AP Calculus AB

AP Calculus AB Practice Test: Practice Test 12

Practice Test 12 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

At $x=-19$ , $g'(-19)=0$ and $g$ is concave down at $x=-19$ ; classify the critical point.

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Question 1

At $x=-19$ , $g'(-19)=0$ and $g$ is concave down at $x=-19$ ; classify the critical point.

Local minimum at $x=-19$ .
Local maximum at $x=-19$ . (correct answer)
Inflection point at $x=-19$ .
Cannot be determined from the information given.
Neither a maximum nor minimum because $g'(-19)=0$ .

Explanation: The Second Derivative Test utilizes concavity information to classify critical points effectively. With $g'(-19)=0$ confirming a critical point and concave down behavior at $x=-19$ (meaning $g''(-19)<0$ ), the function exhibits downward curvature. Downward concavity at a critical point creates a hill-like shape, establishing a local maximum. Choice A could mislead students who connect downward direction with minimums, but concave down specifically refers to the hill formation characteristic of maximums. Ensure both conditions are met: critical point verification and determinable concavity sign.

Question 2

For $G(x)=\dfrac{e^{x}}{\sqrt{x^2+9}}$ , which differentiation rules should be used to compute $G'(x)$ ?

Quotient rule and chain rule (correct answer)
Power rule only
Chain rule only
Implicit differentiation
No rule beyond product rule is needed

Explanation: This problem requires selecting differentiation procedures for $G(x)=\dfrac{e^{x}}{\sqrt{x^2+9}}$ . Since this is a quotient, the quotient rule is necessary. Additionally, the denominator $\sqrt{x^2+9}$ is a composite function where the outer function is the square root and the inner function is $x^2+9$ , requiring the chain rule when differentiating the denominator. The numerator $e^x$ differentiates to itself. The product rule alone wouldn't work since this is a quotient structure, and implicit differentiation isn't needed for this explicit function. When you have a quotient where either part contains a composite function, combine the quotient rule with the chain rule.

Question 3

Let $H(x)=\int_{0}^{x} h(t)\,dt$ . If $h'(x)>0$ for $x<0$ and $h'(x)<0$ for $x>0$ , what is true about $H$ at $x=0$ ?

$H$ can change concavity at $x=0$ (correct answer)
$H$ has a local maximum at $x=0$
$H$ has a local minimum at $x=0$
$H(0)$ must be negative
$H$ is undefined at $x=0$

Explanation: This problem involves understanding how changes in the integrand's derivative affect accumulation function concavity. Since $H'(x) = h(x)$ and $H''(x) = h'(x)$ by differentiation, the concavity of $H$ depends on the sign of $h'(x)$ . Given that $h'(x) > 0$ for $x < 0$ and $h'(x) < 0$ for $x > 0$ , the second derivative $H''(x)$ changes from positive to negative at $x = 0$ . This sign change indicates that $H$ changes from concave up to concave down, creating an inflection point. Choice B would require $h(0) = 0$ with a sign change in $h$ . Changes in the integrand's monotonicity create inflection points in accumulation functions.

Question 4

A car’s fuel-use model includes $\int \left(\ln x\right)\,dx$ ; which technique is most appropriate to antidifferentiate?

Substitution with $u=\ln x$
Integration by parts (correct answer)
Trigonometric substitution
Partial fractions
Reverse the Chain Rule directly

Explanation: This question tests the skill of selecting an appropriate technique for antidifferentiation. The integrand is ln x, a transcendental function without an elementary substitution or direct rule, making integration by parts the best choice. By setting u = ln x and dv = dx, it becomes ∫ u dv = uv - ∫ v du, yielding x ln x - ∫ x (1/x) dx = x ln x - x + C. This method effectively handles the logarithmic function multiplied implicitly by 1. Substitution with u = ln x might seem promising, but it fails because du = (1/x) dx, leaving no x to pair with, resulting in an incomplete transformation. Recognize integration by parts when the integrand is a product where one part is easy to integrate and the other to differentiate.

Question 5

What integral gives the volume if the base is bounded by $x=2$ and $x=2+\cos y$ on $0\le y\le \pi$ , with semicircular cross sections perpendicular to the $y$ -axis?

