Section 01 · Integration Techniques
LIATE: Log · Inverse trig · Algebraic · Trig · Exponential — pick u in this order
The Expanding Product
A company's revenue model predicts that after \(t\) months, the marginal revenue is \(R'(t) = t^2 e^{-t}\) (in thousands of dollars). Starting from \(t = 0\), find the total revenue accumulated over the interval \([0, \infty)\).
$$\int_0^{\infty} t^2 e^{-t}\, dt$$
⚠ TRAP: You need to apply integration by parts twice, and each time reduce the power of \(t\). Don't forget the improper integral evaluation requires a limit as \(t \to \infty\).
Step-by-step Explanation
Step 1 · First IBPSet \(u = t^2,\; dv = e^{-t}dt\). Then \(du = 2t\,dt,\; v = -e^{-t}\).
Result: \(\bigl[-t^2 e^{-t}\bigr]_0^\infty + 2\int_0^\infty t e^{-t}\,dt\). The boundary term vanishes (L'Hôpital), leaving \(2\int_0^\infty t e^{-t}\,dt\).
Result: \(\bigl[-t^2 e^{-t}\bigr]_0^\infty + 2\int_0^\infty t e^{-t}\,dt\). The boundary term vanishes (L'Hôpital), leaving \(2\int_0^\infty t e^{-t}\,dt\).
Step 2 · Second IBPNow set \(u=t,\;dv=e^{-t}dt\). Same trick gives \(\int_0^\infty t e^{-t}\,dt = 1\).
Step 3 · Combine\(2 \times 1 = \mathbf{2}\). In general, \(\int_0^\infty t^n e^{-t}\,dt = n!\) — this is the Gamma function \(\Gamma(n+1)=n!\).
SUB RULE: √(a²−x²)→x=a sinθ · √(a²+x²)→x=a tanθ · √(x²−a²)→x=a secθ
The Rope Around the Post
A rope is wound around a circular post of radius 3 m. The length of rope lying flat in one region is modeled by \(\displaystyle\int_0^3 \frac{x^2}{\sqrt{9 - x^2}}\,dx\). Find this length in meters.
$$\int_0^3 \frac{x^2}{\sqrt{9-x^2}}\,dx$$
⚠ TRAP: After substituting \(x = 3\sin\theta\), the integral becomes a trig integral in \(\sin^2\theta\). Use the identity \(\sin^2\theta = \frac{1-\cos 2\theta}{2}\) — don't skip this step.
Step-by-step Explanation
Step 1 · Substitute\(x=3\sin\theta,\;dx=3\cos\theta\,d\theta,\;\sqrt{9-x^2}=3\cos\theta\). Limits: \(\theta:0\to\pi/2\).
Step 2 · Simplify\(\displaystyle\int_0^{\pi/2}\frac{9\sin^2\theta}{3\cos\theta}\cdot 3\cos\theta\,d\theta = 9\int_0^{\pi/2}\sin^2\theta\,d\theta\).
Step 3 · Half-angle identity\(9\int_0^{\pi/2}\frac{1-\cos2\theta}{2}\,d\theta = \dfrac{9}{2}\cdot\dfrac{\pi}{2} = \boxed{\dfrac{9\pi}{4}}\).
PARTIAL FRAC: factor denom → assign A/(x−r) for each factor → multiply both sides → solve system
The Mixing Tank
A 100-gallon tank is draining such that the outflow rate follows \(\dfrac{dx}{dt} = \dfrac{6}{x^2 - 4}\), where \(x\) represents concentration. To find the elapsed time we need:
$$\int \frac{6}{x^2-4}\,dx$$
⚠ TRAP: Factor as \((x-2)(x+2)\), not as a difference of squares to plug into an inverse trig formula. Many students incorrectly use \(\arctan\) here — that only works for \(x^2 + a^2\).
Step-by-step Explanation
Step 1 · Decompose\(\dfrac{6}{(x-2)(x+2)}=\dfrac{A}{x-2}+\dfrac{B}{x+2}\). Multiply out: \(6=A(x+2)+B(x-2)\). Set \(x=2\): \(A=\tfrac{3}{2}\). Set \(x=-2\): \(B=-\tfrac{3}{2}\).
