AP Calculus BC: Free Response Questions (FRQs)

10 original FRQs covering BC‑only topics. Master the AP Calculus BC exam with rigorous practice covering parametric equations, polar curves, vectors, infinite series, logistic differential equations, improper integrals, and advanced applications of the Fundamental Theorem of Calculus.

🎯 Note: This set contains no AB-only rate or area questions — pure BC focus. Pay attention to the calculator statuses!

Parametric, Polar, and Vector Functions

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FRQ 1 — Parametric Equations

A particle moves in the \(xy\)-plane with position given by \[ x(t) = t^3 - 6t^2 + 9t, \quad y(t) = \frac{1}{3}t^3 - 2t^2 + 3t + 1, \quad t \ge 0 \]

(a) Find the velocity vector at time \(t\).
(b) Find all times \(t > 0\) when the particle is at rest.
(c) Find the slope of the tangent line to the path at \(t = 2\). Then write the equation of that tangent line.
(d) Find the total distance traveled from \(t = 0\) to \(t = 3\). Simplify the integral.
(e) Is the particle speeding up or slowing down at \(t = 1\)? Justify.
Show Detailed Solution
  • (a) Concept: The velocity vector is formed by taking the derivative of the position components with respect to time. This tells us the instantaneous rate of change in both the horizontal and vertical directions.
    \(v(t) = \langle x'(t), y'(t) \rangle = \langle 3t^2 - 12t + 9,\; t^2 - 4t + 3 \rangle\).
  • (b) Concept: A particle is "at rest" only when it has zero velocity in *both* directions simultaneously. If it's moving vertically but not horizontally, it is not at rest.
    Set each component to 0: \(x'(t) = 3(t-1)(t-3)=0\) and \(y'(t) = (t-1)(t-3)=0\). Both equations share the roots \(t=1\) and \(t=3\). Therefore, the particle is at rest at these times.
  • (c) Concept: The slope of a parametric curve at a specific point is given by \(\frac{dy}{dx}\), which can be found using the chain rule: \(\frac{dy/dt}{dx/dt}\).
    At \(t=2\): \(x'(2) = 3(4) - 24 + 9 = -3\) and \(y'(2) = 4 - 8 + 3 = -1\).
    The slope is \(\frac{-1}{-3} = \frac{1}{3}\).
    Find the coordinate point by plugging \(t=2\) into the original equations: \(x(2)=2, y(2)=5/3\).
    Using point-slope form: \(y - 5/3 = \frac{1}{3}(x-2)\).
  • (d) Concept: Total distance traveled is the accumulation (integral) of speed over time. Speed is the magnitude of the velocity vector, derived from the Pythagorean theorem: \(\sqrt{(x'(t))^2 + (y'(t))^2}\).
    Factor the derivatives: \(x' = 3(t-1)(t-3), y' = (t-1)(t-3)\).
    Speed = \(\sqrt{[3(t-1)(t-3)]^2 + [(t-1)(t-3)]^2} = \sqrt{10(t-1)^2(t-3)^2} = \sqrt{10}|(t-1)(t-3)|\).
    To integrate the absolute value over \([0, 3]\), split the integral where the velocity changes direction (at \(t=1\)):
    Distance = \(\sqrt{10}\int_0^3 |(t-1)(t-3)|dt = \sqrt{10} \left[ \int_0^1 (t^2-4t+3) dt - \int_1^3 (t^2-4t+3) dt \right] = \frac{8\sqrt{10}}{3}\).
  • (e) Concept: Speeding up or slowing down is determined by comparing the signs of the velocity and acceleration vectors. However, there's a special case: if a particle is at rest, it can only speed up (or remain at rest).
    At \(t=1\), we found in part (b) that the velocity is zero. Since it is starting from a dead stop, it is instantaneously at rest. Immediately after, it begins moving, so it is neither speeding up nor slowing down *exactly* at \(t=1\), but rather transitioning from rest to motion.
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FRQ 2 — Polar Equations

Consider the polar curve \(r(\theta) = 4 + 3\sin(2\theta)\) for \(0 \le \theta \le 2\pi\).

