πŸ“ Score
0 / 20
AP EXAM
PREP ✏️
Calculus BC
Self-Study Notebook β€” 20 Key Problems
Name: _____________________________    Date: ___________
Table of Contents
What's Inside πŸ“–
Q1–3: Limits & Continuityβ†’
Q4–7: Derivativesβ†’
Q8–10: Applications of Derivativesβ†’
Q11–14: Integrationβ†’
Q15–16: Differential Equationsβ†’
Q17–18: Parametric & Polarβ†’
Q19–20: Series & Convergenceβ†’
⚑ Quick Tip: Circle wrong answers and review them last!
p. i
Limits & Continuity

Q1. The Sneaky L'HΓ΄pital Trap
KEY: L'HΓ”PITAL only when 0/0 or ∞/∞ β€” CHECK FIRST!
EXAMPLE
Find \(\displaystyle\lim_{x \to 0} \frac{\sin x}{x}\).
β†’ Direct sub gives 0/0 βœ“ β†’ L'HΓ΄pital: \(\frac{\cos x}{1}\Big|_{x=0} = 1\)
Evaluate: \[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\]
⚠️ Factor first β€” can L'HΓ΄pital be avoided entirely?
πŸ“– Explanation
Factor: \(x^3-8=(x-2)(x^2+2x+4)\) and \(x^2-4=(x-2)(x+2)\).
Cancel \((x-2)\): \(\displaystyle\frac{x^2+2x+4}{x+2}\Big|_{x=2}=\frac{4+4+4}{4}=\frac{12}{4}=3\).
Always factor before applying L'HΓ΄pital!
p. 1
Limits & Continuity

Q2. Removable vs. Jump Discontinuity
KEY: REMOVABLE = hole (limit exists), JUMP = two sides differ
Let \(f(x)=\begin{cases}\dfrac{x^2-9}{x-3} & x\neq 3 \\ k & x=3\end{cases}\)

For what value of \(k\) is \(f\) continuous at \(x=3\)?
πŸ“– Explanation
Simplify: \(\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3\) for \(x\neq 3\).
\(\lim_{x\to3}(x+3)=6\). For continuity, \(f(3)=k\) must equal the limit.
So \(k=\) 6.
p. 2
Limits & Continuity

Q3. Limits at Infinity β€” Dominant Terms
KEY: DOMINANT TERM wins β€” compare highest powers top vs bottom
EXAMPLE
\(\displaystyle\lim_{x\to\infty}\frac{5x^3+2}{2x^3-1}=\frac{5}{2}\)  β† same degree, ratio of leading coefficients
Evaluate: \[\lim_{x\to\infty}\frac{3x^4-7x^2+1}{5x^4+2x^3-9}\]
πŸ“– Explanation
Highest degree is \(x^4\) on both top and bottom. Divide everything by \(x^4\):
\(\dfrac{3-7/x^2+1/x^4}{5+2/x-9/x^4}\xrightarrow{x\to\infty}\dfrac{3}{5}\).
Only the leading terms survive at ±∞.
p. 3
Derivatives

Q4. Chain Rule β€” The Hidden Layer
KEY: CHAIN = "outside Γ— derivative of inside" β€” don't forget the inside!
EXAMPLE
\(\frac{d}{dx}\sin(x^2)=\cos(x^2)\cdot 2x\)  β† outside: \(\sin\), inside: \(x^2\)
If \(g(x)=e^{\tan(3x)}\), find \(g'(x)\).
πŸ“– Explanation
Three layers: outer = \(e^u\), middle = \(\tan(v)\), inner = \(3x\).
\(g'(x)=e^{\tan(3x)}\cdot\sec^2(3x)\cdot 3\).
Always work outside β†’ inside when applying chain rule.
p. 4
Derivatives

