20 challenging problems covering all key topics. Each question includes a memory tip, worked example, and instant feedback.
20
Problems
5
Topics
★★★
Hard
Q01
Integration
Medium-Hard
A biologist models the rate of bacteria growth as \(r(t) = t \cdot e^{-2t}\) million cells per hour. How many million cells grow from \(t = 0\) to \(t = \infty\)?
$$\int_0^{\infty} t\,e^{-2t}\,dt$$
⚡
Memory Key
IBP = "LIATE" order → Log, Inverse, Algebraic, Trig, Exponential Pick u = first from LIATE, dv = rest
Multiply both sides by \((x+1)(x+3)\): \(1 = A(x+3)+B(x+1)\)
Set \(x=-1\): \(A = 1/2\). Set \(x=-3\): \(B=-1/2\).
So \(\int\frac{dx}{(x+1)(x+3)} = \frac{1}{2}\ln|x+1| - \frac{1}{2}\ln|x+3|+C\)
✅ Explanation
Decompose: \(\frac{5}{(t+1)(t+4)}=\frac{5/3}{t+1}-\frac{5/3}{t+4}\).
\(\int_0^T\left(\frac{5/3}{t+1}-\frac{5/3}{t+4}\right)dt = \frac{5}{3}\left[\ln(t+1)-\ln(t+4)\right]_0^T\)
As \(T\to\infty\): ratio \(\frac{T+1}{T+4}\to1\), so upper term → 0.
Lower: \(-\frac{5}{3}(\ln 1 - \ln 4) = \frac{5}{3}\ln 4\). ✓
Q03
Integration
Hard
An engineer calculates the area of a cross-section using:
\(\int_0^a\sqrt{a^2-x^2}\,dx\) is the area of a quarter-circle of radius \(a\).
Area \(= \frac{1}{4}\pi a^2\).
Always check: can the integral be interpreted geometrically before computing?
✅ Explanation
\(\int_0^3\sqrt{9-x^2}\,dx\) = area of quarter-circle with radius \(r=3\).
\(= \frac{1}{4}\pi(3)^2 = \frac{9\pi}{4}\). ✓
Alternatively via trig-sub \(x=3\sin\theta\): same result.
Q04
Integration
Hard — Common Trap!
A student claims: "Since \(\int_{-1}^{1}\frac{1}{x}\,dx\) is symmetric, the answer is 0."
Evaluate the integral correctly:
$$\int_{-1}^{1} \frac{1}{x}\,dx$$
⚡
Memory Key
IMPROPER = split at singularity FIRST! Never cancel across a discontinuity Both halves must converge independently
📖 The Symmetry Trap▾
\(\int_{-1}^1 \frac{1}{x}dx\) must be split: \(\int_{-1}^0 \frac{dx}{x} + \int_0^1 \frac{dx}{x}\)
Each part → \(\lim_{b\to0^-}[\ln|x|]_{-1}^b = \lim_{b\to0^-}\ln|b| = -\infty\)
Since one part diverges → the whole integral DIVERGES. "Cancellation" is illegal here!
❌ Common Mistake
You CANNOT cancel symmetric improper integrals unless both sides independently converge.
\(\int_{-1}^0\frac{dx}{x} = -\infty\) and \(\int_0^1\frac{dx}{x}=+\infty\).
\(\infty - \infty\) is indeterminate — the integral DIVERGES. The "Cauchy principal value" is 0, but that is NOT the same as convergence!
Q05
Applications
Medium-Hard
A vase is formed by rotating the region between \(y = \sqrt{x}\) and \(y = x^2\) (where \(0 \le x \le 1\)) around the \(x\)-axis. Find its volume.
WASHER = π∫(R² − r²)dx R = outer radius (farther from axis) r = inner radius (closer to axis) DISK = π∫R² dx (no hole)
📖 Disk vs Washer▾
Always sketch! Identify which curve is on the outside (larger y-value).