$\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,(\cos y)^2\,dy$ (correct answer)
$\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,(2+\cos y)^2\,dy$
$\displaystyle \int_{0}^{\pi} \frac{1}{2}\,(\cos y)^2\,dy$
$\displaystyle \int_{0}^{\pi} \frac{\pi}{4}\,(\cos y)^2\,dy$
$\displaystyle \int_{0}^{\pi} \pi\,(\cos y)^2\,dy$

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=2 and x=2+cos y, which is |cos y|. The area of a semicircle is (1/2) π r², where r = |cos y|/2. Therefore, the area simplifies to π (cos y)² / 8 since squaring removes the absolute value. The volume is the integral of this area from y=0 to y=π. A common mistake is to use π (2+cos y)² / 8, as in choice B, which incorrectly takes the distance from x=0 instead of between the curves. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 6

A graph of $f'$ is negative on $(0,2)$ , zero at $x=2$ , positive on $(2,6)$ , and decreasing on $(0,6)$ . Which is true about $f$ ?

Local minimum at $x=2$ ; concave down on $(0,6)$ . (correct answer)
Local maximum at $x=2$ ; concave down on $(0,6)$ .
Local minimum at $x=2$ ; concave up on $(0,6)$ .
No extremum at $x=2$ ; concave down on $(0,6)$ .
Local maximum at $x=2$ ; concave up on $(0,6)$ .

Explanation: This problem involves interpreting a graph of f′ and its behavior to determine properties of f. Since f′ is negative on (0,2), zero at x=2, and positive on (2,6), function f has a local minimum at x=2. The fact that f′ is decreasing on (0,6) means f″<0, so f is concave down throughout. Choice C might be tempting as it correctly identifies the local minimum but incorrectly claims concave up when f′ decreasing indicates f″<0 (concave down). The systematic approach is to use the first derivative test for extrema and f′ decreasing to conclude f″<0.

Question 7

Let $g$ be continuous on $[1,5]$ and have a global minimum at $x=3$ ; which statement must be true?

$x=3$ must be a critical point of $g$ .
$x=3$ cannot be a critical point because it is global.
If $x=3$ lies in $(1,5)$ , then $g'(3)=0$ .
If $x=3$ lies in $(1,5)$ and $g$ is differentiable at $3$ , then $g'(3)=0$ . (correct answer)
A global minimum can only occur at an endpoint.

Explanation: This problem analyzes the conditions under which an interior point with a global minimum must be a critical point. Since $g$ is continuous on $[1,5]$ and has a global minimum at $x=3$ , we need to determine when this interior point must be a critical point. If $x=3$ lies in the interior $(1,5)$ and $g$ is differentiable at $x=3$ , then by Fermat's theorem, we must have $g'(3)=0$ , making it a critical point. However, if $g$ is not differentiable at $x=3$ , then $x=3$ is still a critical point because critical points include both points where the derivative equals zero and points where the derivative doesn't exist. Choice C omits the differentiability condition, while choice A incorrectly assumes all global minima are critical points regardless of location. The precise principle: interior absolute extrema of continuous functions are critical points (either $f'(x)=0$ or $f'(x)$ doesn't exist).

Question 8

For $M(x)=\begin{cases}x^2,&x<3\\9,&x=3\\x^2+1,&x>3\end{cases}$ , is $M$ continuous at $x=3$ , and why?

Yes; $\lim_{x\to3}M(x)=9$ and $M(3)=9$ .
No; $\lim_{x\to3^-}M(x)=9$ and $\lim_{x\to3^+}M(x)=10$ , so the limit does not exist. (correct answer)
No; $\lim_{x\to3}M(x)=10$ but $M(3)=9$ .
No; $M(3)$ is undefined.
Yes; the right-hand limit equals $10$ , so it is continuous.

Explanation: At x=3, M(3)=9, left 9, right 10, DNE, not continuous. Jump. Common: ignoring sides. Discontinuity. Checklist: (1) Defined? (2) Limit exists? (3) Equals?

Question 9

If $f''(x)<0$ on $( -3,1)$ and $f'(x)$ changes from negative to positive at $x=-1$ , what happens to $f$ at $x=-1$ ?

$f$ has a local maximum at $x=-1$ .
$f$ has a local minimum at $x=-1$ . (correct answer)
$f$ has an inflection point at $x=-1$ .
$f$ has neither a local extremum nor an inflection point at $x=-1$ .
$f$ must be increasing and concave up at $x=-1$ .