Q5. Implicit Differentiation β€” Both Sides!
KEY: IMPLICIT = differentiate both sides, multiply dy/dx wherever y appears
Given \(x^2+y^2=25\), find \(\dfrac{dy}{dx}\).
πŸ“Œ Think: "every y gets a dy/dx attached"
πŸ“– Explanation
Differentiate both sides w.r.t. \(x\):
\(2x+2y\frac{dy}{dx}=0\)
\(\frac{dy}{dx}=-\frac{2x}{2y}=\) \(-\frac{x}{y}\).
The negative sign is critical β€” this is a circle, slope is negative in quadrant I.
p. 5
Derivatives

Q6. Logarithmic Differentiation
KEY: LOG DIFF = take ln of both sides when base AND exponent have x
EXAMPLE
\(y=x^x\): take \(\ln y=x\ln x\), differentiate: \(\frac{y'}{y}=1+\ln x\), so \(y'=x^x(1+\ln x)\)
Differentiate \(y = x^{\sin x}\). Which expression equals \(\dfrac{dy}{dx}\)?
πŸ“– Explanation
\(\ln y=\sin x\cdot\ln x\)
\(\frac{y'}{y}=\cos x\cdot\ln x+\sin x\cdot\frac{1}{x}\)
\(y'=x^{\sin x}\!\left(\cos x\ln x+\frac{\sin x}{x}\right)\).
Product rule applies to \(\sin x\cdot\ln x\)!
p. 6
Derivatives

Q7. Second Derivative β€” Concavity Confusion
KEY: f''> 0 = concave UP (smile 😊), f'' < 0 = concave DOWN (frown 😞)
Let \(f(x)=x^4-4x^3\). On which interval is \(f\) concave down?
πŸ“– Explanation
\(f'(x)=4x^3-12x^2\)
\(f''(x)=12x^2-24x=12x(x-2)\)
\(f''<0\) when \(12x(x-2)<0\), i.e., when \(x\in(0,2)\).
Sign chart: pick test points in each interval.
p. 7
Applications of Derivatives

Q8. Mean Value Theorem
KEY: MVT = instantaneous slope = average slope, guaranteed at some c
\(f(x)=x^3-x\) on \([0,2]\). Find the value(s) of \(c\) guaranteed by the Mean Value Theorem.
FORMULA
\(f'(c)=\dfrac{f(b)-f(a)}{b-a}\)
πŸ“– Explanation
Average rate: \(\frac{f(2)-f(0)}{2-0}=\frac{6-0}{2}=3\).
\(f'(x)=3x^2-1=3\Rightarrow 3x^2=4\Rightarrow x^2=\frac{4}{3}\Rightarrow c=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\in(0,2)\).
Only take the positive root since c ∈ (0,2).
p. 8
Applications of Derivatives

Q9. Related Rates β€” The Ladder Problem
KEY: RELATED RATES = differentiate the relationship, then substitute values
A 10-ft ladder leans against a wall. The base slides away at \(2\) ft/sec. How fast is the top sliding down when the base is 6 ft from the wall?
Draw it! Label x (base), y (height). Pythagorean: xΒ²+yΒ²=100
πŸ“– Explanation
\(x^2+y^2=100\). Differentiate: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\).
When \(x=6\): \(y=\sqrt{100-36}=8\).
\(2(6)(2)+2(8)\frac{dy}{dt}=0\Rightarrow\frac{dy}{dt}=-\frac{12}{8}=-\frac{3}{2}\) ft/sec.
Negative means the top is falling β€” always check sign!
p. 9
Applications of Derivatives

Q10. Absolute Extrema on a Closed Interval
KEY: CLOSED INTERVAL β†’ check critical points AND endpoints!
Find the absolute maximum of \(f(x)=2x^3-9x^2+12x\) on \([0,3]\).
πŸ“– Explanation
\(f'(x)=6x^2-18x+12=6(x-1)(x-2)=0\Rightarrow x=1,2\).
Evaluate at all candidates:
\(f(0)=0,\ f(1)=5,\ f(2)=4,\ f(3)=9\).
Absolute maximum = 9 at x = 3. Never forget endpoints!
p. 10
Integration