From \(0\) to \(1\): \(\sqrt{x} \ge x^2\), so \(\sqrt{x}\) is outer, \(x^2\) is inner.
Volume = \(\pi\int_0^1(x - x^4)dx = \pi\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{2}-\frac{1}{5}\right)\)
Use the identity \(1+\sinh^2 x = \cosh^2 x\).
\(L = \int_0^1\cosh x\,dx = \sinh(1) - \sinh(0) = \sinh(1) \approx 1.175\). ✓
Q07
Sequences
Medium-Hard
A drug's concentration decreases each day by a ratio. The concentration on day \(n\) is given by \(a_n = \left(1 + \dfrac{3}{n}\right)^n\).
What does \(a_n\) approach as \(n \to \infty\)?
$$\lim_{n\to\infty}\left(1+\frac{3}{n}\right)^n$$
⚡
Memory Key
GOLDEN LIMIT: \(\lim_{n\to\infty}(1+\frac{k}{n})^n = e^k\) → Take ln, use L'Hôpital, then exponentiate 1∞ form → always take ln first!
📖 Indeterminate 1∞ Form▾
Let \(L = \lim(1+k/n)^n\). Take \(\ln L = \lim n\ln(1+k/n) = \lim\frac{\ln(1+k/n)}{1/n}\).
This is \(\frac{0}{0}\) → L'Hôpital → get \(k\).
So \(\ln L = k\), meaning \(L = e^k\).
✅ Explanation
Using the identity \(\lim_{n\to\infty}(1+k/n)^n = e^k\) with \(k=3\):
\(\lim_{n\to\infty}(1+3/n)^n = e^3\). ✓
This generalizes Euler's number: \(e = \lim(1+1/n)^n\).
Q08
Series
Medium
A ball is dropped from 10 meters. Each bounce reaches 60% of the previous height. What is the total distance the ball travels?
$$D = 10 + 2\sum_{n=1}^{\infty} 10\cdot(0.6)^n$$
⚡
Memory Key
GEO SUM: \(\sum_{n=0}^\infty ar^n = \frac{a}{1-r}\), |r| < 1 Bouncing ball: count DOWN + UP separately Total = first drop + 2 × (geometric sum of bounces)
📖 Bouncing Ball Setup▾
Drop height: 10m. Bounce heights: 6, 3.6, 2.16, …
Each bounce = up + down = \(2 \times\) bounce height.
Total = 10 + 2(6 + 3.6 + 2.16 + …) = 10 + 2 · \(\frac{6}{1-0.6}\) = 10 + 2(15) = 40 m.
✅ Explanation
Sum of bounce heights: \(\sum_{n=1}^\infty 10(0.6)^n = \frac{10 \cdot 0.6}{1-0.6} = \frac{6}{0.4} = 15\) m.
Each bounce = up AND down → ×2. Plus initial drop.
Total = \(10 + 2(15) = 40\) meters. ✓
Q09
Series
Hard — Many Confuse This!
A physics model predicts energy levels proportional to \(\dfrac{1}{n\ln n}\). Does the total energy sum converge?
$$\sum_{n=2}^{\infty} \frac{1}{n \ln n}$$
⚡
Memory Key
INTEGRAL TEST: same behavior as ∫f(x)dx ∫dx/(x ln x) = ln(ln x) → ∞ (diverges!) p-series: 1/nᵖ converges iff p > 1
Since integral diverges → series DIVERGES. Even though terms go to 0!
✅ Explanation
Trap: \(a_n \to 0\) is NECESSARY but NOT sufficient for convergence (harmonic series also goes to 0!).
Integral test: \(\int_2^\infty \frac{dx}{x\ln x} = \ln(\ln x)\Big|_2^\infty = \infty\).
So the series DIVERGES. This is slower than harmonic series but same fate.
