Self-Study Edition

Calculus II
Problem Set

20 word problems covering the core units most likely to trip you up. Work through each — explanations unlock after you answer.

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Chapter 01 — Integration Techniques
Q 01 ★★★ Hard
Integration by Parts
🧠
Memory Key LIATE rule — Log · Inverse trig · Algebraic · Trig · Exponential → pick u left-to-right
The Falling Raindrop — Integration by Parts
A raindrop's velocity (m/s) at time \(t\) seconds is modeled by \(v(t) = t^2 e^{-t}\). The total displacement from \(t = 0\) to \(t = 3\) is \(\displaystyle\int_0^3 t^2 e^{-t}\,dt\).

Using integration by parts twice, which expression correctly represents the antiderivative \(F(t)\) before applying limits?
Quick Example
\(\displaystyle\int t\,e^{-t}\,dt = -te^{-t} + \int e^{-t}\,dt = -te^{-t} - e^{-t} + C\)
Apply same idea twice for \(t^2 e^{-t}\).
A\(-t^2e^{-t} - 2te^{-t} - 2e^{-t} + C\)
B\(-t^2e^{-t} + 2te^{-t} + 2e^{-t} + C\)
C\(-t^2e^{-t} - 2te^{-t} + 2e^{-t} + C\)
D\(t^2e^{-t} + 2te^{-t} - 2e^{-t} + C\)
Explanation

Set \(u = t^2,\ dv = e^{-t}dt\). Then \(du = 2t\,dt,\ v = -e^{-t}\).

First application: \(-t^2e^{-t} + 2\displaystyle\int te^{-t}\,dt\).

Second application on \(\int te^{-t}\,dt\): set \(u=t,\ dv=e^{-t}dt\) → \(-te^{-t} + \int e^{-t}\,dt = -te^{-t} - e^{-t}\).

Combine: \(F(t) = -t^2e^{-t} + 2(-te^{-t} - e^{-t}) = -t^2e^{-t} - 2te^{-t} - 2e^{-t} + C\).

Common trap: Forgetting the factor of 2 from \(du = 2t\,dt\) in the first step.

Q 02 ★★★ Hard
Trigonometric Substitution
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Memory Key √(a²−x²) → x = a·sin θ  |  √(a²+x²) → x = a·tan θ  |  √(x²−a²) → x = a·sec θ
The Arch Bridge — Trig Substitution
An engineer models the cross-section of a bridge arch as \(\displaystyle\int_0^2 \sqrt{4-x^2}\,dx\). After applying the substitution \(x = 2\sin\theta\), the integral transforms. What is the exact value of this definite integral?
Quick Example
\(\displaystyle\int_0^1\sqrt{1-x^2}\,dx = \frac{\pi}{4}\) (quarter-circle of radius 1).
Recognize the geometry: \(\sqrt{4-x^2}\) is the upper semicircle of radius 2.
A\(\pi\)
B\(2\pi\)
C\(\dfrac{\pi}{2}\)
D\(4\pi\)
Explanation

\(\int_0^2\sqrt{4-x^2}\,dx\) is the area of a quarter-circle of radius 2 (from \(x=0\) to \(x=2\), lying above the \(x\)-axis).

Area of full circle = \(\pi r^2 = 4\pi\). Quarter = \(\pi\).

Common trap: Thinking it's a semicircle (half). The limits 0 to 2 cover only a quarter of the circle.

Q 03 ★★★ Hard
Partial Fractions
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Memory Key Degree top ≥ degree bottom? → Long divide FIRST, then partial fractions.
Mixing Tank — Partial Fractions
A chemical mixing tank problem reduces to evaluating \(\displaystyle\int \frac{3x+5}{(x+1)(x+2)}\,dx\). After decomposing into partial fractions \(\dfrac{A}{x+1} + \dfrac{B}{x+2}\), what are the correct values of \(A\) and \(B\)?
Quick Example
\(\dfrac{5}{(x+1)(x-1)} = \dfrac{A}{x+1} + \dfrac{B}{x-1}\).
Multiply both sides by \((x+1)(x-1)\), then cover-up each factor.
A\(A = 2,\; B = 1\)
B\(A = 1,\; B = 2\)
C\(A = 2,\; B = -1\)
D\(A = -1,\; B = 2\)
Explanation

Set \(3x+5 = A(x+2) + B(x+1)\).