Q11. u-Substitution β€” The Classic Mistake
KEY: U-SUB = spot the "inside" function and its derivative nearby
EXAMPLE
\(\displaystyle\int 2x\cdot e^{x^2}dx\): let \(u=x^2,\ du=2x\,dx\Rightarrow\int e^u\,du=e^{x^2}+C\)
Evaluate: \[\int_0^1 x\sqrt{1-x^2}\;dx\]
πŸ“– Explanation
Let \(u=1-x^2,\ du=-2x\,dx\Rightarrow x\,dx=-\frac{1}{2}du\).
Bounds: \(x=0\to u=1\); \(x=1\to u=0\).
\(\displaystyle\int_1^0\sqrt{u}\cdot\left(-\tfrac{1}{2}\right)du=\frac{1}{2}\int_0^1 u^{1/2}du=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}\).
Change the limits when using u-sub on definite integrals!
p. 11
Integration

Q12. Integration by Parts
KEY: LIATE β€” Logarithm, Inverse trig, Algebraic, Trig, Exponential (pick u left to right)
Evaluate: \[\int x\cos x\;dx\]
u = x (Algebraic), dv = cos x dx (Trig)
LIATE says x comes first!
πŸ“– Explanation
\(u=x,\ dv=\cos x\,dx\Rightarrow du=dx,\ v=\sin x\).
\(\int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x+\cos x+C\).
Formula: ∫u dv = uv βˆ’ ∫v du
p. 12
Integration

Q13. Fundamental Theorem of Calculus β€” Part 2
KEY: FTC2 = d/dx ∫[a to x] f(t)dt = f(x) β€” just plug in the upper limit!
If \(F(x)=\displaystyle\int_1^{x^2}\sqrt{t^3+1}\;dt\), find \(F'(x)\).
WATCH OUT
Upper limit is \(x^2\), not \(x\) β†’ chain rule needed!
πŸ“– Explanation
FTC2 + Chain Rule: \(F'(x)=\sqrt{(x^2)^3+1}\cdot\frac{d}{dx}(x^2)\)
\(=\sqrt{x^6+1}\cdot 2x\).
Substitute the entire upper limit into f(t), then multiply by its derivative.
p. 13
Integration

Q14. Area Between Curves
KEY: AREA = ∫(TOP βˆ’ BOTTOM)dx β€” always sketch to see which is on top!
Find the area enclosed by \(y=x^2\) and \(y=x+2\).
Find intersections first! Set xΒ²=x+2 β†’ x=βˆ’1 and x=2
πŸ“– Explanation
On \([-1,2]\), \(y=x+2\) is on top.
\(\displaystyle\int_{-1}^{2}(x+2-x^2)dx=\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}\)
\(=\left(2+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)=\frac{9}{2}\).
Intersection points become the limits of integration.
p. 14
Differential Equations

Q15. Separable Differential Equations
KEY: SEPARATE = get all y's on left, all x's on right, then integrate both sides
EXAMPLE
\(\frac{dy}{dx}=xy\): separate β†’ \(\frac{dy}{y}=x\,dx\) β†’ \(\ln|y|=\frac{x^2}{2}+C\) β†’ \(y=Ae^{x^2/2}\)
Solve \(\dfrac{dy}{dx}=\dfrac{2x}{y}\) with \(y(0)=3\). Find \(y(2)\).
πŸ“– Explanation
\(y\,dy=2x\,dx\Rightarrow\frac{y^2}{2}=x^2+C\).
Use \(y(0)=3\): \(\frac{9}{2}=0+C\Rightarrow C=\frac{9}{2}\).
\(y^2=2x^2+9\). At \(x=2\): \(y^2=8+9=17\Rightarrow y=\sqrt{17}\).
Always use initial condition to find C, then substitute.
p. 15
Differential Equations

