AP Β· CALCULUS AB

πŸ“ Calc AB
Self-Study Notes

20 Key Questions Β· Interactive Quiz Β· Memory Points

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πŸ”΅ Unit 1 β€” Limits & Continuity
Q 1
Easy
KEY SUBSTITUTE β†’ FACTOR β†’ L'HΓ΄pital
Find the limit:
\[ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \]
πŸ“ Example First
\(\displaystyle \lim_{x \to 2} \frac{x^2-4}{x-2}\) β†’ Factor: \(\frac{(x-2)(x+2)}{x-2} = x+2\) β†’ plug in 2 β†’ 4
Plugging in \(x=3\) gives \(\tfrac{0}{0}\) β€” always try factoring first!
πŸ’‘ Explanation
Factor the numerator: \(x^2 - 9 = (x-3)(x+3)\)
Cancel \((x-3)\): limit becomes \(\lim_{x\to3}(x+3) = \mathbf{6}\)
Answer: B β€” 6
Q 2
Medium β€” Often Missed
KEY ONE-SIDED = TWO-SIDED β†’ EXIST
Given: \(f(x) = \begin{cases} x^2 + 1 & x < 2 \\ 3x - 1 & x \geq 2 \end{cases}\)

Does \(\displaystyle\lim_{x \to 2} f(x)\) exist?
Left-hand limit β‰  Right-hand limit β†’ DNE! Students forget to check BOTH sides.
πŸ’‘ Explanation
Left: \(\lim_{x\to2^-}(x^2+1) = 4+1 = 5\)
Right: \(\lim_{x\to2^+}(3x-1) = 6-1 = 5\)
Both sides equal 5 β†’ limit exists and equals 5.
Note: \(f(2) = 5\) too, so it's also continuous here!
Answer: A
Q 3
Tricky ⚠️
KEY sin(x)/x → 1 as x→0 (MEMORIZE!)
Evaluate: \[\lim_{x \to 0} \frac{\sin(5x)}{x}\]
πŸ“ Trig Limit Formula
\(\displaystyle \lim_{x\to0}\frac{\sin(kx)}{x} = k\)  β†  multiply top & bottom by \(k\)
Trick: Rewrite as \(5 \cdot \dfrac{\sin(5x)}{5x}\) β†’ then use the famous limit!
πŸ’‘ Explanation
\(\dfrac{\sin(5x)}{x} = 5 \cdot \dfrac{\sin(5x)}{5x}\)
As \(x\to 0\), \(\dfrac{\sin(5x)}{5x} \to 1\)
So the answer is \(5 \times 1 = \mathbf{5}\)
Answer: C β€” 5
🟠 Unit 2 β€” Derivatives & Rules
Q 4
Easy
KEY POWER RULE: bring down β†’ reduce power
Find \(f'(x)\) if \(f(x) = 3x^4 - 5x^2 + 7\)
πŸ“ Power Rule
\(\dfrac{d}{dx}[x^n] = nx^{n-1}\)   e.g. \(\dfrac{d}{dx}[x^3] = 3x^2\)
πŸ’‘ Explanation
\(\dfrac{d}{dx}[3x^4] = 12x^3\), \(\dfrac{d}{dx}[-5x^2] = -10x\), \(\dfrac{d}{dx}[7] = 0\)
So \(f'(x) = \mathbf{12x^3 - 10x}\)
