πŸŽ“ AP CALC BC
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πŸ“ AP Calculus BC

β˜… Self-Study Key Problems β€” 20 Questions β˜…
Name: _________________________    Date: _____________
Focus Areas: Limits Β· Derivatives Β· Integrals Β· Series Β· Polar Β· Parametric Β· Differential Equations
πŸ“Œ UNIT 1 β€” Limits & Continuity

L'Hôpital's Rule: Only use when you get 0/0 or ∞/∞ !

INDETERMINATE DIFFERENTIATE TOP & BOTTOM REPEAT IF NEEDED
1
LIMITS
πŸ“– EXAMPLE FIRST
Find \(\displaystyle\lim_{x \to 0} \frac{\sin x}{x}\)
β†’ Memorize! This equals 1. It's the most famous limit in calc!
Evaluate: \(\displaystyle\lim_{x \to 0} \dfrac{1 - \cos x}{x^2}\)

⚑ Trick: Use known trig limit identities β€” don't just plug in!

βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Step 1: Direct substitution gives \(\frac{0}{0}\) β€” indeterminate form. Use L'HΓ΄pital's Rule.
Step 2: Differentiate numerator and denominator:
\(\displaystyle\lim_{x \to 0} \frac{\sin x}{2x}\)
Step 3: Still 0/0? Apply L'HΓ΄pital again:
\(\displaystyle\lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}\)
Alt Method: Use the identity \(1 - \cos x \approx \frac{x^2}{2}\) near 0, so \(\frac{x^2/2}{x^2} = \frac{1}{2}\) βœ“
πŸ”‘ KEY: \(\lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac{1}{2}\) β€” memorize this!
2
LIMITS
Find: \(\displaystyle\lim_{x \to \infty} \dfrac{3x^3 - 5x + 1}{7x^3 + 2x^2}\)

⚑ Students often panic at ∞ β€” but there's a slick trick!

βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Rule of Thumb: When degree TOP = degree BOTTOM β†’ limit = ratio of leading coefficients.
Leading coefficient top: 3  |  Leading coefficient bottom: 7
Answer: \(\dfrac{3}{7}\)
πŸ”‘ Three cases for polynomial over polynomial:
β€’ Same degree β†’ leading coeff ratio
β€’ Top degree bigger β†’ \(\pm\infty\)
β€’ Bottom degree bigger β†’ 0
πŸ“Œ UNIT 2 β€” Derivatives & Applications

Chain Rule: "Derivative of outside Γ— derivative of inside"

OUTSIDE FIRST KEEP INSIDE Γ— d(inside)
3
DERIVATIVES
πŸ“– EXAMPLE FIRST
If \(f(x) = \sin(x^2)\), then \(f'(x) = \cos(x^2) \cdot 2x\)
β†’ Chain rule: outside \(\sin\) β†’ \(\cos\), keep inside, times \(d(x^2)=2x\)
Find \(\dfrac{d}{dx}\left[e^{3x^2 - x}\right]\)
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Rule: \(\frac{d}{dx}[e^{u}] = e^u \cdot u'\) β€” keep the exponent, multiply by its derivative.
Here: \(u = 3x^2 - x\), so \(u' = 6x - 1\)
Answer: \((6x-1)e^{3x^2-x}\)
⚠️ Common mistake: Forgetting to multiply by \(u'\) β€” that's the chain rule part!
4
IMPLICIT DIFF
Given \(x^2 + y^2 = 25\), find \(\dfrac{dy}{dx}\).

⚑ Classic trap: Remember \(y\) is a function of \(x\) β€” use chain rule on \(y^2\)!

βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Step 1: Differentiate both sides w.r.t. \(x\):
\(2x + 2y\dfrac{dy}{dx} = 0\)
Step 2: Solve for \(\dfrac{dy}{dx}\):
\(\dfrac{dy}{dx} = -\dfrac{x}{y}\)
πŸ”‘ Key word: IMPLICIT β†’ differentiate both sides, then isolate dy/dx.
5
RELATED RATES
A ladder 10 ft long leans against a wall. The bottom slides away at \(2\) ft/sec. How fast is the top sliding down when the bottom is \(6\) ft from the wall?
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Setup: \(x^2 + y^2 = 100\). Given: \(\frac{dx}{dt} = 2\), find \(\frac{dy}{dt}\) when \(x = 6\).
When \(x=6\): \(y = \sqrt{100-36} = 8\)
Differentiate: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)
Plug in: \(2(6)(2) + 2(8)\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{3}{2}\)
πŸ”‘ Steps: DRAW β†’ EQUATION β†’ DIFFERENTIATE w.r.t. t β†’ SUBSTITUTE
πŸ“Œ UNIT 3 β€” Integration Techniques

Integration by Parts: \(\int u\,dv = uv - \int v\,du\)

LIATE LOG β†’ INV TRIG β†’ ALGEBRAIC β†’ TRIG β†’ EXPO
6
INTEGRATION BY PARTS
πŸ“– EXAMPLE FIRST
\(\int x e^x dx\): Let \(u=x\), \(dv=e^x dx\) β†’ \(xe^x - e^x + C\)
Evaluate: \(\displaystyle\int x \cos x \, dx\)
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
LIATE: \(u = x\) (Algebraic), \(dv = \cos x\,dx\) (Trig)
Then: \(du = dx\), \(v = \sin x\)
Formula: \(\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C\)
⚠️ Note: \(\int \sin x\,dx = -\cos x\), so minus a minus = plus!
7
U-SUBSTITUTION
Evaluate: \(\displaystyle\int \frac{2x}{x^2 + 1}\, dx\)

⚑ Spot the pattern: numerator is the derivative of denominator!

βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Let \(u = x^2 + 1\), then \(du = 2x\,dx\)
Substitutes to: \(\int \frac{du}{u} = \ln|u| + C\)
Back-substitute: \(\ln(x^2+1) + C\) (no abs value needed since \(x^2+1 > 0\) always)
πŸ”‘ KEY PATTERN: \(\int \frac{f'(x)}{f(x)}dx = \ln|f(x)| + C\)
8
DEFINITE INTEGRAL
Evaluate: \(\displaystyle\int_0^{\pi} \sin^2 x \, dx\)

⚑ Can't integrate \(\sin^2 x\) directly β€” use the power-reducing identity!

Power-Reducing: \(\sin^2 x = \dfrac{1 - \cos 2x}{2}\)

HALF-ANGLECOS 2x
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Replace: \(\int_0^\pi \frac{1-\cos 2x}{2}\,dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_0^\pi\)
Evaluate: \(\frac{1}{2}\left[\left(\pi - \frac{\sin 2\pi}{2}\right) - \left(0 - \frac{\sin 0}{2}\right)\right]\)
Since \(\sin 2\pi = \sin 0 = 0\): \(\frac{1}{2}(\pi - 0) = \dfrac{\pi}{2}\)
πŸ“Œ UNIT 4 β€” Infinite Series (Most Tested!)

Convergence Tests Quick Guide:

GEOMETRIC: |r|<1 P-SERIES: p>1 RATIO TEST ALTERNATING
9
SERIES CONVERGENCE
πŸ“– EXAMPLE FIRST
Geometric series \(\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n\) converges because \(|r| = \frac{1}{3} < 1\).
Sum = \(\frac{1}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{3}{2}\)
Does \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}\) converge or diverge?
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
P-Series Rule: \(\sum \frac{1}{n^p}\) converges if \(p > 1\), diverges if \(p \leq 1\)
Here: \(p = 2 > 1\) β†’ CONVERGES
Compare: \(\sum \frac{1}{n}\) (harmonic, \(p=1\)) β†’ DIVERGES. Easy to confuse!
πŸ”‘ Basel Problem: \(\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\) (fun fact!)
10
TAYLOR SERIES
The Taylor series for \(e^x\) centered at \(x = 0\) is: \[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\] Using this, what is the third-degree Taylor polynomial for \(e^{-x^2}\)?
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Substitute \(x \to -x^2\) in the \(e^x\) series:
\(e^{-x^2} = 1 + (-x^2) + \dfrac{(-x^2)^2}{2!} + \cdots = 1 - x^2 + \dfrac{x^4}{2} - \cdots\)
Up to degree-4 terms (counting even powers): \(1 - x^2 + \dfrac{x^4}{2}\)
πŸ”‘ KEY SKILL: Substitute into known series β€” much faster than computing derivatives!
11
RADIUS OF CONVERGENCE
Find the radius of convergence of \(\displaystyle\sum_{n=0}^{\infty} \frac{(2x)^n}{n!}\)

⚑ Use the Ratio Test. This one has a beautiful answer!

βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Ratio Test: \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \frac{(2x)^{n+1}}{(n+1)!} \cdot \frac{n!}{(2x)^n} = \lim_{n\to\infty}\frac{|2x|}{n+1} = 0\)
Since the limit = 0 < 1 for ALL x: Series converges for all real \(x\).
Therefore: \(R = \infty\) β€” this is the \(e^{2x}\) series!
πŸ“Œ UNIT 5 β€” Polar & Parametric

Polar Area Formula: \(A = \dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta\)

HALF Β· r-SQUARED Β· dΞΈ FIND BOUNDS FIRST
12
POLAR AREA
Find the area enclosed by the polar curve \(r = 2\cos\theta\).

⚑ Sketch it first β€” it's actually a circle! Where does it start and end?

βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Recognize: \(r = 2\cos\theta\) is a circle of radius 1 centered at \((1, 0)\). So area should be \(\pi(1)^2 = \pi\).
Using formula: \(A = \frac{1}{2}\int_{-\pi/2}^{\pi/2} (2\cos\theta)^2\,d\theta = \frac{1}{2}\int_{-\pi/2}^{\pi/2} 4\cos^2\theta\,d\theta\)
= \(2\int_{-\pi/2}^{\pi/2} \frac{1+\cos 2\theta}{2}\,d\theta = \left[\theta + \frac{\sin 2\theta}{2}\right]_{-\pi/2}^{\pi/2} = \pi\) βœ“
13
PARAMETRIC SLOPE
For the parametric curve \(x = t^2\), \(y = t^3 - t\), find \(\dfrac{dy}{dx}\).
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Parametric slope formula: \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\)
Compute: \(\dfrac{dx}{dt} = 2t\), \(\quad\dfrac{dy}{dt} = 3t^2 - 1\)
Answer: \(\dfrac{dy}{dx} = \dfrac{3t^2-1}{2t}\)
πŸ”‘ KEY: NEVER just take \(\frac{d}{dt}\) of \(y\) β€” always divide by \(\frac{dx}{dt}\)!
πŸ“Œ UNIT 6 β€” Differential Equations

Separable DE: Move all y's to one side, all x's to other β†’ integrate both sides!

SEPARATE INTEGRATE SOLVE FOR y +C ALWAYS
14
SEPARABLE DE
πŸ“– EXAMPLE FIRST
\(\dfrac{dy}{dx} = xy\) β†’ Separate: \(\dfrac{dy}{y} = x\,dx\) β†’ Integrate: \(\ln|y| = \dfrac{x^2}{2} + C\) β†’ \(y = Ae^{x^2/2}\)
Solve: \(\dfrac{dy}{dx} = \dfrac{x}{y}\), given \(y(0) = 3\).
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Separate: \(y\,dy = x\,dx\)
Integrate: \(\dfrac{y^2}{2} = \dfrac{x^2}{2} + C\)
Apply IC \(y(0)=3\): \(\dfrac{9}{2} = 0 + C \Rightarrow C = \dfrac{9}{2}\)
Solve: \(y^2 = x^2 + 9 \Rightarrow y = \sqrt{x^2 + 9}\)
15
EULER'S METHOD
Use Euler's method with step size \(h = 0.5\) to approximate \(y(1)\), given \(\dfrac{dy}{dx} = y\) and \(y(0) = 1\).