Let \(x = -1\): \(2 = A(1)\) → \(A = 2\).

Let \(x = -2\): \(-1 = B(-1)\) → \(B = 1\).

So \(\int\frac{3x+5}{(x+1)(x+2)}dx = 2\ln|x+1| + \ln|x+2| + C\).

Common trap: Swapping A and B or sign errors when substituting negative values.

Chapter 02 — Improper Integrals
Q 04 ★★★ Hard
Improper Integrals — Convergence
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Memory Key p-integral: \(\int_1^\infty x^{-p}dx\) converges ↔ p > 1.   \(\int_0^1 x^{-p}dx\) converges ↔ p < 1.
Infinite Pipeline — Convergence Test
A pipeline leaks fluid at rate \(r(t) = \dfrac{1}{t^{1.5}}\) liters/hour starting at \(t = 1\). The total fluid leaked forever is \(\displaystyle\int_1^\infty \frac{1}{t^{3/2}}\,dt\).

Does this integral converge, and if so what is its value?
ADiverges to \(\infty\)
BConverges to \(1\)
CConverges to \(2\)
DConverges to \(\dfrac{1}{2}\)
Explanation

\(\int_1^\infty t^{-3/2}\,dt = \lim_{b\to\infty}\left[\frac{t^{-1/2}}{-1/2}\right]_1^b = \lim_{b\to\infty}\left[-\frac{2}{\sqrt{t}}\right]_1^b\).

\(= \lim_{b\to\infty}\left(-\frac{2}{\sqrt{b}} + 2\right) = 0 + 2 = 2\).

Since \(p = 3/2 > 1\), the p-integral converges. Value = 2.

Common trap: Forgetting the \(-1/2\) in the denominator when integrating \(t^{-3/2}\).

Q 05 ★★★ Hard
Comparison Test for Improper Integrals
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Memory Key 0 ≤ f ≤ g  →  g converges ⇒ f converges.   f diverges ⇒ g diverges.
Satellite Signal Decay
A satellite's signal power decays as \(f(t) = \dfrac{e^{-t}}{t^2+1}\) watts for \(t \geq 1\). Using a comparison function, which argument correctly proves the total energy \(\int_1^\infty f(t)\,dt\) converges?
A\(\dfrac{e^{-t}}{t^2+1} \leq e^{-t}\) and \(\displaystyle\int_1^\infty e^{-t}\,dt\) converges → f converges
B\(\dfrac{e^{-t}}{t^2+1} \geq \dfrac{1}{t^2+1}\) so f diverges like \(\arctan\)
C\(\dfrac{e^{-t}}{t^2+1} \leq \dfrac{1}{t^2}\) and \(\int_1^\infty\frac{1}{t^2}\,dt\) diverges, so inconclusive
DThe comparison test does not apply because \(f\) is not monotone
Explanation

For \(t \geq 1\), \(t^2 + 1 \geq 1\), so \(\dfrac{e^{-t}}{t^2+1} \leq e^{-t}\).

Since \(\int_1^\infty e^{-t}\,dt = e^{-1}\) converges, and our function is bounded above by it, the Comparison Test guarantees convergence.

Common trap (B): The inequality in B goes the wrong way — \(e^{-t} \leq 1\) for \(t\geq 0\), so \(\frac{e^{-t}}{t^2+1} \leq \frac{1}{t^2+1}\) not ≥.

Chapter 03 — Sequences & Series
Q 06 ★★★ Hard
Geometric Series
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Memory Key Geo series: \(\sum_{n=0}^\infty ar^n = \dfrac{a}{1-r}\) only when |r| < 1.
Bouncing Ball — Geometric Series
A rubber ball is dropped from 10 m. Each bounce it rises to \(\dfrac{3}{4}\) of the previous height. The total distance traveled (down + up forever) is what value?

