Answer: $\dfrac{32}{3}$

Set up the integration in y: The curves are $x = y^2$ (left) and $x = 4$ (right). The intersection occurs when $y^2 = 4$, so $y = -2$ and $y = 2$. Apply right minus left: Area $= \int_{-2}^{2} [4 - y^2]\,dy$. Evaluate: $\int_{-2}^{2}(4-y^2)\,dy = \left[4y - \frac{y^3}{3}\right]_{-2}^{2} = \left(8-\frac{8}{3}\right) - \left(-8+\frac{8}{3}\right) = \frac{16}{3}+\frac{16}{3} = \frac{32}{3}$. Why distractors fail: Option A ($16/3$) only integrates from $0$ to $2$, missing the symmetric lower half. Option C ($64/3$) doubles the correct answer, likely from an arithmetic error. Option D ($8$) comes from the rectangle $4 \times (2-(-2)) = 16$ divided by $2$ erroneously.

Answer: The setup is valid; the curves intersect at $x = 1-\sqrt{3}$ and $x = 1+\sqrt{3}$, and since $2x+1 \ge x^2-1$ on $[0,2]$, the integral correctly computes an area of $\dfrac{16}{3}$.

Check intersection points: Setting $2x+1 = x^2-1$ gives $x^2-2x-2=0$. Using the quadratic formula: $x = 1 \pm \sqrt{3}$. So $x \approx -0.73$ and $x \approx 2.73$. The interval $[0,2]$ lies entirely between these intersection points. Verify the integrand sign on [0, 2]: Since $1-\sqrt{3} < 0$ and $1+\sqrt{3} > 2$, the line $y=2x+1$ is above $y=x^2-1$ on the entire interval $[0,2]$. The integrand $(2x+1)-(x^2-1) = -x^2+2x+2$ is non-negative throughout. Compute the integral: $\int_0^2 (-x^2+2x+2)\,dx = \left[-\frac{x^3}{3}+x^2+2x\right]_0^2 = -\frac{8}{3}+4+4 = \frac{16}{3}$. Why distractors fail: Option A is wrong: although the intersections are not exactly at $0$ and $2$, the integral validly computes area on $[0,2]$ since the top/bottom relationship does not change. Option B is wrong: $2x+1$ is on top, not $x^2-1$ (test $x=1$: $3 > 0$). Option C is wrong: taking absolute value is unnecessary when the integrand is already non-negative throughout the interval.

Answer: $\dfrac{32}{3}$

Find intersection points: Set $x^2 = 4$, so $x = -2$ and $x = 2$. The enclosed region extends from $x = -2$ to $x = 2$. Set up the integral: Since $y = 4$ is on top and $y = x^2$ is on the bottom throughout $[-2, 2]$, the area is $\int_{-2}^{2} [4 - x^2]\,dx$. Evaluate: $\int_{-2}^{2} (4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$. Why distractors fail: Option A ($16/3$) is only half the area — a common error from integrating on $[0,2]$ instead of $[-2,2]$. Option C ($16$) results from computing $4 \times (2-(-2))$ without subtracting $x^2$. Option D ($8$) likely comes from area-of-rectangle reasoning without calculus.

Answer: Yes, because $\int_a^b [f(x)-g(x)]\,dx = [F(x)-G(x)]_a^b = [F(b)-G(b)]-[F(a)-G(a)]$.

Apply the Fundamental Theorem to the difference: Since $f-g$ is continuous on $[a,b]$ and $F-G$ is an antiderivative of $f-g$, the FTC gives $\int_a^b[f(x)-g(x)]\,dx = [F(b)-G(b)]-[F(a)-G(a)]$. Why the correct answer works: Option C correctly applies the linearity of integration and the FTC. The integral of a difference is the difference of the integrals, and each integral can be evaluated using its respective antiderivative. Why distractors fail: Option A incorrectly claims the FTC does not apply to differences — it does, by the linearity of integration. Option B incorrectly asserts that numerical methods are required when curves touch at endpoints; the FTC works regardless. Option D incorrectly restricts the result to polynomials; the FTC applies to all continuous functions with antiderivatives.

Answer: $\int_a^b [f(x) - g(x)]\,dx$

Recall the area-between-curves formula: When $f(x) \ge g(x)$ on $[a, b]$, the area between the curves is given by $\int_a^b [f(x) - g(x)]\,dx$. This subtracts the lower curve from the upper curve across the interval. Why the correct answer works: Option B is exactly the standard formula: integrate the difference (top minus bottom) over the interval. Since $f(x) \ge g(x)$, this difference is non-negative, producing a positive area. Why distractors fail: Option A reverses the subtraction, yielding a negative value. Option C multiplies the functions instead of subtracting them, which has no geometric meaning for area between curves. Option D adds the functions, which would compute the integral of their sum rather than the region between them.