Q10
Series
Medium-Hard
A computer algorithm's memory usage at step \(n\) is \(\dfrac{n^5}{5^n}\). Does total memory converge?
$$\sum_{n=1}^{\infty} \frac{n^5}{5^n}$$
⚡
Memory Key
RATIO TEST: L = lim|aₙ₊₁/aₙ| L < 1 → Converges, L > 1 → Diverges L = 1 → Inconclusive Best for: factorials, exponentials
AST (Alternating Series Test): bₙ decreasing + bₙ→0 → converges Absolute convergence: ∑|aₙ| converges Conditional: converges but NOT absolutely
📖 Absolute vs Conditional▾
This is the alternating harmonic series = \(\ln 2\).
AST: \(b_n = 1/n\) is decreasing and \(\to 0\) → Converges.
But \(\sum |a_n| = \sum 1/n\) = harmonic series → DIVERGES.
Therefore: conditionally convergent.
✅ Explanation
By AST: \(b_n = 1/n\) decreases to 0 → series converges (= \(\ln 2\)).
\(\sum|a_n| = \sum 1/n\) diverges → NOT absolutely convergent.
Therefore: conditionally convergent. ✓
Q12
Series
Hard
A calculator approximates \(e^{-x^2}\). Using the known Taylor series for \(e^u\), what is the power series for \(e^{-x^2}\)?
\(= 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots\) — this is used in statistics (normal distribution)!
✅ Explanation
Substitute \(u=-x^2\) into \(e^u=\sum u^n/n!\):
\(e^{-x^2} = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}\) ✓
Note: sum starts at \(n=0\), not \(n=1\) (trap in option D!).
Q13
Series
Hard
A signal is modeled by a power series. Find the radius of convergence:
GABRIEL'S HORN paradox: Finite volume but infinite surface area "You can fill it with paint but can't paint it" SA = 2π∫y·ds where ds = √(1+(dy/dx)²)dx
📖 Comparison to Prove Divergence▾
Since \(\sqrt{1+1/x^4} \ge 1\), we have:
\(SA \ge 2\pi\int_1^\infty \frac{1}{x}\,dx = 2\pi[\ln x]_1^\infty = \infty\)
By Comparison Test → SA diverges to infinity!
✅ Explanation — Gabriel's Horn Paradox!
Since \(\sqrt{1+1/x^4}\ge 1\): \(SA \ge 2\pi\int_1^\infty\frac{dx}{x} = 2\pi\ln\infty = \infty\). Paradox: Volume = \(\pi\) (finite), but Surface Area = \(\infty\).
Volume and surface area are independent — finite volume does NOT imply finite surface area!
Q15
Polar & Parametric
Hard
A radar sweep covers the area of one petal of the rose curve \(r = \cos(2\theta)\). Find the area of one petal.
PARAMETRIC ARC = ∫√((dx/dt)²+(dy/dt)²)dt Unit circle: x=cos t, y=sin t → speed=1 Distance = speed × time = 1 × 2π = 2π
📖 Speed on a Parametric Curve▾
\(\frac{dx}{dt}=-\sin t\), \(\frac{dy}{dt}=\cos t\)
Speed \(= \sqrt{\sin^2 t + \cos^2 t} = \sqrt{1} = 1\) (constant!)
\(L = \int_0^{2\pi} 1\,dt = 2\pi\) — circumference of unit circle ✓
✅ Explanation
Speed \(= \sqrt{(-\sin t)^2+(\cos t)^2} = 1\).
\(L = \int_0^{2\pi}1\,dt = 2\pi\) ✓ — equals circumference of unit circle.