Hint: The first drop is 10 m. Then each bounce goes up then down. Σ carefully.
Setup Hint
Total = first drop + 2 × (sum of all bounce heights)
\(= 10 + 2\left(\dfrac{10 \cdot \frac{3}{4}}{1 - \frac{3}{4}}\right)\)
A40 m
B70 m
C80 m
D60 m
Explanation

First drop: 10 m. After 1st bounce, ball rises 7.5 m and falls 7.5 m. This continues geometrically with \(r = 3/4\).

Total = \(10 + 2\sum_{n=1}^\infty 10\left(\frac{3}{4}\right)^n = 10 + 2 \cdot \frac{10 \cdot \frac{3}{4}}{1 - \frac{3}{4}} = 10 + 2 \cdot 30 = 10 + 60 = 70\) m.

Common trap: Forgetting the factor of 2 (ball goes up AND comes back down after each bounce). Many students get 40 m by not doubling.

Q 07 ★★★ Hard
Ratio Test
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Memory Key Ratio Test: L = lim|a_{n+1}/a_n|.   L < 1 → converge, L > 1 → diverge, L = 1 → inconclusive.
Drug Concentration — Ratio Test
A pharmacist models total drug accumulation as \(\displaystyle\sum_{n=1}^\infty \frac{n^2}{3^n}\). Apply the Ratio Test. What is \(L = \displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\), and what does it tell you?
A\(L = 3\); series diverges
B\(L = \dfrac{1}{3}\); series converges
C\(L = 1\); Ratio Test inconclusive
D\(L = \dfrac{1}{9}\); series converges
Explanation

\(\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)^2}{3^{n+1}} \cdot \dfrac{3^n}{n^2} = \dfrac{(n+1)^2}{3n^2}\).

\(\displaystyle\lim_{n\to\infty} \frac{(n+1)^2}{3n^2} = \lim_{n\to\infty} \frac{n^2+2n+1}{3n^2} = \frac{1}{3} < 1\).

Since \(L = 1/3 < 1\), the series converges absolutely.

Common trap: Thinking \(L = 1/9\) by squaring the 3 as well. Only \(n^2\) ratio limits to 1; the \(3^n\) contributes one factor of 3 in the denominator.

Q 08 ★★★ Hard
Alternating Series Test
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Memory Key AST (Leibniz): alternating + b_n decreasing + lim b_n = 0 → converges. Error ≤ first omitted term.
Temperature Oscillation — Alternating Series Error
A thermostat alternately over- and undershoots. The error is modeled as \(\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}\). If you truncate after 3 terms, what is the maximum error of this approximation?
A\(\dfrac{1}{9}\)
B\(\dfrac{1}{16}\)
C\(\dfrac{1}{25}\)
D\(\dfrac{1}{4}\)
Explanation

By the Alternating Series Estimation Theorem, the error is bounded by the absolute value of the first omitted term.

After 3 terms (\(n=1,2,3\)), the first omitted term is \(n=4\): \(\dfrac{1}{4^2} = \dfrac{1}{16}\).

So error \(\leq \dfrac{1}{16}\).

Common trap: Using \(n=3\) (the last included term) instead of \(n=4\) (the first excluded term).

Q 09 ★★★ Hard
Integral Test & p-series
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Memory Key p-series \(\sum 1/n^p\): converge ↔ p > 1. The harmonic series p=1 DIVERGES (very slowly).
Website Load Balancing — Integral Test
A server's queue length after \(n\) requests is proportional to \(\dfrac{1}{n\ln n}\). The total work is \(\displaystyle\sum_{n=2}^\infty \frac{1}{n\ln n}\). Does this series converge or diverge?

Apply the Integral Test using \(u = \ln x\).
AConverges by p-series with \(p = 2\)
BDiverges; \(\int_2^\infty \frac{dx}{x\ln x} = \infty\)
CConverges to \(\ln(\ln 2)\)
DDiverges by Comparison with \(\sum 1/n\)
Explanation

Apply the Integral Test: \(\int_2^\infty \frac{dx}{x\ln x}\). Let \(u = \ln x, du = dx/x\).