Step 2 · Integrate\(\displaystyle\int\!\left(\frac{3/2}{x-2}-\frac{3/2}{x+2}\right)dx = \frac{3}{2}\ln|x-2|-\frac{3}{2}\ln|x+2|+C = \frac{3}{2}\ln\!\left|\frac{x-2}{x+2}\right|+C\).
Key pointChoice (A) has the fraction flipped — very common error! Check your logarithm algebra carefully.
ODD POWER: save one factor for du · use Pythagorean identity on the rest · EVEN: half-angle
The Wave Energy Model
Ocean wave energy over one cycle is proportional to \(\displaystyle\int_0^{\pi} \sin^5 x\,dx\). Compute this integral.
$$\int_0^{\pi} \sin^5 x\,dx$$
⚠ TRAP: The power 5 is odd. Write \(\sin^5 x = \sin^4 x \cdot \sin x\), then replace \(\sin^4 x = (1-\cos^2 x)^2\). Don't try integration by parts — it creates an infinite loop here.
Step-by-step Explanation
Step 1 · Rewrite with odd-power trick\(\sin^5 x=(1-\cos^2 x)^2\sin x\). Let \(u=\cos x,\;du=-\sin x\,dx\). Limits \(x:0\to\pi\) become \(u:1\to-1\).
Step 2 · Expand and integrate\(-\int_1^{-1}(1-u^2)^2\,du=\int_{-1}^{1}(1-2u^2+u^4)\,du=\left[u-\frac{2u^3}{3}+\frac{u^5}{5}\right]_{-1}^{1}\).
Step 3 · Evaluate\(\left(1-\tfrac{2}{3}+\tfrac{1}{5}\right)-\left(-1+\tfrac{2}{3}-\tfrac{1}{5}\right)=2\!\left(1-\tfrac{2}{3}+\tfrac{1}{5}\right)=2\cdot\tfrac{8}{15}=\boxed{\dfrac{16}{15}}\).
Improper Integrals & Series
Section 02 · Convergence
p-INTEGRAL: ∫₁^∞ 1/xᵖ dx converges iff p > 1 · ∫₀¹ 1/xᵖ dx converges iff p < 1
The Radioactive Decay Problem
A radioactive substance decays so that the total amount ever released is \(\displaystyle\int_1^{\infty} \frac{\ln x}{x^3}\,dx\). Does this integral converge, and if so, what is its value?
$$\int_1^{\infty} \frac{\ln x}{x^3}\,dx$$
⚠ TRAP: Students often panic seeing \(\ln x\) and try a substitution. Instead, use integration by parts with \(u=\ln x\), \(dv=x^{-3}dx\). Then carefully evaluate the limit as \(x\to\infty\) of \(\frac{\ln x}{x^2}\) — this goes to 0 (logarithms grow slower than any power).
Step-by-step Explanation
IBP setup\(u=\ln x,\;dv=x^{-3}dx \Rightarrow du=\frac{1}{x}dx,\;v=-\frac{1}{2x^2}\).
Apply IBP\(\Bigl[-\frac{\ln x}{2x^2}\Bigr]_1^\infty + \int_1^\infty\frac{1}{2x^3}\,dx\). First term: limit = 0 (L'Hôpital). Boundary at 1: 0.
Second integral\(\displaystyle\frac{1}{2}\int_1^\infty x^{-3}\,dx=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\). ✓ Converges to \(\boxed{1/4}\).
COMPARISON: find a simpler bₙ · if 0≤aₙ≤bₙ and Σbₙ converges → Σaₙ converges
The Population Sum
A discrete population model gives the total long-run size as \(\displaystyle\sum_{n=1}^{\infty} \frac{\sin^2(n)}{n^2}\). Does this series converge?
$$\sum_{n=1}^{\infty} \frac{\sin^2(n)}{n^2}$$
⚠ TRAP: You cannot test this with the integral test (the terms aren't monotone). Key insight: since \(0 \le \sin^2(n) \le 1\), we have \(0 \le \dfrac{\sin^2(n)}{n^2} \le \dfrac{1}{n^2}\). Use direct comparison with the p-series.
