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 \(u=t,\ dv=e^{-t}dt\) → \(-te^{-t} + \int e^{-t}\,dt = -te^{-t} - e^{-t}\).
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?
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 \(r = 3/4\).
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?
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.
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?
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
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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.
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?
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
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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.
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 \(-\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}\).
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 \(-\infty/+\infty\)).