\arctan\!\left(\dfrac{x+3}{2}\right)+C$ (correct answer)
$\arctan\!\left(\dfrac{x+3}{2}\right)+C$
$\dfrac{1}{2}\ln\left(x^2+6x+13\right)+C$
$\dfrac{1}{4}\arctan\!\left(\dfrac{x+3}{4}\right)+C$
$-\dfrac{1}{2(x^2+6x+13)}+C$

Explanation: This integral requires completing the square as preparatory algebra for integration. We rewrite $x^2+6x+13 = (x+3)^2+4$ by completing the square: add and subtract $(6/2)^2 = 9$ to get $(x+3)^2+4$ . This transforms the integral into the standard arctangent form: $\int\frac{1}{(x+3)^2+4}dx = \int\frac{1}{4[(x+3)^2/4+1]}dx = \frac{1}{2}\arctan\left(\frac{x+3}{2}\right)+C$ . Without completing the square, you might incorrectly attempt a logarithmic antiderivative (choice C) since the denominator is quadratic. For integrals with irreducible quadratic denominators, always complete the square to reveal the arctangent structure.

Question 21

Given $r(x)=\begin{cases}x+4,&x< -1\\2,&x=-1\\x^2,&x>-1\end{cases}$ , is $r$ continuous at $x=-1$ , and why?

Yes; $\lim_{x\to-1}r(x)$ exists and equals $r(-1)$ .
No; $\lim_{x\to-1^-}r(x)=3$ and $\lim_{x\to-1^+}r(x)=1$ , so the limit does not exist. (correct answer)
No; $r(-1)$ is undefined.
No; the limit exists and equals $2$ , but $r(-1)=0$ .
Yes; both pieces are polynomials, so the function must be continuous.

Explanation: For continuity at x = -1, we check: (1) r(-1) exists, (2) lim[x→-1] r(x) exists, and (3) they are equal. From the piecewise definition, r(-1) = 2, satisfying condition (1). For the limit, we compute one-sided limits: lim[x→-1⁻] r(x) = lim[x→-1⁻] (x + 4) = -1 + 4 = 3, and lim[x→-1⁺] r(x) = lim[x→-1⁺] (x²) = (-1)² = 1. Since the one-sided limits differ (3 ≠ 1), the two-sided limit does not exist, violating condition (2). A common error is assuming piecewise functions with polynomial pieces are automatically continuous, but we must check that the pieces "match up" at transition points. Apply the continuity checklist: (1) Function value exists? (2) Limit exists? (3) Are they equal?

Question 22

A cyclist’s speed is $s(t)$ miles/hour at time $t$ hours. What does $\int_{2}^{4} s(t)\,dt$ represent?

The cyclist’s speed at $t=4$ , in miles per hour
The cyclist’s total distance traveled from $t=2$ to $t=4$ , in miles (correct answer)
The cyclist’s average speed from $t=2$ to $t=4$ , in miles per hour
The time per mile from $t=2$ to $t=4$ , in hours per mile
The change in speed from $t=2$ to $t=4$ , in miles per hour

Explanation: This problem requires interpreting definite integrals as accumulation in motion contexts. Since s(t) represents speed in miles per hour, the definite integral ∫₂⁴ s(t)dt accumulates these speeds over the 2-hour period to give the total distance traveled from t=2 to t=4. The units verify this: (miles/hour) × (hours) = miles, which measures distance. Choice C incorrectly suggests this is an average speed, but the integral gives total distance, not average speed (which would require dividing by 2). When speed is always positive, integrating speed gives total distance traveled; this differs from displacement when dealing with velocity that can be negative.

Question 23

For $u(x)=\begin{cases}1,&x<2\\frac{1}{x-2},&x>2\\0,&x=2\end{cases}$ , what type of discontinuity is at $x=2$ ?

Removable discontinuity
Jump discontinuity
Infinite discontinuity (correct answer)
No discontinuity (continuous)
Corner discontinuity

Explanation: This function has an infinite discontinuity at x = 2. The left-hand limit is lim(x→2⁻) 1 = 1, but the right-hand limit is lim(x→2⁺) 1/(x-2) = +∞, and u(2) = 0. Since one of the one-sided limits is infinite, this creates an infinite discontinuity. The function exhibits vertical asymptote behavior as x approaches 2 from the right. Students might focus on the fact that the left limit exists, but an infinite discontinuity occurs when at least one one-sided limit is infinite. To identify infinite discontinuities: check if either one-sided limit approaches ±∞, regardless of the other limit's behavior.

Question 24

A signal is $S(t)=(e^{-t})(\sin t)$ . What is $S'(t)$ ?

$e^{-t}\cos t$
$-e^{-t}\sin t$
$-e^{-t}\sin t+e^{-t}\cos t$ (correct answer)
$e^{-t}(\sin t+\cos t)$
$-e^{-t}(\sin t+\cos t)$

Explanation: S(t) = $e^{-t} \sin t$ is a product: $(-e^{-t} \sin t) + e^{-t} \cos t$ , or $-e^{-t} \sin t + e^{-t} \cos t$ . Note chain for exponential. Error: Forgetting chain, like B. Rule with chain is correct. Strategy: Exponential-trig products need both rules.

Question 25

A particle’s velocity is $v(t)$ with antiderivative $V(t)=t^3-6t$ . Find $\int_{-2}^{1} v(t)\,dt$ .

$V(-2)-V(1)$
$V(1)-V(-2)$ (correct answer)
$V(1)+V(-2)$
$v(1)-v(-2)$
$V(2)-V(1)$

Explanation: This problem uses the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral of velocity. Since V(t) is an antiderivative of v(t) (meaning V'(t) = v(t)), and V(t) = t³ - 6t, we can find ∫[-2 to 1] v(t)dt by calculating V(1) - V(-2). We compute V(1) = 1³ - 6(1) = 1 - 6 = -5 and V(-2) = (-2)³ - 6(-2) = -8 + 12 = 4. Therefore, ∫[-2 to 1] v(t)dt = V(1) - V(-2) = -5 - 4 = -9. Choice A (V(-2) - V(1)) reverses the order, which would give the negative of the correct answer. When applying FTC Part 2, always subtract the antiderivative at the lower bound from the antiderivative at the upper bound.

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