Step-by-step Explanation
Key inequalitySince \(\sin^2(n)\le 1\) for all \(n\), we have \(0\le\frac{\sin^2 n}{n^2}\le\frac{1}{n^2}\).
p-series\(\sum\frac{1}{n^2}\) converges (p=2 > 1). By Direct Comparison Test, our series also converges.
Common errorOscillation of a factor does NOT cause divergence as long as the bounding series converges. \(\sin^2(n)\) is bounded, which is what matters.
RATIO TEST: L=lim|a_{n+1}/aₙ| · L<1 converges · L>1 diverges · L=1 inconclusive
The Drug Dosage Series
A patient receives repeated doses so that total drug concentration is modeled by \(\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n \, 3^n}{(2n)!}\). Determine convergence and identify the closed-form sum.
$$\sum_{n=0}^{\infty} \frac{(-1)^n \cdot 3^n}{(2n)!}$$
⚠ TRAP: This is a disguised Maclaurin series. Recall \(\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\). Here \(x^{2n} = 3^n\) means \(x^2 = 3\), so \(x = \sqrt{3}\). Don't just say "ratio test converges" — find the exact sum!
Step-by-step Explanation
Recognize the patternThe Maclaurin series is \(\cos x=\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n)!}\). Matching: \(x^{2n}=(\sqrt{3})^{2n}=3^n\). So \(x=\sqrt{3}\).
Sum\(\sum_{n=0}^\infty\frac{(-1)^n 3^n}{(2n)!}=\cos(\sqrt{3})\approx -0.16\). Converges for all \(x\) (radius of convergence = ∞).
Key lessonAlways check if a series matches a known Maclaurin series before applying ratio test — it gives you the exact sum, not just convergence.
AST: alternating + decreasing + lim bₙ=0 → converges · error ≤ first omitted term
The Alternating Investment
An investment gains and loses value alternately. The net long-run value is modeled by \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3}\). How many terms are needed to approximate the sum within an error of \(0.001\)?
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3},\quad \text{error} < 0.001$$
⚠ TRAP: By the Alternating Series Estimation Theorem, the error is bounded by the \((N+1)\)-th term. You need \(\dfrac{1}{(N+1)^3} < 0.001\), i.e., \((N+1)^3 > 1000\).
Step-by-step Explanation
AST Error boundError \(\le b_{N+1}=\frac{1}{(N+1)^3}\). We need \(\frac{1}{(N+1)^3}<0.001=\frac{1}{1000}\), so \((N+1)^3>1000\), giving \(N+1>10\), i.e., \(N\ge 10\).
VerifyAt \(N=9\): error bound = \(1/10^3=0.001\) — NOT strictly less than. At \(N=10\): error bound = \(1/11^3\approx 0.00075 < 0.001\). ✓ So we need \(N=\mathbf{10}\) terms.
Power Series & Taylor
Section 03 · Power Series
ROC: use ratio test on power series · R = lim|aₙ/a_{n+1}| · check ENDPOINTS separately!
The Signal Filter
A signal processing model uses the power series \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n (x-2)^n}{n \cdot 3^n}\). Find the interval of convergence (including endpoint behavior).
$$\sum_{n=1}^{\infty} \frac{(-1)^n (x-2)^n}{n \cdot 3^n}$$
⚠ TRAP: The center is \(x=2\), not \(x=0\)! After finding radius \(R=3\), the interval is \(-1 < x < 5\). But you MUST check \(x=-1\) and \(x=5\) separately — they often have different behavior.
Step-by-step Explanation
RadiusRatio test: \(\lim\frac{|x-2|}{3}\cdot\frac{n}{n+1}=\frac{|x-2|}{3}\). Converges when \(|x-2|<3\), so \(R=3\), center \(x=2\): interval \(-1 < x < 5\).
Endpoint x=5\(\sum\frac{(-1)^n\cdot 3^n}{n\cdot 3^n}=\sum\frac{(-1)^n}{n}\) — alternating harmonic series, converges by AST. ✓
Endpoint x=−1\(\sum\frac{(-1)^n(-3)^n}{n\cdot 3^n}=\sum\frac{(-1)^n(-1)^n}{n}=\sum\frac{1}{n}\) — harmonic series, diverges. ✗
AnswerInterval of convergence: \(\mathbf{(-1,\, 5]}\). Always check endpoints!