Explanation: This problem uses multi-representation reasoning to analyze critical points and concavity. Since f'(x) changes from negative to positive at x = -1, and assuming f' is continuous, we have f'(-1) = 0 with f increasing after this point. This sign change in f' from negative to positive indicates a local minimum at x = -1. The condition f''(x) < 0 on (-3,1) tells us f is concave down at x = -1, which doesn't prevent the minimum. Choice A incorrectly suggests a maximum, which would require f' to change from positive to negative. When f' changes from negative to positive at a critical point, there's always a local minimum regardless of concavity.

Question 10

A control system requires $\int \frac{\sin x}{1+\cos x}\,dx$ ; which technique is most appropriate?

Partial fraction decomposition
Integration by parts
Substitution with $u=1+\cos x$ (correct answer)
Trigonometric substitution
Polynomial long division

Explanation: This question tests the skill of selecting an appropriate technique for antidifferentiation. The integrand $\sin x / (1 + \cos x)$ has $\sin x$ in the numerator, which is the negative derivative of $\cos x$ in the denominator, suggesting substitution. Letting $u = 1 + \cos x$ , $du = -\sin x \, dx$ , gives $-\int \frac{du}{u} = -\ln|u| + C$ . This simplifies the trig rational directly. Partial fractions might be tempting if misreading as rational in x, but the trig functions make it inapplicable. Identify substitutions in trig rationals where the numerator matches the derivative of part of the denominator.

Question 11

Let $y=\cos x$ and $y=-\cos x$ on $[0,2\pi]$ . Which setup gives total area between curves?

$\displaystyle \int_{0}^{2\pi}(\cos x-(-\cos x))\,dx$
$\displaystyle \int_{0}^{2\pi}((-\cos x)-\cos x)\,dx$
$\displaystyle \int_{0}^{2\pi}(\cos x+(-\cos x))\,dx$
$\displaystyle \int_{0}^{\pi/2}(\cos x-(-\cos x))dx+\int_{\pi/2}^{3\pi/2}((-\cos x)-\cos x)dx+\int_{3\pi/2}^{2\pi}(\cos x-(-\cos x))dx$ (correct answer)
$\displaystyle \int_{0}^{2\pi}(\cos^2 x)\,dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=π/2 and x=3π/2 because those are where the order changes for y=cos x and y=-cos x. From 0 to π/2, y=cos x is above y=-cos x. From π/2 to 3π/2, y=-cos x is above y=cos x in parts. From 3π/2 to 2π, y=cos x is above y=-cos x. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 12

Find the washer-method integral when the region between $y=\sqrt{x}+2$ and $y=\tfrac{x}{3}+1$ on $[1,4]$ is revolved about the x-axis.

$\pi\int_{1}^{4}\left[\left(\tfrac{x}{3}+1\right)^2-(\sqrt{x}+2)^2\right]dx$
$\pi\int_{1}^{4}\left[(\sqrt{x}+2)^2-\left(\tfrac{x}{3}+1\right)^2\right]dx$ (correct answer)
$\pi\int_{1}^{4}(\sqrt{x}+2)^2dx$
$\pi\int_{1}^{4}\left[(\sqrt{x}+2)-\left(\tfrac{x}{3}+1\right)\right]^2dx$
$\pi\int_{1}^{4}\left[(\sqrt{x}+2)^2+\left(\tfrac{x}{3}+1\right)^2\right]dx$

Explanation: The washer method is used to find the volume of a solid formed by revolving a region between two curves around an axis, creating cross-sections that are washers. For revolution about the x-axis, the outer radius is the y-value of the upper curve, and the inner radius is the y-value of the lower curve. Here, y=√x+2 is above y=(x/3)+1 on [1,4], so the outer radius is √x+2 and the inner radius is (x/3)+1. Thus, the volume integral is π ∫ from 1 to 4 of [(√x+2)^2 - ((x/3)+1)^2] dx. A tempting distractor like choice A subtracts wrongly, giving negative volume. For washer setups, checklist: confirm the axis, select integration variable, set limits, identify outer (farther) and inner (closer) distances to axis, and subtract inner squared from outer squared.

Question 13

For $f(x)=\int_0^x g(t)\,dt$ , where $g$ is decreasing on $(-2,3)$ and increasing on $(3,6)$ , where is $f$ concave down?