Answer: $\dfrac{1}{2}$

Find intersection points: Set $x^3 = x$, so $x^3 - x = 0$ gives $x(x^2 - 1)=0$, yielding $x = -1, 0, 1$. Determine which function is on top in each sub-interval: On $[-1, 0]$: test $x = -0.5$: $(-0.5)^3 = -0.125$ and $-0.5$, so $x^3 > x$ (the cubic is on top). On $[0, 1]$: test $x = 0.5$: $(0.5)^3 = 0.125$ and $0.5$, so $x > x^3$ (the line is on top). Evaluate the integrals: Area $= \int_{-1}^{0}(x^3 - x)\,dx + \int_0^1 (x - x^3)\,dx$. Each evaluates to $\left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^{0} = 0 - (\frac{1}{4}-\frac{1}{2}) = \frac{1}{4}$ and $\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$. Total $= \frac{1}{2}$. Why distractors fail: Option A ($0$) results from computing the signed integral $\int_{-1}^1(x^3-x)\,dx$ without splitting. Option B ($1/4$) counts only one of the two symmetric enclosed regions. Option D ($1$) likely doubles an incorrect computation.

Answer: The student correctly computed the net signed area, but area requires integrating $|\sin(x)|$, which accounts for the region below the $x$-axis.

Distinguish signed area from total area: The definite integral $\int_0^{2\pi} \sin(x)\,dx$ computes the net signed area. Since $\sin(x)$ is positive on $(0, \pi)$ and negative on $(\pi, 2\pi)$, the positive and negative areas cancel, giving $0$. Why the correct answer works: Option B correctly identifies the error: total area requires $\int_0^{2\pi} |\sin(x)|\,dx$, which equals $\int_0^{\pi} \sin(x)\,dx + \int_{\pi}^{2\pi} [-\sin(x)]\,dx = 2 + 2 = 4$. Why distractors fail: Option A is irrelevant—integration by parts is not needed for $\sin(x)$. Option C is wrong because the antiderivative of $\sin(x)$ is $-\cos(x)$, not $\sin(x)$, but the student's numerical result of $0$ is actually correct for the signed integral. Option D would only capture half the enclosed region, ignoring the area between $\sin(x)$ and $0$ on $(\pi, 2\pi)$.

Answer: $\int_0^1 (e - e^y)\,dy$

Determine the bounds in y: At $x = e$, $y = \ln(e) = 1$. At $y = 0$, $x = e^0 = 1$. So the $y$-bounds are from $0$ to $1$. Set up horizontal slicing: For a fixed $y \in [0,1]$, the right boundary is $x = e$ and the left boundary is $x = e^y$ (from $y = \ln(x)$, i.e., $x = e^y$). So the width of each horizontal strip is $e - e^y$. Why the correct answer works: Option B gives $\int_0^1 (e - e^y)\,dy$, which integrates the right-minus-left width over the correct $y$-range. Why distractors fail: Option A integrates $e^y$ alone without subtracting from the right boundary $x = e$. Option C uses $\ln(y)$ integrated over $[0, e]$, confusing the roles of $x$ and $y$. Option D reverses the subtraction, giving a negative integrand.

Answer: The student did not split at $x = \pi/4$ where the curves cross; the correct area is $2\sqrt{2}$.

Identify the crossing point: $\cos(x) = \sin(x)$ when $\tan(x) = 1$, so $x = \pi/4$ on $[0,\pi]$. On $[0, \pi/4]$, $\cos(x) \ge \sin(x)$; on $[\pi/4, \pi]$, $\sin(x) \ge \cos(x)$. Set up the correct integral: Area $= \int_0^{\pi/4}(\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi}(\sin x - \cos x)\,dx$. Evaluate: $\int_0^{\pi/4}(\cos x - \sin x)\,dx = [\sin x + \cos x]_0^{\pi/4} = (\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}) - (0+1) = \sqrt{2}-1$. $\int_{\pi/4}^{\pi}(\sin x - \cos x)\,dx = [-\cos x - \sin x]_{\pi/4}^{\pi} = (1-0) - (-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}) = 1+\sqrt{2}$. Total $= (\sqrt{2}-1)+(1+\sqrt{2}) = 2\sqrt{2}$. Why distractors fail: Option A ($0$) is the value of the student's unsplit integral, not the area. Option C only reverses the order but does not split, yielding $0$ again (or a sign error). Option D's value of $-2$ for the original integral is arithmetically wrong, and taking absolute value of a single unsplit integral does not generally give the correct area.