(a) Find the area of the region enclosed by the curve in the first quadrant (\(0 \le \theta \le \frac{\pi}{2}\)).
(b) Find the slope of the tangent line to the curve at \(\theta = \frac{\pi}{4}\).
(c) Set up an integral for the total arc length of the curve.
(d) At \(\theta = \frac{\pi}{3}\), is a particle moving along the curve getting closer to or further from the origin? Justify your answer.
Show Detailed Solution
  • (a) Concept: The area of a polar region is accumulated by sweeping out infinitesimally small circular sectors. The area of a sector is \(\frac{1}{2}r^2\theta\), leading to the integral formula \(\frac{1}{2}\int r^2 d\theta\).
    Area = \(\frac{1}{2}\int_0^{\pi/2} (4+3\sin(2\theta))^2 d\theta\).
    Using a calculator to evaluate: \(\frac{41\pi}{8} + 12 \approx 28.102\).
  • (b) Concept: The slope of a tangent line is always \(\frac{dy}{dx}\). In polar coordinates, we must convert to Cartesian using \(y = r\sin\theta\) and \(x = r\cos\theta\), and then use the product rule to find \(dy/d\theta\) and \(dx/d\theta\).
    \(\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}\).
    Evaluate at \(\theta = \pi/4\): \(r(\pi/4) = 4 + 3(1) = 7\), and \(r'(\pi/4) = 6\cos(\pi/2) = 0\).
    Substituting these values: \(\frac{dy}{dx} = \frac{0 + 7(\sqrt{2}/2)}{0 - 7(\sqrt{2}/2)} = -1\).
  • (c) Concept: Arc length for polar curves is derived from the parametric arc length formula, replacing \(x'\) and \(y'\) with polar equivalents, which simplifies beautifully to \(\int \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta\).
    First, find \(\frac{dr}{d\theta} = 6\cos(2\theta)\).
    Arc length = \(\int_0^{2\pi} \sqrt{(4+3\sin(2\theta))^2 + (6\cos(2\theta))^2}\, d\theta\).
  • (d) Concept: Distance from the origin is strictly represented by the absolute value of \(r\). If \(r\) is positive and its derivative \(dr/d\theta\) is negative, the distance is shrinking.
    Evaluate \(r\) and \(r'\) at \(\pi/3\):
    \(r(\pi/3) = 4 + 3\sin(2\pi/3) = 4 + \frac{3\sqrt{3}}{2}\), which is positive.
    \(\frac{dr}{d\theta} = 6\cos(2\pi/3) = 6(-1/2) = -3\), which is negative.
    Because \(r > 0\) and \(\frac{dr}{d\theta} < 0\), the value of \(r\) is decreasing towards zero. Therefore, the particle is moving closer to the origin.
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FRQ 3 — Vector-Valued Functions

A particle moves with position \(\mathbf{r}(t) = \langle 2\cos t, 3\sin t \rangle\) for \(t \ge 0\).

(a) Find the velocity and acceleration vectors.
(b) Find the speed of the particle at \(t = \frac{\pi}{3}\).
(c) Find the equation of the tangent line to the path at \(t = \frac{\pi}{2}\).
(d) Set up an integral for the total distance traveled from \(t=0\) to \(t=2\pi\).
Show Detailed Solution
  • (a) Concept: Velocity is the first derivative of position, and acceleration is the second derivative. The rules of calculus apply component-by-component.
    \(v(t) = \mathbf{r}'(t) = \langle -2\sin t, 3\cos t \rangle\).
    \(a(t) = v'(t) = \langle -2\cos t, -3\sin t \rangle\).
  • (b) Concept: Speed is a scalar quantity representing the magnitude of the velocity vector.
    Evaluate \(v(\pi/3) = \langle -2\sin(\pi/3), 3\cos(\pi/3) \rangle = \langle -\sqrt{3}, 1.5 \rangle\).
    Speed = \(|v(\pi/3)| = \sqrt{(-\sqrt{3})^2 + (1.5)^2} = \sqrt{3 + 2.25} = \sqrt{5.25} \approx 2.291\).
  • (c) Concept: The tangent line requires a point and a slope. The slope is the ratio of the velocity components \(v_y / v_x\).
    Point at \(t=\pi/2\): \(\mathbf{r}(\pi/2) = \langle 2(0), 3(1) \rangle = (0, 3)\).
    Velocity at \(t=\pi/2\): \(v(\pi/2) = \langle -2(1), 3(0) \rangle = \langle -2, 0 \rangle\).
    Slope = \(0 / -2 = 0\). A slope of zero means the tangent line is completely horizontal.
    Equation: \(y - 3 = 0(x - 0) \implies y = 3\).
  • (d) Concept: Total distance is the accumulation of speed over a given time interval.
    Distance = \(\int_0^{2\pi} |v(t)|\, dt = \int_0^{2\pi} \sqrt{(-2\sin t)^2 + (3\cos t)^2}\, dt = \int_0^{2\pi} \sqrt{4\sin^2 t + 9\cos^2 t}\, dt\).