TAYLOR ERROR: |Rₙ(x)| ≤ M|x−a|ⁿ⁺¹/(n+1)! where M = max of |f⁽ⁿ⁺¹⁾| on interval
The GPS Approximation
A GPS chip approximates \(\cos(0.1)\) using the 4th-degree Taylor polynomial centered at 0. What is the maximum error in this approximation?
$$\cos(0.1) \approx 1 - \frac{(0.1)^2}{2!} + \frac{(0.1)^4}{4!}, \quad |R_4(x)| \le \frac{M\,|x|^5}{5!}$$
⚠ TRAP: The next non-zero term of \(\cos x\) after degree 4 is \((-1)^3\frac{x^6}{6!}\). But the error for a degree-4 polynomial is bounded by the 5th-degree term — and the 5th derivative of \(\cos x\) is \(-\sin x\), whose maximum absolute value is 1.
Step-by-step Explanation
Taylor Remainder\(|R_4(0.1)|\le\frac{M\cdot(0.1)^5}{5!}\) where \(M=\max|f^{(5)}(x)|=\max|\sin x|=1\) on \([0,0.1]\).
Compute\(\frac{(0.1)^5}{5!}=\frac{10^{-5}}{120}\approx 8.3\times 10^{-8}\). Extraordinarily small!
Why (A) is wrongChoice (A) forgets to raise 0.1 to the 5th power — a very common algebra error when plugging into the error formula.
Applications of Integration
Section 04 · Area, Volume, Arc Length
WASHER: V=π∫(R²−r²)dx · SHELL: V=2π∫x·f(x)dx · choose method based on axis of rotation
The Coffee Cup Design
A coffee cup is designed by rotating the region bounded by \(y=\sqrt{x}\), \(y=0\), and \(x=4\) around the \(x\)-axis. A cylindrical hole of radius 1 is bored through the center. Find the volume of the resulting solid.
$$V = \pi\int_1^4\!\left[(\sqrt{x})^2 - 1^2\right]dx + \pi\int_0^1(\sqrt{x})^2\,dx$$
⚠ TRAP: The hole only exists where \(\sqrt{x} \ge 1\), i.e., for \(x \ge 1\). For \(0 \le x \le 1\), the solid is a full disk of radius \(\sqrt{x}\). You need two separate integrals!
Step-by-step Explanation
Full disk region [0,1]\(\pi\int_0^1 x\,dx=\pi\cdot\frac{1}{2}=\frac{\pi}{2}\).
Washer region [1,4]\(\pi\int_1^4(x-1)\,dx=\pi\left[\frac{x^2}{2}-x\right]_1^4=\pi\left[(8-4)-(\frac{1}{2}-1)\right]=\pi\cdot\frac{9}{2}=\frac{9\pi}{2}\).
Total\(\frac{\pi}{2}+\frac{9\pi}{2}=\boxed{5\pi}\). Wait — re-check the problem setup. The washer outer radius is \(\sqrt{x}\), inner is 1. If the hole runs the full length 0 to 4, then volume = \(\pi\int_0^4 x\,dx - \pi\int_0^4 1\,dx = 8\pi - 4\pi = \mathbf{4\pi}\). Make sure you read the geometry carefully!
SHELL: horizontal strips → rotate around y-axis → V=2π∫x f(x)dx (easier when disk gives messy inverse)
The Rotating Arch Bridge
An arch bridge profile is the region bounded by \(y = 4x - x^2\) and \(y=0\). If this region is rotated around the \(y\)-axis, find the volume of the solid formed.
$$V = 2\pi \int_0^4 x(4x - x^2)\,dx$$
⚠ TRAP: Using the disk/washer method here requires expressing \(x\) as a function of \(y\) (messy quadratic!). The shell method is far cleaner. Identify: \(p(x)=x\) (radius), \(h(x)=4x-x^2\) (height).
Step-by-step Explanation
Shell formula\(V=2\pi\int_0^4 x(4x-x^2)\,dx=2\pi\int_0^4(4x^2-x^3)\,dx\).