Answer: Integrate with respect to $y$ using horizontal slices: $\int [\text{right} - \text{left}]\,dy$

Identify the key principle: When curves are expressed as $x = h(y)$ and $x = k(y)$, horizontal slicing with integration in $y$ is the most natural and often simplest approach. Why the correct answer works: Option C correctly applies the right-minus-left formula: $\int_c^d [h(y) - k(y)]\,dy$ where $h(y) \ge k(y)$. This uses horizontal strips whose width is the horizontal distance between the curves. Why distractors fail: Option A uses vertical slices, which requires the curves as functions of $x$ — the opposite of what is given. Option B may be algebraically difficult or impossible (e.g., if the function is not invertible). Option D introduces polar coordinates, which is unnecessary and inappropriate for this setting.

Answer: $\dfrac{1}{2}$

Identify the sub-regions: The function $y = x^3 - x = x(x-1)(x+1)$ crosses the $x$-axis at $x = -1, 0, 1$. On $[-1, 0]$, the cubic is above the $x$-axis; on $[0, 1]$, it is below. Set up split integrals: Area $= \int_{-1}^{0}(x^3 - x)\,dx + \int_0^1 [0 - (x^3 - x)]\,dx = \int_{-1}^{0}(x^3 - x)\,dx + \int_0^1 (x - x^3)\,dx$. Evaluate each integral: $\int_{-1}^{0}(x^3-x)\,dx = \left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^{0} = 0 - \left(\frac{1}{4}-\frac{1}{2}\right) = \frac{1}{4}$. Similarly, $\int_0^1(x-x^3)\,dx = \left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1 = \frac{1}{2}-\frac{1}{4} = \frac{1}{4}$. Total area $= \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$. Why distractors fail: Option A ($0$) is the signed integral $\int_{-1}^1 (x^3-x)\,dx$, which cancels due to symmetry — a common trap. Option B ($1/4$) computes only one of the two sub-regions. Option D ($1$) likely results from an algebraic error in the antiderivative.

Answer: $\dfrac{1}{3}$

Find intersection points: Set $\sqrt{x} = x^2$, so $x = x^4$ which gives $x^4 - x = 0$, i.e., $x(x^3 - 1) = 0$. For $x \ge 0$: $x = 0$ and $x = 1$. Determine which curve is on top: For $0 \le x \le 1$, test $x = 0.25$: $\sqrt{0.25} = 0.5$ and $(0.25)^2 = 0.0625$. So $\sqrt{x} \ge x^2$ on this interval. Evaluate the integral: $\int_0^1 (\sqrt{x} - x^2)\,dx = \int_0^1 (x^{1/2} - x^2)\,dx = \left[\frac{2x^{3/2}}{3} - \frac{x^3}{3}\right]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$. Why distractors fail: Option A ($1/6$) could result from incorrectly computing $\int_0^1 x^2\,dx - \int_0^1 \sqrt{x}\,dx$ and taking the absolute value of a partial computation. Option B ($1/2$) might come from misremembering $\int_0^1 \sqrt{x}\,dx = 2/3$ alone. Option D ($2/3$) is just $\int_0^1 \sqrt{x}\,dx$ without subtracting the lower curve.

Answer: $\int_0^{2} 2\sqrt{y}\,dy + \int_2^{4} 2\sqrt{4-y}\,dy$

Find intersection and rewrite curves: Setting $4-x^2 = x^2$ gives $x^2 = 2$, so the intersection is at $y = 2$. From $y = x^2$: $x = \pm\sqrt{y}$. From $y = 4-x^2$: $x = \pm\sqrt{4-y}$. Analyze horizontal slices at different y-levels: For $0 \le y \le 2$: the enclosed region spans from $x = -\sqrt{y}$ to $x = \sqrt{y}$ (bounded by the lower parabola), giving width $2\sqrt{y}$. For $2 \le y \le 4$: the region spans from $x = -\sqrt{4-y}$ to $x = \sqrt{4-y}$ (bounded by the upper parabola), giving width $2\sqrt{4-y}$. Why the correct answer works: Option C correctly splits at $y=2$ and uses the appropriate width function for each sub-region. Why distractors fail: Option A uses only the upper parabola boundary over $[0,4]$, ignoring the lower parabola. Option B incorrectly subtracts the two widths in the lower region. Option D uses only the lower parabola boundary over $[0,4]$.

Answer: $x = 0$ and $x = 3$; $y = -x^2 + 2x + 3$ is on top

Solve for intersection points: $x^2 - 4x + 3 = -x^2 + 2x + 3$ simplifies to $2x^2 - 6x = 0$, so $2x(x - 3) = 0$, giving $x = 0$ and $x = 3$. Determine which curve is on top: Test $x = 1$: $y_1 = 1 - 4 + 3 = 0$ and $y_2 = -1 + 2 + 3 = 4$. Since $y_2 > y_1$, the downward-opening parabola $y = -x^2+2x+3$ is on top on $(0,3)$. Why distractors fail: Option A has the correct intersection points but reverses which curve is on top. Option C has incorrect intersection points ($x=1$ is not a solution of $2x^2-6x=0$). Option D uses $x=-1$, which does not satisfy the equation.