\(= \int_{\ln 2}^\infty \frac{du}{u} = [\ln u]_{\ln 2}^\infty = \infty\).

The integral diverges, so the series diverges. Note: (D) is also true, but the integral test is the rigorous method asked here, and (B) is the correct statement of what the integral test shows.

Common trap: Thinking \(1/(n\ln n)\) decays "fast enough" because of the \(\ln n\). It does not — it diverges (just barely, like \(\ln\ln n \to \infty\)).

Chapter 04 — Power Series & Taylor Series
Q 10 ★★★ Hard
Radius of Convergence
🧠
Memory Key Ratio Test on power series → R = lim|a_n/a_{n+1}|. Always CHECK endpoints separately.
Signal Frequency Model — Radius of Convergence
A signal is modeled as \(\displaystyle\sum_{n=1}^\infty \frac{(-1)^n x^n}{n \cdot 4^n}\). Find the interval of convergence, including a check at the endpoints.
A\((-4, 4)\) — open interval
B\([-4, 4)\) — left endpoint included
C\((-4, 4]\) — right endpoint included
D\([-4, 4]\) — both endpoints included
Explanation

Ratio Test: \(\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x}{4}\cdot\frac{n}{n+1}\right| \to \frac{|x|}{4} < 1\) → \(R = 4\).

At \(x = 4\): \(\sum \frac{(-1)^n 4^n}{n \cdot 4^n} = \sum \frac{(-1)^n}{n}\) = alternating harmonic series → converges.

At \(x = -4\): \(\sum \frac{(-1)^n(-4)^n}{n\cdot 4^n} = \sum \frac{(-1)^n(-1)^n}{n} = \sum \frac{1}{n}\) = harmonic series → diverges.

Interval of convergence: \((-4, 4]\).

Common trap: Forgetting to substitute the \((-1)^n\) already in the series when plugging in negative endpoints.

Q 11 ★★★ Hard
Taylor Series Approximation
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Memory Key Key series: \(e^x = \sum x^n/n!\), \(\sin x = \sum (-1)^n x^{2n+1}/(2n+1)!\), \(\cos x = \sum (-1)^n x^{2n}/(2n)!\)
GPS Error Estimation — Taylor Polynomial
A GPS chip approximates \(\cos(\theta)\) by a 4th-degree Taylor polynomial centered at 0 for a small angle \(\theta = 0.1\) radians.

The 4th-degree Taylor polynomial for \(\cos x\) is \(P_4(x) = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}\). What is \(P_4(0.1)\) to 6 decimal places?
A0.995004
B0.990000
C0.995000
D0.998000
Explanation

\(P_4(0.1) = 1 - \dfrac{(0.1)^2}{2} + \dfrac{(0.1)^4}{24}\)

\(= 1 - \dfrac{0.01}{2} + \dfrac{0.0001}{24} = 1 - 0.005 + 0.0000041\overline{6}\)

\(\approx 0.995004\).

Actual \(\cos(0.1) \approx 0.995004165\ldots\) — remarkably accurate!

Common trap: Stopping at the \(x^2/2\) term and getting 0.995000, missing the tiny but non-zero \(x^4/24\) correction.

Q 12 ★★★ Hard
Taylor Remainder Theorem
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Memory Key Lagrange remainder: |R_n(x)| ≤ M·|x−a|^{n+1} / (n+1)! where M = max|f^{(n+1)}| on interval.
Structural Beam Vibration — Remainder Bound
An engineer uses \(P_3(x) = x - \dfrac{x^3}{6}\) (the degree-3 Taylor polynomial for \(\sin x\)) to model a beam's deflection for \(|x| \leq 0.5\).

Using the Lagrange Remainder, what is an upper bound on the error \(|\sin x - P_3(x)|\) for \(|x| \leq 0.5\)?
A\(\dfrac{(0.5)^4}{24}\)
B\(\dfrac{(0.5)^5}{120}\)
C\(\dfrac{(0.5)^4}{4!}\)
D\(\dfrac{(0.5)^3}{6}\)
Explanation

The remainder after \(P_3\) is \(R_3(x) = \dfrac{f^{(4)}(c)}{4!}x^4\) for some \(c\) between 0 and \(x\).