Infinite Series & Taylor Polynomials

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FRQ 4 — Infinite Series (Interval of Convergence)

Consider the series \(\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1} (x-1)^n}{n \cdot 2^n}\).

(a) Find the radius and interval of convergence. Be sure to test the endpoints.
(b) Let \(f(x)\) be the function represented by this series. Find \(f^{(3)}(1)\).
(c) Write the third-degree Taylor polynomial \(P_3(x)\) centered at \(x=1\) for \(f\).
(d) Use the alternating series error bound to estimate the maximum error when \(P_3(1.5)\) is used to approximate \(f(1.5)\).
(e) Does the series converge absolutely, conditionally, or diverge at \(x=3\)? Justify your answer.
Show Detailed Solution
  • (a) Concept: The Ratio Test determines the open interval where a series converges absolutely. Because the Ratio Test is inconclusive when the limit equals exactly 1, we must manually test the endpoints of the interval.
    Set up the Ratio Test: \(\lim_{n \to \infty} \left| \frac{(x-1)^{n+1}}{(n+1)2^{n+1}} \cdot \frac{n 2^n}{(x-1)^n} \right| = \frac{|x-1|}{2}\).
    For convergence, set the limit < 1: \(\frac{|x-1|}{2} < 1 \implies |x-1| < 2\). The radius is \(R=2\), and the open interval is \((-1, 3)\).
    Test \(x=-1\): The series becomes \(\sum \frac{(-1)^{n+1}(-2)^n}{n2^n} = \sum \frac{-1}{n}\). This is the negative Harmonic Series, which diverges by the p-test.
    Test \(x=3\): The series becomes \(\sum \frac{(-1)^{n+1}2^n}{n2^n} = \sum \frac{(-1)^{n+1}}{n}\). This is the Alternating Harmonic Series, which converges by the Alternating Series Test.
    Interval of convergence: \((-1, 3]\).
  • (b) Concept: The coefficients of a Taylor series are defined by \(c_n = \frac{f^{(n)}(a)}{n!}\). We can extract a specific derivative by matching the general term of our series to the definition.
    For \(n=3\), the term is \(\frac{(-1)^4 (x-1)^3}{3 \cdot 2^3} = \frac{1}{24}(x-1)^3\).
    Set the coefficient equal to the Taylor formula: \(\frac{f^{(3)}(1)}{3!} = \frac{1}{24}\).
    Solve for the derivative: \(f^{(3)}(1) = 6 \cdot \frac{1}{24} = \frac{1}{4}\).
  • (c) Concept: A Taylor polynomial is simply the partial sum of the Taylor series up to degree \(n\).
    Write out the terms for \(n=1, 2, 3\):
    \(P_3(x) = \frac{(x-1)^1}{1 \cdot 2^1} - \frac{(x-1)^2}{2 \cdot 2^2} + \frac{(x-1)^3}{3 \cdot 2^3} = \frac{x-1}{2} - \frac{(x-1)^2}{8} + \frac{(x-1)^3}{24}\).
  • (d) Concept: For a convergent alternating series whose terms decrease in magnitude, the error (remainder) of a partial sum is strictly less than the magnitude of the first omitted term.
    We used \(P_3(x)\), so the first omitted term is for \(n=4\).
    Error \(\le |a_4(1.5)| = \left| \frac{(-1)^5 (1.5-1)^4}{4 \cdot 2^4} \right| = \left| \frac{(0.5)^4}{64} \right| = \frac{1/16}{64} = \frac{1}{1024}\).
  • (e) Concept: Absolute convergence means the series converges even if all terms are made positive. Conditional convergence means it converges as is, but diverges if you take the absolute value.
    At \(x=3\), the series is \(\sum \frac{(-1)^{n+1}}{n}\). Taking the absolute value yields \(\sum \frac{1}{n}\), which is the divergent harmonic series. Because the alternating version converges but the absolute version diverges, the series converges conditionally.
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FRQ 5 — Taylor / Maclaurin Series

