\(\frac{b^x}{\ln b} + C\)
14.) \(\int \frac{1}{1+x^2}\,dx\)
\(\arctan x + C\)
15.) \(\int \frac{1}{\sqrt{1-x^2}}\,dx\)
\(\arcsin x + C\)
*16.) \(\int \frac{1}{|x|\sqrt{x^2-1}}\,dx\)
\(\operatorname{arcsec} x + C\)
Key • Page 1 of 4
10 Points (2 pts each)
17.) \(\int (x^3 - 6x)\,dx\) Correct: (D)
• Power Rule of Integration: \(\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\) for \(n \ne -1\).
• \(\int x^3\,dx = \frac{x^4}{4}\) and \(\int 6x\,dx = 6\left(\frac{x^2}{2}\right) = 3x^2\).
• Combining terms gives \(\frac{x^4}{4} - 3x^2 + C\).
Distractor Analysis: (A) took the derivative; (B) failed to divide \(6x\) by 2; (C) divided \(x^4\) by 3.
18.) Trapezoidal Sum Approximation Correct: (B)
• Trapezoid formula for unequal subintervals: \(\text{Area}_i = \frac{R(t_{i-1}) + R(t_i)}{2} \cdot \Delta t_i\)
\([0, 3]\): \(\frac{5+9}{2}(3) = 7(3) = 21\)
\([3, 7]\): \(\frac{9+11}{2}(4) = 10(4) = 40\)
\([7, 10]\): \(\frac{11+15}{2}(3) = 13(3) = 39\)
• Total \(= 21 + 40 + 39 = 100\) gallons.
Distractor Analysis: (A) 84 is left Riemann sum; (D) 120 is right Riemann sum; (C) 112 is midpoint or arithmetic error.
19.) \(\int_4^0 \sqrt{16 - x^2}\,dx\) Correct: (A)
• \(y = \sqrt{16 - x^2}\) represents the upper semicircle of radius \(r = 4\) centered at \((0,0)\).
• On \([0, 4]\), the area represents a quarter circle: \(\frac{1}{4}\pi r^2 = \frac{1}{4}\pi(16) = 4\pi\).
• Reversing the limits switches the sign: \(\int_4^0 \sqrt{16-x^2}\,dx = -\int_0^4 \sqrt{16-x^2}\,dx = -4\pi\).
Distractor Analysis: (B) forgot the negative sign from reversed limits; (C) used semicircle instead of quarter circle.
20.) \(\int \left(e^x + \frac{4}{x}\right)\,dx\) Correct: (C)
• Sum rule: \(\int e^x\,dx + 4\int \frac{1}{x}\,dx = e^x + 4\ln|x| + C = 4\ln|x| + e^x + C\).
• Essential rule: \(\int x^{-1}\,dx = \ln|x| + C\). The absolute value bars are mandatory.
Distractor Analysis: (A) treated \(4/x\) as \(4x\); (B) swapped coefficient 4 onto \(e^x\); (D) used power rule incorrectly (\(x^0/0\)).
21.) Total Accumulation from Rate Model Correct: (D)
• By the Net Change Theorem, the definite integral of a rate of change \(r(t)\) over an interval \([a, b]\) gives the net accumulation: \(\text{Total} = \int_0^{10} r(t)\,dt\).
Distractor Analysis: (A) only gives the instantaneous rate at \(t=10\); (B) and (C) apply to the integral of the derivative \(r'(t)\), which gives change in rate rather than total emissions.
Key • Page 2 of 4
10 Points (2 pts each)
22.) \(\int \frac{7}{1+x^2}\,dx\) Correct: (D)
• Constant multiple rule: \(\int \frac{7}{1+x^2}\,dx = 7 \int \frac{1}{1+x^2}\,dx = 7\arctan(x) + C\).
Distractor Analysis: (A) took derivative using quotient rule; (B) and (E) mistakenly used logarithm without recognizing inverse trig form (which would require \(2x\) in numerator).
23.) \(\int_{-3}^3 (4x + 3)\,dx\) Correct: (A)
• Method 1 (Symmetry): \(4x\) is odd over symmetric interval \([-3, 3]\), so \(\int_{-3}^3 4x\,dx = 0\). The constant \(3\) gives \(3 \times (3 - (-3)) = 3 \times 6 = 18\).
• Method 2 (FTC): \(\left[2x^2 + 3x\right]_{-3}^3 = \left(2(9) + 9\right) - \left(2(9) - 9\right) = 27 - 9 = 18\).
Distractor Analysis: (C) 0 assumes the entire integrand is odd; (B) sign error; (E) 36 doubled without subtracting.
24.) \(\int_{-2}^4 \frac{|x|}{x}\,dx\) Correct: (C)
• Definition of \(\frac{|x|}{x}\): \(-1\) for \(x < 0\) and \(+1\) for \(x > 0\) (removable discontinuity at \(x = 0\)).