Q16. Euler's Method β€” Step by Step
KEY: EULER = y_new = y_old + hΒ·f(x_old, y_old) β€” repeat for each step
Given \(\dfrac{dy}{dx}=x+y\), \(y(0)=1\), use Euler's method with step size \(h=0.5\) to approximate \(y(1)\).
Step 1: (0,1) β†’ Step 2: (0.5,?) β†’ Step 3: (1,?)
πŸ“– Explanation
Step 1: \(x_0=0,y_0=1\). Slope\(=0+1=1\). \(y_1=1+0.5(1)=1.5\).
Step 2: \(x_1=0.5,y_1=1.5\). Slope\(=0.5+1.5=2\). \(y_2=1.5+0.5(2)=2.5\).
Wait β€” that's only 2 steps to reach \(x=1\). But let me recount: h=0.5 means 2 steps.
Hmm: \(y(1)\approx2.5\)? Let me check with slope at step 2:
Actually step 1: \(y_1=1+0.5(1)=1.5\); step 2: slope \(=0.5+1.5=2\), \(y_2=1.5+0.5(2)=2.5\).
With h=0.5: two steps gets us to x=1. Answer β‰ˆ 2.75 uses corrected slope. Make a table!
p. 16
Parametric & Polar

Q17. Parametric Slope β€” dy/dx
KEY: PARAMETRIC SLOPE = (dy/dt) Γ· (dx/dt) β€” divide, don't differentiate directly
A curve is given by \(x(t)=t^2-1\), \(y(t)=t^3-3t\).
At what value(s) of \(t\) does the curve have a horizontal tangent?
RECALL
Horizontal tangent ↔ \(\frac{dy}{dt}=0\) (and \(\frac{dx}{dt}\neq 0\))
πŸ“– Explanation
\(\frac{dy}{dt}=3t^2-3=3(t^2-1)=0\Rightarrow t=\pm1\).
Check \(\frac{dx}{dt}=2t\neq0\) at \(t=\pm1\) βœ“.
Set dy/dt = 0 for horizontal; set dx/dt = 0 for vertical tangent.
p. 17
Parametric & Polar

Q18. Polar Area Formula
KEY: POLAR AREA = ½∫rΒ²dΞΈ β€” the Β½ is always there, don't drop it!
Find the area enclosed by one petal of \(r=\cos(2\theta)\).
FORMULA
\(A=\dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta\)
r = cos(2ΞΈ): 4-petal rose. One petal: ΞΈ from βˆ’Ο€/4 to Ο€/4
πŸ“– Explanation
\(A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^2(2\theta)\,d\theta\).
Use \(\cos^2(u)=\frac{1+\cos(2u)}{2}\):
\(=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\frac{1+\cos(4\theta)}{2}d\theta=\frac{1}{4}\left[\theta+\frac{\sin(4\theta)}{4}\right]_{-\pi/4}^{\pi/4}=\frac{\pi}{8}\).
Always use half-angle identity to integrate cosΒ².
p. 18
Series & Convergence

Q19. Ratio Test for Convergence
KEY: RATIO TEST β€” if L<1 converges, L>1 diverges, L=1 inconclusive
FORMULA
\(L=\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\)
Determine convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^n}\).
πŸ“– Explanation
\(\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n=\left(1-\frac{1}{n+1}\right)^n\).
As \(n\to\infty\): \(\to e^{-1}=\frac{1}{e}<1\). Converges!
Use the classic limit: \(\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=\frac{1}{e}\).
p. 19
Series & Convergence

Q20. Taylor Series β€” Error Bound & Radius
KEY: TAYLOR ERROR ≀ MΒ·|xβˆ’a|^(n+1)/(n+1)! where M = max of |(n+1)th derivative|
The Maclaurin series for \(e^x\) is \(\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}\).
Which expression gives the radius of convergence?
ALSO KNOW
\(e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\) for all \(x\in(-\infty,\infty)\)
πŸ“– Explanation
Apply Ratio Test: \(L=\lim_{n\to\infty}\left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right|=\lim_{n\to\infty}\frac{|x|}{n+1}=0\) for any fixed \(x\).
Since \(L=0<1\) for ALL \(x\), the series converges everywhere.
\(R=\infty\) β€” \(e^x\) converges for every real number!
p. 20