Q17
Applications
Medium-Hard
A spring with spring constant \(k = 6\) N/m requires work to stretch it from its natural length. How much work (in Joules) is done stretching it from \(x = 0\) to \(x = 0.5\) m?
$$W = \int_0^{0.5} 6x\,dx$$
⚡
Memory Key
HOOKE'S LAW: F = kx Work = ∫F dx = ∫kx dx = ½kx² Units: N·m = Joules Don't forget: stretch FROM natural length
📖 Hooke's Law Setup▾
Spring force: \(F(x) = kx = 6x\)
Work to stretch from 0 to 0.5:
\(W = \int_0^{0.5}6x\,dx = 6\cdot\frac{x^2}{2}\Big|_0^{0.5} = 3(0.25) = 0.75\) J
ROOT TEST: L = lim ⁿ√|aₙ| = lim |aₙ|^(1/n) L < 1 → Converges, L > 1 → Diverges Best for: (something)ⁿ — just remove the exponent n!
📖 Root Test Technique▾
\(a_n = \left(\frac{2n+1}{3n+2}\right)^n\)
\(\sqrt[n]{a_n} = \frac{2n+1}{3n+2} \to \frac{2}{3}\) as \(n\to\infty\)
Since \(L = 2/3 < 1\) → Converges absolutely.
✅ Explanation
Root Test: \(L = \lim_{n\to\infty}\sqrt[n]{a_n} = \lim\frac{2n+1}{3n+2} = \frac{2}{3} < 1\)
→ Converges absolutely. ✓
The Root Test is ideal when \(a_n\) is already in the form \((\cdots)^n\).
Q19
Sequences
Hard — L'Hôpital Trap
A population model uses \(a_n = \dfrac{\ln n}{n}\). Find \(\lim_{n \to \infty} a_n\).
$$\lim_{n\to\infty}\frac{\ln n}{n}$$
Now find: does \(\displaystyle\sum_{n=2}^{\infty} \frac{\ln n}{n}\) converge or diverge?
⚡
Memory Key
aₙ→0 does NOT guarantee convergence! Comparison: ln n / n > 1/n for large n Since ∑1/n diverges → ∑(ln n)/n diverges
📖 L'Hôpital for Sequences▾
Treat as a function of \(x\): \(\lim_{x\to\infty}\frac{\ln x}{x}\) is \(\infty/\infty\).
L'Hôpital: \(\lim \frac{1/x}{1} = \lim\frac{1}{x} = 0\).
So \(a_n \to 0\). BUT for the series: since \(\frac{\ln n}{n} > \frac{1}{n}\) for \(n \ge 3\), by comparison with harmonic series → DIVERGES.
✅ Explanation — Classic Trap!
L'Hôpital: \(\lim\frac{\ln n}{n} = 0\) ✓
BUT Divergence Test only rules out convergence if limit ≠ 0. Limit = 0 tells us nothing!
Comparison: \(\frac{\ln n}{n} > \frac{1}{n}\) for \(n\ge 3\) → by comparison with harmonic series → series DIVERGES. ✓
Q20
Applications
Hard — Boss Level
A population \(P(t)\) satisfies the logistic equation \(\dfrac{dP}{dt} = 2P(1-P)\) with \(P(0) = \dfrac{1}{3}\). Using partial fractions to integrate, find \(P(t)\).
$$\int\frac{dP}{P(1-P)} = \int 2\,dt$$
⚡
Memory Key
LOGISTIC: separate variables → partial fractions ∫dP/[P(1−P)] = ln|P| − ln|1−P| = ln|P/(1−P)| Solve for P: P = Ce²ᵗ/(1+Ce²ᵗ) Apply IC to find C
After partial fractions and integration: \(\frac{P}{1-P} = Ae^{2t}\).
IC \(P(0)=1/3\): \(\frac{1/3}{2/3} = A = 1/2\).
Solve for \(P\): \(P = \frac{e^{2t}/2}{1+e^{2t}/2} = \frac{e^{2t}}{2+e^{2t}}\) ✓
Note: option B is equivalent! \(\frac{1}{1+2e^{-2t}} = \frac{e^{2t}}{e^{2t}+2} = \frac{e^{2t}}{2+e^{2t}}\). Both A and B are correct — good trap!