⚑ Euler's method: \(y_{n+1} = y_n + h \cdot f(x_n, y_n)\)

βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Step 1: \(x_0=0, y_0=1\). Slope \(= y_0 = 1\).
\(y_1 = 1 + 0.5 \times 1 = 1.5\)
Step 2: \(x_1=0.5, y_1=1.5\). Slope \(= y_1 = 1.5\).
\(y_2 = 1.5 + 0.5 \times 1.5 = 2.25\)
Exact answer: \(y(1) = e^1 \approx 2.718\) β€” Euler's method underestimates here because \(y=e^x\) is concave up.
πŸ”‘ Concave up β†’ Euler UNDERESTIMATES; Concave down β†’ OVERESTIMATES
πŸ“Œ UNIT 7 β€” Advanced Topics (BC Only)
16
ARC LENGTH
The arc length formula for \(y = f(x)\) from \(a\) to \(b\) is:
\[L = \int_a^b \sqrt{1 + [f'(x)]^2}\, dx\] Find the arc length of \(y = \dfrac{x^2}{2}\) from \(x=0\) to \(x=1\). (Set up only β€” do NOT evaluate.)
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Find \(f'(x)\): If \(y = \dfrac{x^2}{2}\), then \(f'(x) = x\)
Substitute into formula: \(L = \displaystyle\int_0^1 \sqrt{1 + x^2}\,dx\)
πŸ”‘ Steps: Find derivative β†’ Square it β†’ Add 1 β†’ Square root β†’ Integrate
17
IMPROPER INTEGRAL
Evaluate: \(\displaystyle\int_1^{\infty} \frac{1}{x^2}\, dx\)

⚑ Replace ∞ with a limit variable β€” then evaluate!

βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Replace ∞: \(\lim_{b\to\infty}\int_1^b x^{-2}\,dx = \lim_{b\to\infty}\left[-x^{-1}\right]_1^b\)
= \(\lim_{b\to\infty}\left(-\dfrac{1}{b} + 1\right) = 0 + 1 = 1\)
πŸ”‘ Improper integral converges ↔ the limit exists and is finite!
18
ALTERNATING SERIES ERROR
The alternating series \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) is approximated by its first 3 terms. What is the maximum error?

⚑ Alternating Series Estimation Theorem: error ≀ |first omitted term|

βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
First 3 terms: \(1 - \frac{1}{2} + \frac{1}{3}\)
First omitted term (4th): \(\dfrac{1}{4}\)
By AST: |Error| ≀ |4th term| = \(\dfrac{1}{4}\)
πŸ”‘ ALTERNATING SERIES TEST: terms decrease β†’ 0, alternating signs β†’ CONVERGES
19
FTC PART 2
Let \(F(x) = \displaystyle\int_0^x \sin(t^2)\,dt\). Find \(F'(x)\).

⚑ Fundamental Theorem of Calculus Part 1: the answer is simpler than you think!

FTC Part 1: \(\dfrac{d}{dx}\!\displaystyle\int_a^x f(t)\,dt = f(x)\)

PLUG IN UPPER BOUNDCHAIN RULE IF NEEDED
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
FTC Part 1: \(\dfrac{d}{dx}\displaystyle\int_0^x f(t)\,dt = f(x)\)
Here: \(f(t) = \sin(t^2)\), plug in \(t = x\) β†’ \(F'(x) = \sin(x^2)\)
No chain rule needed because the upper limit is just \(x\) (not \(x^2\) or something else).
⚠️ If upper bound were \(x^2\): \(F'(x) = \sin(x^4) \cdot 2x\) β€” THEN use chain rule!
20
MACLAURIN SERIES
Using the Maclaurin series, what is the exact value of: \[\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\]

⚑ Recognize this as a famous series evaluated at a specific point!

Know these cold: \(\sin x = \displaystyle\sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)    \(\cos x = \displaystyle\sum \frac{(-1)^n x^{2n}}{(2n)!}\)

sin: ODD POWERScos: EVEN POWERS
βœ… Correct! Great job!
πŸ“ FULL EXPLANATION
Maclaurin for sin: \(\sin x = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)
Given series: \(\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\) β€” this is \(\sin x\) with \(x=1\)!
Answer: \(\sin(1)\)
πŸ”‘ KEY SKILL: Match the series to a known formula β€” this is tested heavily on the AP exam!