On $(-2,3)$ only (correct answer)
On $(3,6)$ only
On $(-2,6)$
On $(-2,3)\cup(3,6)$
Nowhere on $(-2,6)$

Explanation: This question tests understanding of concavity for a function defined as an integral. Since f(x) = ∫₀ˣ g(t)dt, we have f'(x) = g(x) by the Fundamental Theorem of Calculus, and thus f''(x) = g'(x). The function f is concave down where f''(x) < 0, which means where g'(x) < 0, occurring when g is decreasing. Since g is decreasing on (-2,3) and increasing on (3,6), we have f concave down on (-2,3) only. Students might confuse this with where g itself is negative, but concavity of f depends on the monotonicity of g. Key insight: for f(x) = ∫ₐˣ g(t)dt, f is concave up/down where g is increasing/decreasing.

Question 14

A Norman window is a rectangle topped by a semicircle with fixed perimeter 12 m; what objective is optimized?

Minimize total perimeter, $P$ , subject to area fixed.
Maximize rectangle width, $w$ , subject to perimeter 12.
Minimize semicircle radius, $r$ , subject to perimeter 12.
Maximize total area of the window, $A$ , subject to perimeter 12 m. (correct answer)
Maximize total perimeter, $P$ , subject to perimeter 12.

Explanation: This problem optimizes a Norman window design with fixed perimeter. A Norman window consists of a rectangle (width w, height h) topped by a semicircle (radius r = w/2), giving total perimeter P = 2h + w + πw/2 = 12 m. To maximize natural light, we want to maximize the total area A = wh + πw²/8. Option E incorrectly suggests maximizing perimeter when it's already constrained at 12 m. The optimization principle here is maximizing benefit (window area for light) while respecting the constraint (fixed frame material).

Question 15

A particle has acceleration $a(t)=4$ (m/s $^2$ ), initial velocity $v(0)=-3$ , and $x(0)=10$ . What is $x(2)$ ?

$4\text{ m}$
$8\text{ m}$
$12\text{ m}$ (correct answer)
$16\text{ m}$
$20\text{ m}$

Explanation: This problem requires two integrations: first from acceleration to velocity, then from velocity to position. First, v(t) = v(0) + ∫[0 to t] a(s) ds = -3 + ∫[0 to t] 4 ds = -3 + 4t. Then x(t) = x(0) + ∫[0 to t] v(s) ds = 10 + ∫[0 to t] (-3 + 4s) ds = 10 + [-3s + 2s²]₀^t = 10 - 3t + 2t². At t = 2: x(2) = 10 - 6 + 8 = 12. A common mistake is using a(t) = 4 directly without first finding velocity, which would incorrectly give x(2) = 10 + 8 = 18. The strategy: when given acceleration, integrate twice—once for velocity, once for position—always including initial conditions.

Question 16

The region bounded by $y=\cos x$ and the x-axis on $0\le x\le \frac{\pi}{2}$ is revolved about the x-axis. Which integral gives the volume?

$\pi\int_{0}^{\pi/2} (\cos x)^2\,dx$ (correct answer)
$\pi\int_{0}^{\pi/2} \cos x\,dx$
$\pi\int_{0}^{\pi} (\cos x)^2\,dx$
$\pi\int_{0}^{\pi/2} (\cos x)^2\,dy$
$\pi\int_{0}^{\pi/2} (\sin x)^2\,dx$

Explanation: This problem uses the disc method for revolution about the x-axis. Since the region is bounded by y = cos x and the x-axis with no inner boundary, we apply discs rather than washers. The radius of each disc at position x is the distance from the x-axis to the curve: r(x) = cos x. The volume formula becomes V = π∫[a to b] [r(x)]² dx = π∫[0 to π/2] (cos x)² dx. Choice B is tempting but fails because it doesn't square the radius function, which is required by the disc method. Use the disc method when the region extends from the axis of revolution to exactly one bounding curve.

Question 17

A rectangular garden uses 80 m of edging on all four sides. What objective quantity should be maximized?