Let \(g(x) = e^{x^2}\).

(a) Write the Maclaurin series for \(g(x)\) in sigma notation.
(b) Find the radius of convergence for the Maclaurin series of \(g(x)\).
(c) Use the Maclaurin series to find \(g^{(4)}(0)\).
(d) Write the fourth-degree Taylor polynomial, \(P_4(x)\), for \(g(x)\) centered at \(x=0\).
(e) Use the Maclaurin series for \(g(x)\) to evaluate \(\lim_{x \to 0} \frac{g(x) - 1 - x^2}{x^4}\).
Show Detailed Solution
  • (a) Concept: Instead of computing derivatives from scratch (which becomes very messy for \(e^{x^2}\)), we can use known Maclaurin series and substitute the argument. This is a crucial time-saving technique on the AP Exam.
    The known Maclaurin series for \(e^x\) is \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\).
    Substitute \(x^2\) for \(x\): \(g(x) = e^{x^2} = \sum_{n=0}^{\infty} \frac{(x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}\).
  • (b) Concept: The Ratio Test tells us where the series converges.
    \(\lim_{n \to \infty} \left| \frac{x^{2(n+1)}}{(n+1)!} \cdot \frac{n!}{x^{2n}} \right| = \lim_{n \to \infty} \left| \frac{x^{2n+2}}{x^{2n}} \cdot \frac{1}{n+1} \right| = \lim_{n \to \infty} \left| \frac{x^2}{n+1} \right| = 0\).
    Since the limit is \(0\), which is always less than 1 regardless of the value of \(x\), the series converges everywhere. The radius of convergence is \(R = \infty\).
  • (c) Concept: Because a function's power series representation is unique, the coefficients we generated by substitution must perfectly match the formal Taylor definition \(c_n = \frac{g^{(n)}(0)}{n!}\).
    Write out the first few terms: \(g(x) = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots\)
    The coefficient of the \(x^4\) term is \(\frac{1}{2!}\).
    Equating this to the Taylor definition: \(\frac{g^{(4)}(0)}{4!} = \frac{1}{2} \implies g^{(4)}(0) = \frac{24}{2} = 12\).
  • (d) Concept: A Taylor polynomial of degree \(k\) includes all terms up to and including \(x^k\).
    \(P_4(x) = 1 + x^2 + \frac{x^4}{2}\).
  • (e) Concept: Series can be used to evaluate tricky limits as an alternative to L'Hospital's Rule, by expanding the function and canceling terms algebraically.
    Substitute the expanded series into the limit:
    \(\lim_{x \to 0} \frac{\left(1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \dots\right) - 1 - x^2}{x^4}\).
    The \(1\) and \(x^2\) terms cancel out in the numerator:
    \(\lim_{x \to 0} \frac{\frac{x^4}{2} + \frac{x^6}{6} + \dots}{x^4}\).
    Divide every term by \(x^4\):
    \(\lim_{x \to 0} \left(\frac{1}{2} + \frac{x^2}{6} + \dots\right)\).
    As \(x \to 0\), all remaining terms containing \(x\) approach zero, leaving exactly \(\frac{1}{2}\).

Differential Equations & Logistic Growth

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FRQ 6 — Logistic Differential Equation

Consider the logistic differential equation \(\frac{dy}{dt} = 0.08y\left(1 - \frac{y}{500}\right)\) with initial condition \(y(0)=100\).