For \(\sin x\), all derivatives are bounded by 1 in absolute value, so \(M = 1\).

\(|R_3(x)| \leq \dfrac{1 \cdot (0.5)^4}{4!} = \dfrac{(0.5)^4}{24}\).

Note: Options A and C are identical in value — \(4! = 24\). B is for \(n=4\) (fifth-degree polynomial), not third.

Common trap: Using \(n+1 = 3\) instead of \(n+1 = 4\) (since \(P_3\) means degree 3, the NEXT derivative is the 4th).

Chapter 05 — Applications of Integration
Q 13 ★★★ Hard
Arc Length
🧠
Memory Key Arc length = \(\int_a^b \sqrt{1 + [f'(x)]^2}\,dx\). Simplify under the root FIRST — many problems are designed so it simplifies nicely.
Cable Suspension — Arc Length
A suspension cable hangs along the curve \(y = \dfrac{x^2}{4} - \dfrac{\ln x}{2}\) from \(x = 1\) to \(x = e\).

Compute \(f'(x)\), then simplify \(1 + [f'(x)]^2\) to find the arc length integral in its simplest form. Which expression is the arc length?
A\(\displaystyle\int_1^e \left(\frac{x}{2} + \frac{1}{2x}\right)dx\)
B\(\displaystyle\int_1^e \sqrt{1 + \frac{x^2}{4} + \frac{1}{4x^2}}\,dx\)
C\(\displaystyle\int_1^e \left(\frac{x}{2} - \frac{1}{2x}\right)dx\)
D\(\displaystyle\int_1^e x\,dx\)
Explanation

\(f'(x) = \dfrac{x}{2} - \dfrac{1}{2x}\).

\([f'(x)]^2 = \dfrac{x^2}{4} - \dfrac{1}{2} + \dfrac{1}{4x^2}\).

\(1 + [f'(x)]^2 = \dfrac{x^2}{4} + \dfrac{1}{2} + \dfrac{1}{4x^2} = \left(\dfrac{x}{2} + \dfrac{1}{2x}\right)^2\).

\(\sqrt{\left(\frac{x}{2} + \frac{1}{2x}\right)^2} = \frac{x}{2} + \frac{1}{2x}\) (positive for \(x > 0\)).

Common trap (C): Taking \(\sqrt{[f']^2} = f'\) directly without computing \(1 + [f']^2\) first.

Q 14 ★★★ Hard
Volume by Washer Method
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Memory Key Washer = π∫(R²−r²)dx. R = outer radius, r = inner radius. NEVER do π(R−r)² — that's wrong!
Hollow Turbine Blade — Washer Method
A turbine blade cross-section is formed by rotating the region between \(y = \sqrt{x}\) (outer) and \(y = x\) (inner) for \(0 \leq x \leq 1\) around the \(x\)-axis.

Which integral gives the volume of the hollow solid?
A\(\pi\displaystyle\int_0^1 (\sqrt{x} - x)^2\,dx\)
B\(\pi\displaystyle\int_0^1 (x - x^2)\,dx\)
C\(2\pi\displaystyle\int_0^1 x(\sqrt{x} - x)\,dx\)
D\(\pi\displaystyle\int_0^1 (\sqrt{x})^2\,dx - \pi\displaystyle\int_0^1 x^2\,dx\)
Explanation

Washer method: \(V = \pi\int_0^1\left[R(x)^2 - r(x)^2\right]dx\) where \(R = \sqrt{x},\ r = x\).

\(V = \pi\int_0^1\left(x - x^2\right)dx\).

Option D is equivalent: \(\pi\int(\sqrt{x})^2dx - \pi\int x^2\,dx = \pi\int x\,dx - \pi\int x^2\,dx = \pi\int(x-x^2)dx\). So B = D in value, but B is more compact.

Common trap (A): Squaring the difference \((\sqrt{x}-x)^2 \neq x - x^2\). You must square each term separately.

Option C is the Shell Method — also correct in value, but not what "washer method" asks for.