Answer: B
Q 5
Tricky ⚠️
KEY PRODUCT RULE: (uv)β€² = uβ€²v + uvβ€²
Find \(\dfrac{d}{dx}[x^2 \sin x]\)
Students forget which part to differentiate. Write \(u\) and \(v\) FIRST before computing!
πŸ“ Product Rule
Let \(u = x^2\), \(v = \sin x\)
\((uv)' = u'v + uv' = 2x\sin x + x^2\cos x\)
πŸ’‘ Explanation
\(u = x^2 \Rightarrow u' = 2x\),   \(v = \sin x \Rightarrow v' = \cos x\)
\((uv)' = 2x \cdot \sin x + x^2 \cdot \cos x\)
Answer: C
Q 6
Tricky ⚠️
KEY CHAIN RULE: outerβ€²(inner) Γ— innerβ€²
Find \(\dfrac{d}{dx}[\sin(3x^2)]\)
πŸ“ Chain Rule
\(\dfrac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)
Here: outer = \(\sin(\square)\), inner = \(3x^2\)
The most common mistake: forgetting to multiply by the derivative of the inside!
πŸ’‘ Explanation
Outer: \(\sin \to \cos\), Inner: \(3x^2 \to 6x\)
Result: \(\cos(3x^2) \cdot 6x = \mathbf{6x\cos(3x^2)}\)
Answer: B
Q 7
Medium
KEY IMPLICIT: d/dx both sides, solve for dy/dx
If \(x^2 + y^2 = 25\), find \(\dfrac{dy}{dx}\).
πŸ“ Implicit Differentiation
Differentiate both sides: \(2x + 2y\dfrac{dy}{dx} = 0\)
Solve: \(\dfrac{dy}{dx} = -\dfrac{x}{y}\)
πŸ’‘ Explanation
Differentiate: \(2x + 2y\dfrac{dy}{dx} = 0\)
\(2y\dfrac{dy}{dx} = -2x\)
\(\dfrac{dy}{dx} = -\dfrac{x}{y}\)
Answer: A
🟒 Unit 3 β€” Applications of Derivatives
Q 8
Easy
KEY CRITICAL POINT: fβ€²(x) = 0 or undefined
Find the critical numbers of \(f(x) = x^3 - 6x^2 + 9x\)
πŸ“ How to Find Critical Numbers
Step 1: Find \(f'(x) = 3x^2 - 12x + 9\)
Step 2: Set \(f'(x) = 0\) β†’ \(3(x^2-4x+3) = 0\) β†’ \((x-1)(x-3)=0\)
πŸ’‘ Explanation
\(f'(x) = 3x^2-12x+9 = 3(x-1)(x-3) = 0\)
Critical numbers: \(x=1\) and \(x=3\)
Answer: D
Q 9
Tricky ⚠️
KEY 1st DERIV TEST: +β†’βˆ’ = LOCAL MAX, βˆ’β†’+ = LOCAL MIN
For \(f(x) = x^3 - 6x^2 + 9x + 1\), at \(x=1\), the function has a:
Students confuse local max/min with global max/min. On an OPEN interval, there may be NO absolute max/min!
πŸ“ First Derivative Test
Sign of \(f'(x)\):   + (increasing) β†’ 0 β†’ βˆ’ (decreasing) = Local MAX