(a) What is the carrying capacity of the population? What is the maximum growth rate?
(b) Use Euler’s method, starting at \(t=0\) with two steps of equal size \(h=1\), to approximate \(y(2)\).
(c) Solve the differential equation exactly to find \(y(t)\).
(d) Find \(\lim_{t\to\infty} y(t)\).
(e) At what time \(t\) is the population growing the fastest?
Show Detailed Solution
  • (a) Concept: The standard logistic differential equation is \(\frac{dy}{dt} = ky(1 - \frac{y}{K})\), where \(K\) is the carrying capacity. The growth rate is a downward-facing parabola that reaches its vertex (maximum) when the population is exactly half the carrying capacity (\(y = K/2\)).
    Carrying capacity \(K=500\).
    The population grows fastest when \(y = 500/2 = 250\).
    The maximum growth rate is found by plugging 250 back into the DE: \(\frac{dy}{dt} = 0.08(250)(1 - 250/500) = 20(0.5) = 10\).
  • (b) Concept: Euler's method uses localized tangent lines to step forward iteratively. The formula is \(y_{new} = y_{old} + h \cdot y'_{old}\).
    Step 1 (Find \(y_1\) at \(t=1\)): \(y_1 = 100 + 1 \cdot [0.08(100)(1 - 100/500)] = 100 + 1(8 \cdot 0.8) = 106.4\).
    Step 2 (Find \(y_2\) at \(t=2\)): \(y_2 = 106.4 + 1 \cdot [0.08(106.4)(1 - 106.4/500)] \approx 106.4 + 6.701 = 113.101\).
  • (c) Concept: While you can solve logistic equations via partial fraction decomposition, on the AP exam it is advantageous to memorize the solution structure \(y = \frac{K}{1+Ae^{-kt}}\), where \(A = \frac{K - y_0}{y_0}\).
    \(A = \frac{500 - 100}{100} = \frac{400}{100} = 4\).
    Solution: \(y(t)=\frac{500}{1+4e^{-0.08t}}\).
  • (d) Concept: Evaluating the limit to infinity verifies the carrying capacity analytically.
    As \(t \to \infty\), the exponent \(-0.08t\) becomes a massive negative number, meaning \(e^{-0.08t} \to 0\).
    \(\lim_{t\to\infty} \frac{500}{1 + 4(0)} = \frac{500}{1} = 500\).
  • (e) Concept: We established in part (a) that the population grows fastest when \(y = 250\). We just need to find the time \(t\) when this occurs using our exact solution.
    Set \(y(t) = 250\): \(\frac{500}{1+4e^{-0.08t}} = 250\).
    Divide both sides into 500: \(1+4e^{-0.08t} = 2\).
    Solve for the exponential: \(4e^{-0.08t} = 1 \implies e^{-0.08t} = 0.25\).
    Take the natural log: \(-0.08t = \ln(0.25)\).
    \(t = \frac{\ln(0.25)}{-0.08} = \frac{-\ln 4}{-0.08} \approx 17.329\).
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FRQ 7 — Differential Eq (Separation & Slope Field)

Consider the differential equation \(\frac{dy}{dx} = \frac{xy}{x^2+1}\) with the initial condition \(y(0)=2\).