The garden’s area (correct answer)
The garden’s perimeter
The garden’s length
The garden’s width
The difference between length and width

Explanation: This question tests the skill of setting up optimization problems by identifying the quantity to be maximized with a fixed perimeter for a garden. The rectangular garden uses 80 m of edging on all four sides, so the objective is to maximize the garden's area to get the most space for planting. The perimeter constraint is 2(length + width) = 80, allowing us to express area as length times width in terms of one variable. This setup comes from efficiently using limited edging to enclose the largest possible area. A tempting distractor is choice B, the garden’s perimeter, but this is fixed at 80 m and not what we maximize. In optimization problems with fixed perimeters, model the area as the objective function and use the perimeter constraint to reduce variables for finding maxima.

Question 18

A farm’s egg production $E(d)$ increases by 21 eggs when feed increases by 7 lb. What is $\Delta E/\Delta d$ ?

3 eggs/lb (correct answer)
$\tfrac{1}{3}$ lb/egg
28 eggs/lb
14 eggs/lb
3 lb/egg

Explanation: This question tests understanding of rate calculation when relating two different quantities. Egg production increases by 21 eggs when feed increases by 7 lb, so $\Delta E / \Delta d$ = 21 eggs / 7 lb = 3 eggs/lb. This represents the production efficiency: how many additional eggs result from each additional pound of feed. Students might calculate the reciprocal (lb per egg) or make arithmetic errors. The units eggs/lb show the relationship between feed input and egg output. For agricultural efficiency rates, calculate output change divided by input change to determine productivity per unit of resource invested.

Question 19

The concentration of a dye is $c(t)$ mg/L, where $t$ is minutes after mixing begins; interpret $c'(4)$ .

The dye concentration 4 minutes after mixing begins, in mg/L
The instantaneous rate the concentration is changing at 4 minutes, in mg/L per minute (correct answer)
The average concentration over the first 4 minutes, in mg/L
The instantaneous rate minutes are changing with respect to concentration at 4 minutes, in minutes per mg/L
The total amount of dye present at 4 minutes, in mg

Explanation: This problem requires interpreting c'(4) in a chemistry context. Since c(t) represents dye concentration in mg/L as a function of time in minutes, c'(4) measures how rapidly the concentration is changing at exactly 4 minutes after mixing begins, with units of mg/L per minute. Students often mistakenly think c'(4) gives the concentration value at 4 minutes (that's c(4), choice A) or confuse concentration with total amount (choice E). The derivative captures the instantaneous rate of change of concentration, not the concentration itself or total quantities. When interpreting derivatives in scientific contexts, always identify what's changing (concentration) with respect to what (time), and express the rate with compound units that reflect 'output per input.'

Question 20

A signal’s total energy is $\int_0^{\pi/2} \sin(3x)\cos(3x)\,dx$ ; which technique is most appropriate?

Integration by parts
Partial fraction decomposition
Algebraic manipulation using a trig identity (correct answer)
Trigonometric substitution
Long division then integrate

Explanation: This question tests the skill of selecting an appropriate technique for antidifferentiation. The integrand is a product of sine and cosine with the same argument, which suggests using a trigonometric identity to simplify before integrating. The double-angle identity sin(2θ) = 2 sin θ cos θ allows rewriting sin(3x) cos(3x) as (1/2) sin(6x), leading to a straightforward integral of (1/2) ∫ sin(6x) dx. This algebraic manipulation fits perfectly for products of like trig functions. Integration by parts might be tempting for the product form, but it would lead to more complex terms without simplification, whereas the identity reduces it efficiently. Look for trig products that match known identities to simplify the integrand before choosing other methods.

Question 21

For $y=\arcsin(\tfrac{2}{3}x+1)$ , what is $\dfrac{dy}{dx}$ ?

$\dfrac{2/3}{\sqrt{1-(\tfrac{2}{3}x+1)^2}}$ (correct answer)
$\dfrac{1}{\sqrt{1-(\tfrac{2}{3}x+1)^2}}$
$\dfrac{2}{\sqrt{1-(\tfrac{2}{3}x+1)^2}}$
$\dfrac{2/3}{\sqrt{1-(\tfrac{2}{3}x+1)}}$
$\cos\!\left(\tfrac{2}{3}x+1\right)$

Explanation: This question tests the skill of differentiating inverse trigonometric functions, specifically the inverse sine. The derivative of arcsin(u) is 1/√(1 - u²) times du/dx. u = (2/3)x + 1, du/dx = 2/3, positive. Sine inverse. Result: (2/3) / √(1 - ((2/3)x + 1)²). Choice C doubles coefficient, wrong. Pattern: arcsin positive scaled by coefficient.