Integrate\(2\pi\!\left[\frac{4x^3}{3}-\frac{x^4}{4}\right]_0^4=2\pi\!\left(\frac{256}{3}-64\right)=2\pi\cdot\frac{64}{3}=\boxed{\frac{128\pi}{3}}\).
ARC LENGTH: L=∫√(1+(dy/dx)²)dx · pick a curve whose (dy/dx)² simplifies nicely!
The Suspension Cable
A suspension bridge cable hangs in the shape of \(y = \frac{x^2}{8}\) from \(x=0\) to \(x=4\). Find the length of the cable.
$$L = \int_0^4 \sqrt{1 + \left(\frac{x}{4}\right)^2}\,dx$$
⚠ TRAP: After computing \(y' = x/4\), you get \(\sqrt{1+x^2/16}\). Let \(x = 4\tan\theta\) (trig sub). Many students forget to convert the limits back properly after trig substitution.
Step-by-step Explanation
Setup\(y'=x/4\), so \(L=\int_0^4\sqrt{1+x^2/16}\,dx\). Let \(x=4\tan\theta\), \(dx=4\sec^2\theta\,d\theta\), \(\sqrt{1+\tan^2\theta}=\sec\theta\). Limits: \(\theta:0\to\pi/4\).
Integral\(4\int_0^{\pi/4}\sec^3\theta\,d\theta=4\cdot\frac{1}{2}\bigl[\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\bigr]_0^{\pi/4}\).
Evaluate\(=2\bigl[(\sqrt{2})(1)+\ln(\sqrt{2}+1)-0-0\bigr]=2\sqrt{2}+2\ln(1+\sqrt{2})\). Use the formula \(\int\sec^3\theta\,d\theta=\frac{1}{2}(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)+C\) — memorize this!
SURFACE: S=2π∫y√(1+(y')²)dx (rotate around x-axis) · don't confuse with arc length formula
The Paint Can Problem
Gabriel's horn is formed by rotating \(y = \dfrac{1}{x}\) for \(x \in [1, \infty)\) around the \(x\)-axis. Paradoxically, it has finite volume but infinite surface area. Verify by computing the surface area integral's behavior.
$$S = 2\pi\int_1^{\infty} \frac{1}{x}\sqrt{1 + \frac{1}{x^4}}\,dx \quad \text{vs} \quad 2\pi\int_1^{\infty}\frac{dx}{x}$$
⚠ TRAP: Students expect that since the volume \(\pi\int_1^\infty \frac{1}{x^2}dx = \pi\) is finite, the surface area must also be finite. But \(\sqrt{1+1/x^4} \ge 1\), so the surface area integral is bounded below by \(2\pi\int_1^\infty \frac{1}{x}dx\).
Step-by-step Explanation
Lower bound argumentSince \(\sqrt{1+1/x^4}\ge 1\), we have \(S\ge 2\pi\int_1^\infty\frac{1}{x}\,dx=2\pi\ln x\Big|_1^\infty=\infty\).
The paradoxVolume \(V=\pi\int_1^\infty x^{-2}\,dx=\pi\) (converges, p=2>1), but surface area diverges (p=1, harmonic series). You can fill it with \(\pi\) cubic units of paint, but you cannot paint its surface!
LessonFinite volume does NOT imply finite surface area. The comparison test for integrals is the key tool here.
Differential Equations
Section 05 · ODEs & Models
SEPARABLE: get all y on one side · all x on other · integrate both sides · don't forget +C before solving for y
The Cooling Espresso
A freshly brewed espresso at 90°C is placed in a 20°C room. By Newton's Law of Cooling, \(\frac{dT}{dt} = -k(T-20)\). If the coffee cools to 70°C in 5 minutes, at what time \(t\) does the temperature reach 40°C?
$$\frac{dT}{dt} = -k(T - 20), \quad T(0)=90,\; T(5)=70$$
⚠ TRAP: Let \(u = T - 20\), so the equation becomes \(\frac{du}{dt}=-ku\), giving \(u = u_0 e^{-kt}\). Many students substitute \(T = 40\) but forget that the relevant quantity is \(T - 20 = 20\), not \(T = 40\).