(a) Sketch the slope field for the given differential equation at the points \((0,2), (1,2), (2,2), (0,3), (1,3)\).
(b) Use Euler’s method, starting at \(x=0\) with two steps of equal size (\(h=0.5\)), to approximate \(y(1)\).
(c) Find the particular solution \(y = f(x)\) to the given differential equation with the initial condition \(y(0)=2\).
(d) Find \(\lim_{x\to\infty} y(x)\).
(e) Does \(y(x)\) have a relative minimum, relative maximum, or neither at \(x=0\)? Justify your answer.
Show Detailed Solution
  • (a) Concept: A slope field visualizes the differential equation by plotting short line segments whose slopes are determined by evaluating \(dy/dx\) at specific coordinates.
    Evaluate \(\frac{dy}{dx} = \frac{xy}{x^2+1}\):
    At (0,2): \(0/1 = 0\) (horizontal segment)
    At (1,2): \(2/2 = 1\) (segment angled at 45 degrees)
    At (2,2): \(4/5 = 0.8\) (slightly less steep than 1)
    At (0,3): \(0/1 = 0\) (horizontal segment)
    At (1,3): \(3/2 = 1.5\) (steeper segment).
  • (b) Concept: Step forward using \(y_{new} = y_{old} + h(dy/dx)_{old}\).
    Step 1 (\(x=0.5\)): \(y(0.5) \approx 2 + 0.5\left(\frac{(0)(2)}{0^2+1}\right) = 2 + 0.5(0) = 2\).
    Step 2 (\(x=1.0\)): \(y(1) \approx 2 + 0.5\left(\frac{(0.5)(2)}{(0.5)^2+1}\right) = 2 + 0.5\left(\frac{1}{1.25}\right) = 2 + 0.5(0.8) = 2.4\).
  • (c) Concept: Separation of variables requires using algebra to get all \(y\)'s on the \(dy\) side, and all \(x\)'s on the \(dx\) side before integrating.
    Divide by \(y\) and multiply by \(dx\): \(\frac{1}{y} dy = \frac{x}{x^2+1} dx\).
    Integrate both sides (use u-substitution \(u=x^2+1, du=2x dx\) on the right):
    \(\int \frac{1}{y} dy = \frac{1}{2} \int \frac{2x}{x^2+1} dx \implies \ln|y| = \frac{1}{2}\ln(x^2+1) + C\).
    Use initial condition \(y(0)=2\) to solve for \(C\): \(\ln 2 = \frac{1}{2}\ln(1) + C \implies C = \ln 2\).
    Isolate \(y\): \(\ln|y| = \ln((x^2+1)^{1/2}) + \ln 2 = \ln(2\sqrt{x^2+1})\).
    Exponentiating both sides yields the exact solution: \(y = 2\sqrt{x^2+1}\).
  • (d) Concept: Evaluate the behavior of the solution as \(x\) grows infinitely large.
    \(\lim_{x\to\infty} 2\sqrt{x^2+1} = \infty\). The curve increases without bound.
  • (e) Concept: The First Derivative Test checks how the slope changes around a critical point. A transition from negative slope to positive slope indicates a local minimum.
    At the point \(x=0, y=2\), the derivative is \(\frac{(0)(2)}{1} = 0\), establishing a critical point.
    For \(x < 0\) (and \(y>0\)), the numerator \(xy\) is negative, so \(\frac{dy}{dx} < 0\).
    For \(x > 0\) (and \(y>0\)), the numerator \(xy\) is positive, so \(\frac{dy}{dx} > 0\).
    Because the derivative changes from negative to positive at \(x=0\), \(y(x)\) has a relative minimum.

Advanced Integration & FTC

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FRQ 8 — Improper Integrals & Comparison

Evaluate each expression or determine whether it converges or diverges.