Q 15 ★★☆ Medium
Work — Variable Force
🧠
Memory Key Work = ∫F(x)dx. Hooke's Law: F = kx (spring). Find k first from a given condition.
Stretching a Spring — Work Integral
A spring requires 9 J of work to stretch it from its natural length to 3 m beyond. Using Hooke's Law (\(F = kx\)), find the work done in stretching the spring from 1 m to 2 m beyond natural length.
Find k First
\(9 = \int_0^3 kx\,dx = \dfrac{k \cdot 9}{2}\) → \(k = 2\) N/m. Then compute \(\int_1^2 2x\,dx\).
A3 J
B5 J
C6 J
D4 J
Explanation

From given info: \(9 = \int_0^3 kx\,dx = k \cdot \frac{9}{2}\) → \(k = 2\).

Work from 1 to 2: \(\int_1^2 2x\,dx = [x^2]_1^2 = 4 - 1 = 3\) J.

Common trap: Forgetting to find \(k\) first and assuming \(F = x\), giving wrong answers.

Chapter 06 — Differential Equations
Q 16 ★★★ Hard
Separable Differential Equations
🧠
Memory Key Separate → Integrate both sides → Solve for y → Apply initial condition.
Population Growth — Logistic vs Exponential
A bacterial culture grows according to \(\dfrac{dy}{dt} = 3y\) with initial condition \(y(0) = 500\).

At what time \(t\) does the population reach 4000?
A\(t = \dfrac{\ln 8}{3}\)
B\(t = \ln 8\)
C\(t = \dfrac{\ln(4000/500)}{3}\) — same as A
D\(t = \dfrac{8}{3}\)
Explanation

Separate: \(\dfrac{dy}{y} = 3\,dt\) → \(\ln|y| = 3t + C\) → \(y = Ae^{3t}\).

IC: \(500 = Ae^0\) → \(A = 500\).

Set \(500e^{3t} = 4000\) → \(e^{3t} = 8\) → \(3t = \ln 8\) → \(t = \dfrac{\ln 8}{3}\).

Options A and C are the same value; both are correct. The answer is \(\frac{\ln 8}{3} \approx 0.693\) hours.

Common trap (D): Computing \(4000/(500 \cdot 3) = 8/3\) — forgetting to take the natural log.

Q 17 ★★★ Hard
Logistic Differential Equation
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Memory Key Logistic: dy/dt = ky(M−y). Solution: \(y = \frac{M}{1+Ae^{-kMt}}\). Inflection at y = M/2.
Island Ecosystem — Logistic Growth
An island ecosystem supports at most \(M = 1000\) foxes (carrying capacity). The population grows logistically with \(k = 0.002\) and initial population \(y(0) = 100\).

At what population size does the growth rate \(\dfrac{dy}{dt}\) reach its maximum?
AWhen \(y = 1000\) (carrying capacity)
BWhen \(y = 100\) (initial population)
CWhen \(y = 500\) (half the carrying capacity)
DWhen \(y = 200\)
Explanation

\(\dfrac{dy}{dt} = ky(M-y)\) is a quadratic in \(y\), maximized at \(y = M/2 = 500\).

Verify: \(\dfrac{d}{dy}[y(M-y)] = M - 2y = 0 \Rightarrow y = M/2\).

This is the inflection point of the logistic curve — where population grows fastest.

Common trap (A): At \(y = M\), growth rate = 0 (population plateaus). At \(y = 0\), growth rate also = 0.

Chapter 07 — Parametric & Polar Curves
Q 18 ★★★ Hard
Parametric Curves — Area
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Memory Key Parametric area: \(\int y\,dx = \int y(t)\cdot x'(t)\,dt\). Watch the direction — if x decreases, the sign flips.
Robotic Arm Path — Parametric Area
A robotic arm traces the curve \(x = t^2,\ y = t^3\) for \(0 \leq t \leq 2\). The area under the parametric curve above the \(x\)-axis is \(\displaystyle\int_0^2 y\,\frac{dx}{dt}\,dt\).