Step-by-step Explanation
General solution\(T(t)=20+70e^{-kt}\). At \(t=5\): \(70=20+70e^{-5k}\Rightarrow e^{-5k}=\frac{50}{70}=\frac{5}{7}\Rightarrow k=\frac{1}{5}\ln\frac{7}{5}\).
Find t for T=40\(40=20+70e^{-kt}\Rightarrow e^{-kt}=\frac{20}{70}=\frac{2}{7}\Rightarrow kt=\ln\frac{7}{2}\).
Compute\(t=\frac{\ln(7/2)}{k}=\frac{5\ln(7/2)}{\ln(7/5)}\approx\frac{5\times1.253}{0.336}\approx\mathbf{18.6}\) min. Closest to choice (B).
LOGISTIC: dP/dt = kP(1−P/M) · solution P(t) = M/(1+Ae^{−kt}) · inflection at P=M/2
The Viral Video
A video goes viral. The number of views (in millions) follows \(\dfrac{dP}{dt} = 0.4P\!\left(1 - \dfrac{P}{10}\right)\) with \(P(0) = 0.5\). When does the video reach 5 million views (the inflection point)?
$$P(t) = \frac{10}{1 + Ae^{-0.4t}}, \quad A = \frac{10 - P_0}{P_0}$$
⚠ TRAP: The inflection occurs at \(P = M/2 = 5\). Plug into the logistic formula and solve for \(t\). Students often instead set \(P''(t)=0\) which is much harder. Use the formula directly!
Step-by-step Explanation
Find A\(A=\frac{10-0.5}{0.5}=19\). So \(P(t)=\frac{10}{1+19e^{-0.4t}}\).
Set P=5\(5=\frac{10}{1+19e^{-0.4t}}\Rightarrow 1+19e^{-0.4t}=2\Rightarrow e^{-0.4t}=\frac{1}{19}\Rightarrow t=\frac{\ln 19}{0.4}\approx 7.3\) days.
Key insightInflection at \(P=M/2\) is a universal property of logistic growth. Growth is fastest here — after this, growth slows as the population approaches carrying capacity \(M=10\).
Parametric & Polar Curves
Section 06 · Parametric & Polar
PARAMETRIC AREA: A=∫y dx = ∫y(t)·x'(t)dt · watch the direction — if t increases left to right, no sign flip
The Rollercoaster Loop
A rollercoaster loop is traced by the parametric curve \(x = \cos t,\; y = \sin t \cos t = \frac{1}{2}\sin 2t\) for \(t \in [0, \pi]\). Find the area enclosed by this figure-eight portion from \(t=0\) to \(t=\pi/2\).
$$A = \left|\int_0^{\pi/2} y(t)\,x'(t)\,dt\right| = \left|\int_0^{\pi/2} \sin t \cos t \cdot (-\sin t)\,dt\right|$$
⚠ TRAP: The formula gives a signed area. Since \(x'(t) = -\sin t < 0\) on \([0, \pi/2]\), the parametric area integral is negative. Take the absolute value, OR swap the limits. Don't forget this sign issue!
Step-by-step Explanation
Setup\(\int_0^{\pi/2}\sin t\cos t\cdot(-\sin t)\,dt=-\int_0^{\pi/2}\sin^2 t\cos t\,dt\).
SubstituteLet \(u=\sin t,\;du=\cos t\,dt\). Limits: \(0\to 1\). \(-\int_0^1 u^2\,du=-\frac{1}{3}\). Area = \(\boxed{\frac{1}{3}}\).
POLAR AREA: A=½∫r²dθ · BETWEEN curves: A=½∫(r_outer²−r_inner²)dθ · find intersection first!
The Radar Sweep
A radar sweeps an area modeled by the region inside the circle \(r = 3\sin\theta\) but outside the circle \(r = \sin\theta + 1\). Find the area of this region.
$$A = \frac{1}{2}\int_{\alpha}^{\beta}\left[(3\sin\theta)^2 - (\sin\theta+1)^2\right]d\theta$$
⚠ TRAP: First find where \(3\sin\theta = \sin\theta + 1\), i.e., \(\sin\theta = 1/2\), so \(\theta = \pi/6\) and \(\theta = 5\pi/6\). Many students forget to find these limits and just integrate from 0 to \(\pi\).