πŸ’‘ Explanation
\(f'(x) = 3x^2-12x+9 = 3(x-1)(x-3)\)
For \(x < 1\): \(f'>0\) (increasing); For \(1<x<3\): \(f'<0\) (decreasing)
Sign changes + β†’ βˆ’ at \(x=1\) β†’ Local MAX
Answer: B
Q 10
Medium β€” Often Missed
KEY CONCAVE UP: fβ€³ > 0 Β· CONCAVE DOWN: fβ€³ < 0
Where is \(f(x) = x^4 - 4x^3\) concave up?
πŸ“ Concavity Test
Find \(f''(x)\), solve \(f''(x) > 0\)
\(f'(x)=4x^3-12x^2\), \(f''(x)=12x^2-24x=12x(x-2)\)
πŸ’‘ Explanation
\(f''(x) = 12x(x-2)\)
\(f''(x) > 0\) when both factors same sign:
β†’ \(x<0\) (both negative) or \(x>2\) (both positive)
Answer: C
Q 11
Tricky ⚠️
KEY ROLLE'S: f(a)=f(b) β†’ βˆƒc where fβ€²(c)=0
Using the Mean Value Theorem, find \(c\) in \([1,3]\) for \(f(x)=x^2\).
\[\text{MVT: } f'(c) = \frac{f(b)-f(a)}{b-a}\]
MVT guarantees at least one \(c\) β€” but students forget to CHECK that f is continuous & differentiable on the interval first!
πŸ’‘ Explanation
\(\dfrac{f(3)-f(1)}{3-1} = \dfrac{9-1}{2} = 4\)
\(f'(x) = 2x\), set \(2c = 4 \Rightarrow c = 2\)
Answer: B β€” c = 2
🟣 Unit 4 β€” Integrals & Antiderivatives
Q 12
Easy
KEY POWER RULE ∫: add 1 to power β†’ divide by new power
Evaluate: \[\int (4x^3 - 6x + 2)\, dx\]
πŸ“ Antiderivative Power Rule
\(\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C\)   (don't forget +C!)
πŸ’‘ Explanation
\(\int 4x^3\,dx = x^4\), \(\int -6x\,dx = -3x^2\), \(\int 2\,dx = 2x\)
Answer: \(x^4 - 3x^2 + 2x + C\)
Note: Option D is wrong β€” missing the constant \(C\)!
Answer: A
Q 13
Tricky ⚠️
KEY u-SUB: pick u = inside function, find du
Evaluate: \[\int 2x\cos(x^2)\, dx\]
πŸ“ u-Substitution
Let \(u = x^2\), then \(du = 2x\,dx\)
Integral becomes: \(\int \cos(u)\,du = \sin(u)+C = \sin(x^2)+C\)
Students often forget to substitute back in terms of \(x\) at the end!
πŸ’‘ Explanation
\(u=x^2 \Rightarrow du=2x\,dx\)
\(\int \cos(u)\,du = \sin(u) + C = \sin(x^2)+C\)
Answer: C
Q 14
Medium β€” Often Missed
KEY FTC Part 1: d/dx[βˆ«β‚Λ£ f(t)dt] = f(x)
Fundamental Theorem of Calculus, Part 1:

\[\frac{d}{dx}\left[\int_1^x (t^2 + 3t)\,dt\right] = ?\]
Many students try to evaluate the integral first β€” don't! Just replace \(t\) with \(x\)!
πŸ’‘ Explanation
FTC Part 1: \(\dfrac{d}{dx}\left[\int_a^x f(t)\,dt\right] = f(x)\)
Just replace \(t\) with \(x\): answer is \(x^2+3x\)
B is the full antiderivative β€” NOT what's asked!
Answer: A
Q 15
Easy
KEY DEFINITE ∫: F(b) βˆ’ F(a) (plug in TOP minus BOTTOM)
Evaluate: \[\int_0^2 (3x^2 + 1)\, dx\]
πŸ“ FTC Part 2
\(\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)\)
Find antiderivative, then subtract: F(2) βˆ’ F(0)
πŸ’‘ Explanation
\(F(x) = x^3 + x\)
\(F(2) - F(0) = (8+2) - (0+0) = 10\)
Answer: D β€” 10
πŸ”΄ Unit 5 β€” Applications of Integrals
Q 16
Medium
KEY AREA BETWEEN CURVES: ∫[top βˆ’ bottom]dx
Find the area between \(y = x^2\) and \(y = x\) on \([0,1]\).
πŸ“ Area Setup
On \([0,1]\): \(x \geq x^2\) (top βˆ’ bottom)
\(A = \displaystyle\int_0^1 (x - x^2)\,dx\)
Always check which function is on TOP by plugging in a test point!
πŸ’‘ Explanation
\(\int_0^1(x-x^2)\,dx = \left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2}-\dfrac{1}{3} = \dfrac{1}{6}\)
Answer: B β€” 1/6
Q 17
Tricky ⚠️
KEY TOTAL DISTANCE β‰  DISPLACEMENT Β· use |v(t)|
A particle has velocity \(v(t) = t^2 - 4\) for \(0 \leq t \leq 3\).
Find the total distance traveled.
Displacement = \(\int v(t)\,dt\) (can be negative!)
Distance = \(\int |v(t)|\,dt\) β€” split at zeros of \(v(t)\)!
πŸ“ Key Step
\(v(t) = 0\) at \(t = 2\). Split: \([0,2]\) and \([2,3]\)
On \([0,2]\): \(v < 0\), so distance = \(\left|\int_0^2(t^2-4)\,dt\right|\)
πŸ’‘ Explanation
\(v=0\) at \(t=2\). On \([0,2]\): \(v\leq0\); On \([2,3]\): \(v\geq0\)
\(\left|\int_0^2(t^2-4)\,dt\right| = \left|\left[\frac{t^3}{3}-4t\right]_0^2\right| = \left|\frac{8}{3}-8\right| = \frac{16}{3}\)
\(\int_2^3(t^2-4)\,dt = \left[\frac{t^3}{3}-4t\right]_2^3 = (9-12)-(\frac{8}{3}-8) = \frac{1}{3}\cdot\ldots\)
Total = \(\frac{16}{3}+\frac{1}{3} = \mathbf{\frac{17}{3}}\)
Answer: C β€” 17/3
Q 18
Medium β€” Often Missed
KEY DISK METHOD: V = Ο€βˆ«[f(x)]Β² dx
The region bounded by \(y = \sqrt{x}\), \(x=4\), and the \(x\)-axis is revolved about the \(x\)-axis.
Find the volume.
πŸ“ Disk Method
\(V = \pi\displaystyle\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx\)
πŸ’‘ Explanation
\(V = \pi\int_0^4 x\,dx = \pi\left[\dfrac{x^2}{2}\right]_0^4 = \pi\cdot 8 = 8\pi\)
Answer: C β€” 8Ο€
⚑ Unit 6 β€” Mixed Challenge
Q 19
Tricky ⚠️
KEY L'HΓ”PITAL: 0/0 or ∞/∞ β†’ differentiate TOP & BOTTOM separately
Evaluate using L'HΓ΄pital's Rule: \[\lim_{x \to 0} \frac{e^x - 1}{x}\]
Apply L'Hôpital only when you have 0/0 or ∞/∞ form. Always verify first!
πŸ“ L'HΓ΄pital's Rule
If \(\lim\frac{f}{g} = \frac{0}{0}\), then \(\lim\frac{f}{g} = \lim\frac{f'}{g'}\)
πŸ’‘ Explanation
Form is \(\frac{0}{0}\) β†’ apply L'HΓ΄pital:
\(\dfrac{d}{dx}[e^x-1] = e^x\), \(\dfrac{d}{dx}[x] = 1\)
\(\lim_{x\to0}\dfrac{e^x}{1} = e^0 = 1\)
Answer: B β€” 1
Q 20
Boss Level πŸ”₯
KEY RELATED RATES: draw β†’ label β†’ differentiate w.r.t. t β†’ plug in
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? \[x^2 + y^2 = 100\]
πŸ“ Related Rates Setup
Given: \(\dfrac{dx}{dt} = 2\), find \(\dfrac{dy}{dt}\) when \(x = 6\)
Differentiate: \(2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0\)
Find y first using Pythagoras: \(6^2+y^2=100 \Rightarrow y=8\). Students often forget this step!
πŸ’‘ Explanation
\(y = \sqrt{100-36} = 8\)
\(2(6)(2) + 2(8)\dfrac{dy}{dt} = 0\)
\(24 + 16\dfrac{dy}{dt} = 0 \Rightarrow \dfrac{dy}{dt} = -\dfrac{3}{2}\) ft/sec
Negative = sliding down βœ“
Answer: C β€” βˆ’3/2 ft/sec

πŸŽ‰ Quiz Complete!

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