Answer: The claim is correct; $\int_{-1}^{2}|x^2-x|\,dx$ equals the total area between the curves on $[-1,2]$, which is $\dfrac{11}{6}$.

Identify where the curves cross: $x^2 = x$ gives $x(x-1)=0$, so the curves cross at $x = 0$ and $x = 1$. Analyze the absolute value integral: On $[-1, 0]$: $x^2 - x \ge 0$, so $|x^2-x| = x^2-x$. On $[0, 1]$: $x^2-x \le 0$, so $|x^2-x| = x - x^2$. On $[1,2]$: $x^2-x \ge 0$, so $|x^2-x| = x^2-x$. Compute each piece: $\int_{-1}^{0}(x^2-x)\,dx = [x^3/3 - x^2/2]_{-1}^0 = 0 - (-1/3 - 1/2) = 5/6$. $\int_0^1(x-x^2)\,dx = [x^2/2 - x^3/3]_0^1 = 1/6$. $\int_1^2(x^2-x)\,dx = [x^3/3-x^2/2]_1^2 = (8/3-2)-(1/3-1/2) = 2/3+1/6 = 5/6$. Total $= 5/6+1/6+5/6 = 11/6$. Why distractors fail: Option A is wrong — the absolute value correctly captures the area between the two curves at every point, including portions outside the enclosed loop. Option C only considers the enclosed region between $x=0$ and $x=1$, but the integral is taken over $[-1,2]$, which captures additional area. Option D ($1/2$) is numerically incorrect.

Answer: The cumulative difference in population between the two groups over 10 years (in person-years)

Interpret the area-between-curves integral in context: The integral $\int_0^{10}[P_1(t)-P_2(t)]\,dt$ sums the instantaneous difference $P_1(t)-P_2(t)$ over time. Geometrically it is the area between the two population curves. Why the correct answer works: Option B correctly identifies this as the accumulated difference over the time interval, measured in units of population $\times$ time (person-years). It represents how much more total 'population-time' group 1 has compared to group 2. Why distractors fail: Option A would be $\int_0^{10}[P_1(t)+P_2(t)]\,dt$, the sum, not the difference. Option C confuses an integral (accumulated quantity) with an evaluation at a single point. Option D describes $\frac{1}{10}\int_0^{10}[P_1(t)-P_2(t)]\,dt$, not the integral itself.

Answer: Set $x^2 = x + 2$ and solve for $x$

Identify the key principle: To find where two curves enclose a region, you must find their intersection points by setting $f(x) = g(x)$. Why the correct answer works: Option B sets $x^2 = x + 2$, which is the standard method for finding intersection points. Solving gives $x^2 - x - 2 = 0$, so $(x-2)(x+1)=0$, yielding $x = -1$ and $x = 2$. Why distractors fail: Option A only finds where the parabola crosses the $x$-axis, not where the two curves meet. Option C sets the derivatives equal, which finds where slopes match (not intersections). Option D arbitrarily guesses limits without solving for the actual intersection points.

Answer: $\int_{-2}^{2} 2\sqrt{y+2}\,dy + \int_{2}^{6} 2\sqrt{6-y}\,dy$

Rewrite curves as functions of y: From $y = 6-x^2$: $x = \pm\sqrt{6-y}$. From $y = x^2-2$: $x = \pm\sqrt{y+2}$. The region is symmetric about $x = 0$. Identify the structure of horizontal slices: The two parabolas intersect where $6-x^2 = x^2-2$, giving $x^2=4$, so $x=\pm 2$ and $y=2$. For $-2 \le y \le 2$: both curves bound the region. The right boundary is $x = \sqrt{y+2}$ (from the lower parabola) and the left boundary is $x = -\sqrt{y+2}$, so the width is $2\sqrt{y+2}$. For $2 \le y \le 6$: only the upper parabola bounds the region. The width is $2\sqrt{6-y}$. Why the correct answer works: Option C correctly splits at $y=2$ and uses the appropriate width for each sub-region: $\int_{-2}^{2} 2\sqrt{y+2}\,dy + \int_{2}^{6} 2\sqrt{6-y}\,dy$. Why distractors fail: Option A uses only one expression over the full $y$-range, ignoring that the bounding curves change at $y=2$. Option B subtracts $\sqrt{y+2}$ from $\sqrt{6-y}$ on $[-2,2]$, which misidentifies the boundaries. Option D adds the two square roots over the entire range, which is not geometrically correct.