Step-by-step Explanation
Intersection\(3\sin\theta=\sin\theta+1\Rightarrow\sin\theta=\frac{1}{2}\Rightarrow\theta=\frac{\pi}{6},\frac{5\pi}{6}\).
Area integral\(\frac{1}{2}\int_{\pi/6}^{5\pi/6}[9\sin^2\theta-(\sin\theta+1)^2]\,d\theta=\frac{1}{2}\int_{\pi/6}^{5\pi/6}[8\sin^2\theta-2\sin\theta-1]\,d\theta\).
Expand using half-angle\(8\sin^2\theta=4(1-\cos2\theta)\). Then integrate term by term over \([\pi/6,5\pi/6]\). After careful evaluation: \(A=\pi+\frac{\sqrt{3}}{2}\).
L'HOPITAL: only for 0/0 or ∞/∞ · for 0·∞ or 1^∞ rewrite first! · apply limit of ln, then exponentiate
The Compound Interest Limit
A bank compounds interest continuously. To find what happens as compounding frequency \(n\to\infty\), we evaluate the indeterminate form:
$$\lim_{x \to 0^+} x^x$$
⚠ TRAP: This is the \(0^0\) indeterminate form, NOT equal to 1. Take \(\ln L = \lim_{x\to0^+} x\ln x\), which is \(0 \cdot (-\infty)\). Rewrite as \(\dfrac{\ln x}{1/x}\) to get \(-\infty/\infty\), then apply L'Hôpital.
Step-by-step Explanation
Take lnLet \(L=\lim x^x\). Then \(\ln L=\lim_{x\to0^+}x\ln x=\lim_{x\to0^+}\frac{\ln x}{1/x}\). This is \(-\infty/+\infty\).
L'Hôpital\(\lim\frac{1/x}{-1/x^2}=\lim(-x)=0\). So \(\ln L=0\Rightarrow L=e^0=\mathbf{1}\).
Key point\(0^0\) is indeterminate — the answer is 1 in this specific case, but not in general. Always check via the logarithm trick.
WORK=∫F·dx · PUMP: W=∫(weight density)·(distance lifted)·(slice volume)dh · slice at depth h, lifted to top
The Water Tower
A conical water tower (vertex at bottom) is 10 m tall and 6 m in diameter at the top. It is full of water (density \(1000\) kg/m³, \(g=9.8\) m/s²). Find the work required to pump all the water to the top of the tank.
$$W = \int_0^{10} \rho g \cdot \pi \left(\frac{3h}{10}\right)^{\!2} \cdot (10 - h)\,dh$$
⚠ TRAP: At height \(h\), the radius of the cone is \(r = \frac{3h}{10}\) (by similar triangles, since radius goes from 0 at bottom to 3 at top). Water at height \(h\) is lifted \((10-h)\) meters. Students often write \(h\) instead of \((10-h)\) for the lift distance.
Step-by-step Explanation
Slice at height hRadius \(r=\frac{3h}{10}\), thickness \(dh\), volume \(dV=\pi r^2\,dh=\frac{9\pi h^2}{100}\,dh\).
Weight and liftWeight of slice \(=\rho g\,dV=1000\cdot9.8\cdot\frac{9\pi h^2}{100}\,dh\). Lift distance \(=(10-h)\).
Integrate\(W=\frac{9800\cdot9\pi}{100}\int_0^{10}h^2(10-h)\,dh=\frac{88200\pi}{100}\int_0^{10}(10h^2-h^3)\,dh\).
Compute integral\(\int_0^{10}(10h^2-h^3)\,dh=\left[\frac{10h^3}{3}-\frac{h^4}{4}\right]_0^{10}=\frac{10000}{3}-2500=\frac{2500}{3}\).
Final\(W=\frac{88200\pi}{100}\cdot\frac{2500}{3}=\frac{88200\pi\cdot25}{3}=\frac{2{,}205{,}000\pi}{3}=735{,}000\pi\approx\mathbf{2.31\times10^6}\) J. This is closest to choice (B) in order of magnitude — always check units and powers of 10!