Question 22

Which slope field corresponds to $\dfrac{dy}{dx}=\sin x$ ?

Slopes depend only on $y$ and repeat periodically as $y$ increases.
Slopes depend only on $x$ and alternate sign periodically in vertical columns; $0$ at multiples of $\pi$ . (correct answer)
Slopes are always positive but approach $0$ as $x$ increases.
Slopes are $0$ along $y=\sin x$ ; positive above the curve and negative below.
Slopes are constant along diagonals $y-x=\text{constant}$ .

Explanation: This question requires sketching a slope field for dy/dx = sin x, where slopes depend only on the x-coordinate. All points in the same vertical column share identical slopes since the equation involves only x. Slopes are zero when sin x = 0, which occurs at x = kπ for integer k. Slopes alternate between positive and negative as x increases, following the periodic pattern of sine. Between zeros, slopes reach maximum magnitude of 1 at x = π/2 + kπ. Choice A incorrectly suggests slopes depend on y, but sin x involves only the x-variable. For trigonometric differential equations, identify the zeros and extrema of the trig function to understand the periodic slope pattern.

Question 23

Does $y=xe^{x}$ satisfy the differential equation $\dfrac{dy}{dx}=y+e^{x}$ for all $x$ ?

No, because $y'=e^{x}$ but $y+e^{x}=xe^{x}+e^{x}$ .
Yes, because $y'=xe^{x}+e^{x}$ and $y+e^{x}=xe^{x}+e^{x}$ . (correct answer)
No, because $y'=xe^{x}$ but $y+e^{x}=xe^{x}+e^{x}$ .
Yes, because $y'=e^{x}(x-1)$ equals $y+e^{x}$ .
No, because $y'=x e^{x-1}$ but $y+e^{x}=xe^{x}+e^{x}$ .

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 = xe^x, we find y' = e^x + xe^x = e^x(1 + x) using the product rule. The differential equation dy/dx = y + e^x requires that y' equals y + e^x. We compute y + e^x = xe^x + e^x = e^x(x + 1). Since y' = e^x(1 + x) and y + e^x = e^x(x + 1), both sides are equal. Choice C incorrectly states y' = xe^x, forgetting to apply the product rule completely. When differentiating products, use the product rule: d/dx[uv] = u'v + uv'.

Question 24

Define $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f$ is increasing on $(0,8)$ , which statement about $A$ is correct on $(0,8)$ ?

$A$ is concave up on $(0,8)$ (correct answer)
$A$ is concave down on $(0,8)$
$A$ is increasing on $(0,8)$
$A$ is decreasing on $(0,8)$
$A(x)=f(x)$ on $(0,8)$

Explanation: This problem tests understanding of accumulation function concavity based on integrand monotonicity. Since $A'(x) = f(x)$ and $A''(x) = f'(x)$ by differentiation, the concavity of $A$ depends on whether $f$ is increasing or decreasing. Given that $f$ is increasing on $(0,8)$ , we have $f'(x) > 0$ , making $A''(x) > 0$ . This means $A$ is concave up on $(0,8)$ . The monotonicity of the integrand directly determines the concavity of the accumulation function. Choice B would be incorrect since concave down requires $f$ to be decreasing. When the integrand is increasing, the accumulation function is concave up.

Question 25

A runner’s speed is $s(t)$ miles/hour at time $t$ hours. What does $\int_{0.5}^{1.5} s(t)\,dt$ represent?

The runner’s average speed from $t=0.5$ to $t=1.5$ , in miles per hour
The runner’s change in speed from $t=0.5$ to $t=1.5$ , in miles per hour
The total distance the runner travels from $t=0.5$ to $t=1.5$ , in miles (correct answer)
The runner’s speed at $t=1.5$ , in miles per hour
The time needed to run 1 mile, in hours

Explanation: This problem involves interpreting accumulation in a running context. Since s(t) represents the runner's speed in miles per hour, the definite integral ∫₀.₅¹·⁵ s(t)dt accumulates all the instantaneous speeds over the 1-hour interval from t=0.5 to t=1.5. This gives the total distance traveled during this time period, measured in miles. Choice A might seem reasonable mentioning average speed, but that would require dividing the integral by 1. Remember: speed integrated over time equals distance, with units (miles/hour) × (hours) = miles.

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