(a) \(\displaystyle \int_1^{\infty} \frac{1}{x^2+3x+2}\,dx\)
(b) Use a comparison test to determine convergence: \(\displaystyle \int_2^{\infty} \frac{3+\cos x}{x^2}\,dx\).
(c) \(\displaystyle \int_0^{3} \frac{1}{\sqrt{9-x^2}}\,dx\)
(d) \(\displaystyle \lim_{x\to0^+} x\ln x\)
Show Detailed Solution
  • (a) Concept: Infinity is a concept, not a number, so we must replace it with a variable (like \(b\)) and take the limit as \(b \to \infty\). To integrate this rational function, we must first decompose it into partial fractions.
    Decompose: \(\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \implies A=1, B=-1\).
    Set up the limit: \(\lim_{b \to \infty} \int_1^b (\frac{1}{x+1} - \frac{1}{x+2}) dx\).
    Integrate: \(\lim_{b \to \infty} \left[ \ln|x+1| - \ln|x+2| \right]_1^b = \lim_{b \to \infty} \left[ \ln\left|\frac{x+1}{x+2}\right| \right]_1^b\).
    Evaluate bounds: \(\lim_{b \to \infty} \ln\left(\frac{b+1}{b+2}\right) - \ln\left(\frac{2}{3}\right)\).
    The limit of the fraction as \(b \to \infty\) is 1, and \(\ln(1)=0\). The final value is \(-\ln(2/3)\), which equals \(\ln(3/2)\). The integral converges.
  • (b) Concept: The Direct Comparison Test asks us to find a larger, simpler function that we *know* converges. If the "ceiling" converges, the smaller function trapped beneath it must also converge.
    Because cosine oscillates between -1 and 1 (\(-1 \le \cos x \le 1\)), the numerator \(3+\cos x\) is bounded between 2 and 4.
    Therefore, \(\frac{3+\cos x}{x^2} \le \frac{4}{x^2}\) for all \(x \ge 2\).
    We know that \(\int_2^\infty \frac{4}{x^2} dx\) converges because it is a \(p\)-integral with \(p=2\), and \(2 > 1\).
    Since the larger integral converges, our original integral converges by the Direct Comparison Test.
  • (c) Concept: This integral is improper because the denominator equals zero at the upper bound \(x=3\), creating a vertical asymptote. We must use a one-sided limit.
    Set up the limit: \(\lim_{b \to 3^-} \int_0^b \frac{1}{\sqrt{9-x^2}} dx\).
    Recognize the integral as the derivative of inverse sine: \(\int \frac{1}{\sqrt{a^2-x^2}} dx = \arcsin(x/a)\).
    \(\lim_{b \to 3^-} \left[ \arcsin\left(\frac{x}{3}\right) \right]_0^b = \arcsin(1) - \arcsin(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}\). The integral converges.
  • (d) Concept: This limit currently yields the indeterminate form \(0 \cdot -\infty\). To use L'Hospital's Rule, we must algebraically force it into a fraction yielding \(0/0\) or \(\infty/\infty\).
    Rewrite as a fraction: \(\lim_{x\to0^+} \frac{\ln x}{1/x}\). As \(x \to 0\), this is \(-\infty/\infty\).
    Apply L'Hospital's Rule (take derivatives of top and bottom): \(\lim_{x\to0^+} \frac{1/x}{-1/x^2}\).
    Simplify: \(\lim_{x\to0^+} (-x) = 0\).
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FRQ 9 — Integration Techniques

Evaluate each integral.

(a) \(\displaystyle \int x e^{2x}\,dx\) (by parts).
(b) \(\displaystyle \int \frac{5x+7}{(x-1)(x+2)}\,dx\) (partial fractions).
(c) \(\displaystyle \int \ln x\,dx\).
(d) \(\displaystyle \int_0^{\pi} x\cos x\,dx\).
Show Detailed Solution
  • (a) Concept: Use Integration by Parts (\(\int u\,dv = uv - \int v\,du\)). Follow the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) to choose \(u\). Algebraic (\(x\)) comes before Exponential (\(e^{2x}\)).
    Let \(u=x\) and \(dv=e^{2x}dx\).
    Take the derivative to find \(du\): \(du=dx\).
    Take the antiderivative to find \(v\): \(v=\frac{1}{2}e^{2x}\).
    Apply the formula: \(x(\frac{1}{2}e^{2x}) - \int \frac{1}{2}e^{2x} dx = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C\).
  • (b) Concept: Decompose the rational expression into simpler fractions that yield natural log integrals.
    \(\frac{5x+7}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}\).
    Multiply by the common denominator: \(5x+7 = A(x+2) + B(x-1)\).
    Use the "cover-up" method: Let \(x=1 \implies 12 = A(3) \implies A=4\). Let \(x=-2 \implies -3 = B(-3) \implies B=1\).
    The integral becomes \(\int \left(\frac{4}{x-1} + \frac{1}{x+2}\right)dx = 4\ln|x-1| + \ln|x+2| + C\).
  • (c) Concept: A classic trick for integrating a lone \(\ln x\) is to use Integration by Parts by forcing \(dv = 1 dx\).
    Let \(u=\ln x\) and \(dv=dx\).
    Then \(du=\frac{1}{x}dx\) and \(v=x\).
    Apply the formula: \(uv - \int v\,du = x\ln x - \int x(\frac{1}{x}) dx = x\ln x - \int 1 dx = x\ln x - x + C\).
  • (d) Concept: Use Integration by Parts with definite bounds. The LIATE rule dictates \(u=x\) (Algebraic) and \(dv=\cos x dx\) (Trig).
    Let \(u=x \implies du=dx\). Let \(dv=\cos x dx \implies v=\sin x\).
    Apply formula with bounds: \([x\sin x]_0^\pi - \int_0^\pi \sin x dx\).
    Evaluate the first part: \((\pi\sin\pi - 0\sin 0) = 0 - 0 = 0\).
    Integrate the second part: \(-[-\cos x]_0^\pi = [\cos x]_0^\pi = \cos(\pi) - \cos(0) = -1 - 1 = -2\).
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FRQ 10 — Advanced FTC & Derivatives