What is this area?
A\(\dfrac{64}{5}\)
B\(16\)
C\(\dfrac{32}{5}\)
D\(8\)
Explanation

\(\dfrac{dx}{dt} = 2t\). So the integral becomes \(\int_0^2 t^3 \cdot 2t\,dt = 2\int_0^2 t^4\,dt = 2\left[\dfrac{t^5}{5}\right]_0^2 = 2 \cdot \dfrac{32}{5} = \dfrac{64}{5}\).

Common trap: Forgetting to compute \(dx/dt = 2t\) and instead integrating \(y\,dt\) directly, getting \(\int_0^2 t^3\,dt = 4\).

Q 19 ★★★ Hard
Polar Coordinates — Area
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Memory Key Polar area: \(A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta\). For regions between curves: \(\frac{1}{2}\int(r_{outer}^2 - r_{inner}^2)d\theta\).
Radar Sweep — Polar Area
A radar dish sweeps through the region inside the cardioid \(r = 1 + \cos\theta\) but outside the circle \(r = 1\) for \(-\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}\).

Which integral gives this area?
A\(\dfrac{1}{2}\displaystyle\int_{-\pi/2}^{\pi/2}(1+\cos\theta)^2\,d\theta\)
B\(\dfrac{1}{2}\displaystyle\int_{-\pi/2}^{\pi/2}\left[(1+\cos\theta)^2 - 1\right]d\theta\)
C\(\dfrac{1}{2}\displaystyle\int_{-\pi/2}^{\pi/2}(cos\theta)^2\,d\theta\)
D\(\displaystyle\int_{-\pi/2}^{\pi/2}(1+\cos\theta - 1)\,d\theta\)
Explanation

Area between two polar curves: \(A = \dfrac{1}{2}\int_\alpha^\beta\left[r_{outer}^2 - r_{inner}^2\right]d\theta\).

Here \(r_{outer} = 1 + \cos\theta\), \(r_{inner} = 1\), so:

\(A = \dfrac{1}{2}\int_{-\pi/2}^{\pi/2}\left[(1+\cos\theta)^2 - 1^2\right]d\theta\).

Common trap (A): Only subtracting the circle area AFTER integrating (equivalent to B but set up incorrectly inside the integral).

Common trap (C): Incorrectly simplifying \((1+\cos\theta)^2 - 1\) as \(\cos^2\theta\). The expansion is \(1 + 2\cos\theta + \cos^2\theta - 1 = 2\cos\theta + \cos^2\theta\).

Q 20 ★★★ Hard
L'Hôpital's Rule — Indeterminate Forms
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Memory Key 0·∞, ∞−∞, 1^∞, 0^0, ∞^0 → rewrite to get 0/0 or ∞/∞, THEN apply L'Hôpital.
Rocket Thrust Limit — L'Hôpital's Rule (Tricky Form)
A rocket's thrust efficiency as speed \(x \to 0^+\) is modeled by \(L = \displaystyle\lim_{x\to 0^+} x^x\).

This is an indeterminate form \(0^0\). Use the substitution \(y = x^x\), take \(\ln y\), and evaluate the limit. What is \(L\)?
Strategy
\(\ln y = x \ln x\) as \(x \to 0^+\) is \(0 \cdot (-\infty)\). Rewrite as \(\dfrac{\ln x}{1/x}\) to get \(\dfrac{-\infty}{+\infty}\), then apply L'Hôpital.
A\(L = 0\)
B\(L = e\)
C\(L = \infty\)
D\(L = 1\)
Explanation

Let \(y = x^x\). Then \(\ln y = x\ln x = \dfrac{\ln x}{1/x}\) (form \(-\infty/+\infty\)).

L'Hôpital: \(\lim_{x\to 0^+}\dfrac{\ln x}{1/x} = \lim_{x\to 0^+}\dfrac{1/x}{-1/x^2} = \lim_{x\to 0^+}(-x) = 0\).

So \(\lim \ln y = 0\), meaning \(\lim y = e^0 = 1\).

Common trap (A): Since \(x \to 0\), students guess \(x^x \to 0\). But the exponent \(x\) also → 0, keeping the result at 1.

Common trap (B): Confusing \(e^0 = e\) (it equals 1, not e).