• \(\int_{-2}^0 (-1)\,dx + \int_0^4 (1)\,dx = (-1)(0 - (-2)) + (1)(4 - 0) = -2 + 4 = 2\).
Distractor Analysis: (E) believes jump discontinuity prevents integration; (A) missed positive interval; (B) calculated for \([-1, 2]\).
25.) Antiderivative of \(6x^2 + \sqrt[3]{x} + \sin x\) Correct: (B)
• Rewrite with rational exponents: \(f(x) = 6x^2 + x^{1/3} + \sin x\).
• Integrate each term: \(\int 6x^2\,dx = 2x^3\); \(\int x^{1/3}\,dx = \frac{x^{4/3}}{4/3} = \frac{3\sqrt[3]{x^4}}{4}\); \(\int \sin x\,dx = -\cos x\).
• Result: \(F(x) = 2x^3 + \frac{3\sqrt[3]{x^4}}{4} - \cos x + C\).
Distractor Analysis: (A) and (C) differentiated \(6x^2\) instead of integrating; (C) had wrong sign for \(\cos x\).
26.) \(\int_0^4 |x - 1|\,dx\) Correct: (D)
• Vertex of \(V\)-shape is at \(x = 1\), splitting into two right triangles:
• On \([0, 1]\): base \(= 1\), height \(= |0 - 1| = 1 \implies \text{Area}_1 = \frac{1}{2}(1)(1) = 0.5\).
• On \([1, 4]\): base \(= 3\), height \(= |4 - 1| = 3 \implies \text{Area}_2 = \frac{1}{2}(3)(3) = 4.5\).
• Total \(= 0.5 + 4.5 = 5\).
Distractor Analysis: (A) integrated without absolute value: \([x^2/2 - x]_0^4 = 8 - 4 = 4\); (B) calculation error on triangle 2.
Key • Page 3 of 4
FRQ Scoring Rubric
Calculator Permitted
Part (a): Interpretation & Right Riemann Sum 2 Points
Interpretation [1 pt]: "\(\int_{60}^{135} f(t)\,dt\) is the total volume of water in gallons pumped into the cistern from time \(t = 60\) seconds to \(t = 135\) seconds." (Must include context, units [gallons], and time interval).
Right Riemann Sum [1 pt]: \[ \begin{aligned} \int_{60}^{135} f(t)\,dt &\approx (90 - 60)f(90) + (120 - 90)f(120) + (135 - 120)f(135) \\ &= 30(0.18) + 30(0.12) + 15(0.06) = 5.4 + 3.6 + 0.9 = \mathbf{9.9\text{ gallons}} \end{aligned} \]
Part (b): Existence of Horizontal Tangent \(f'(c) = 0\) 2 Points
Continuity & Differentiability [1 pt]: State that \(f\) is differentiable on \([0, 150]\), which guarantees \(f\) is continuous on \([60, 120]\) and differentiable on \((60, 120)\).
MVT / Rolle's Justification [1 pt]: Since \(f(60) = 0.12\) and \(f(120) = 0.12\), the average rate of change is: \[\frac{f(120) - f(60)}{120 - 60} = \frac{0.12 - 0.12}{60} = 0\] By Rolle's Theorem (or the Mean Value Theorem), there must exist at least one \(c \in (60, 120)\) such that \(f'(c) = 0\).
Part (c): Integral of Model \(g(t)\) & Riemann Over/Underestimate 3 Points
Definite Integral Value [1 pt]: \[\int_0^{80} g(t)\,dt = \int_0^{80} \left(\frac{t}{400}\right) \cos\left(\left(\frac{t}{120}\right)^2\right)\,dt = 18\sin\left(\frac{4}{9}\right) \approx \mathbf{7.737\text{ gallons}}\]
Comparison & Justification [2 pts]: The value \(\int_0^{80} g(t)\,dt\) is more than the left Riemann sum approximation [1 pt].
Justification [1 pt]: On the interval \([0, 80]\), \(\left(\frac{t}{120}\right)^2 \in [0, 4/9] \subset [0, \pi/2]\), where \(g'(t) > 0\), meaning \(g(t)\) is strictly increasing. Because the function is strictly increasing, any left Riemann sum provides an underestimate, meaning the true integral is strictly greater.
Part (d): Derivative Evaluation & Physical Meaning 2 Points
Value [1 pt]: Using product and chain rules (or graphing calculator numerical derivative \(\left.\frac{d}{dt}g(t)\right|_{t=140}\)): \[g'(140) \approx \mathbf{-0.006\text{ (or } -0.00613\text{)}} \text{ gallons/second}^2\]
Contextual Interpretation [1 pt]: "At time \(t = 140\) seconds, the rate of water flow into the cistern is decreasing at a rate of \(0.006\) gallons per second per second (gal/sec\(^2\))." (Must specify decreasing or negative rate, timestamp, and correct squared units).
Key • Page 4 of 4