Evaluate the following advanced derivatives.

(a) Let \(F(x) = \int_{0}^{x^2} \sin(t^2)\,dt\). Find \(F'(x)\).
(b) Find \(\frac{d}{dx} \int_{x}^{x^3} e^{t^2}\,dt\).
(c) A curve is defined by parametric equations \(x=t^2\) and \(y=t^3\). Find \(\frac{d^2y}{dx^2}\).
(d) A curve is defined by the polar equation \(r=2\theta\). Find \(\frac{dy}{dx}\) at \(\theta=\frac{\pi}{2}\).
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  • (a) Concept: The Fundamental Theorem of Calculus (Part 1) states that the derivative of an accumulation function simply evaluates the integrand at the upper bound. However, because the bound is a function of \(x\) (it's \(x^2\), not just \(x\)), we must multiply by the derivative of that bound (the Chain Rule).
    \(F'(x) = \sin((x^2)^2) \cdot \frac{d}{dx}(x^2) = \sin(x^4) \cdot 2x = 2x\sin(x^4)\).
  • (b) Concept: When taking the derivative of an integral with variables in both the upper and lower bounds, you apply the FTC and Chain Rule to the upper bound, and subtract the FTC and Chain Rule evaluation of the lower bound.
    Upper bound evaluation: \(e^{(x^3)^2} \cdot \frac{d}{dx}(x^3) = e^{x^6} \cdot 3x^2\).
    Lower bound evaluation: \(e^{(x)^2} \cdot \frac{d}{dx}(x) = e^{x^2} \cdot 1\).
    Combined: \(3x^2 e^{x^6} - e^{x^2}\).
  • (c) Concept: The second derivative of a parametric equation is highly tested. It is NOT simply the second derivative of \(y\) over the second derivative of \(x\). You must take the derivative of \(dy/dx\) with respect to \(t\), and divide that whole expression by \(dx/dt\) again.
    First derivative: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3}{2}t\).
    Second derivative: \(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt} = \frac{\frac{d}{dt}(\frac{3}{2}t)}{2t} = \frac{3/2}{2t} = \frac{3}{4t}\).
    At \(t=1\), this evaluates to \(\frac{3}{4}\).
  • (d) Concept: To find the Cartesian slope of a polar curve, convert to parametric equations using \(\theta\) as the parameter: \(x = r\cos\theta\) and \(y = r\sin\theta\). Substitute the given \(r\) into these formulas before differentiating.
    \(x = 2\theta\cos\theta\) and \(y = 2\theta\sin\theta\).
    Apply the product rule to find the derivatives:
    \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2\sin\theta + 2\theta\cos\theta}{2\cos\theta - 2\theta\sin\theta}\).
    Evaluate at \(\theta = \frac{\pi}{2}\). Remember that \(\sin(\pi/2) = 1\) and \(\cos(\pi/2) = 0\):
    \(\frac{dy}{dx} = \frac{2(1) + 2(\pi/2)(0)}{2(0) - 2(\pi/2)(1)} = \frac{2 + 0}{0 - \pi} = -\frac{2}{\pi}\).

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