📐 Calculus AB — Study Notes & Quiz

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📐 Calculus AB

Self-Study Notes & Quiz — 20 Core Problems

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UNIT 1 Limits & Continuity

L-E-C: "Limit → Exists → Continuous"

For f to be continuous at x=a:
① Limit exists ② f(a) Exists ③ They're Connected: lim = f(a)

✏️ EXAMPLE

\( \displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3} \) — Factor first!
\( = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3}(x+3) = 6 \)

1 Evaluate: \(\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\) ⚠️ TRAP: Don't just plug in 2!

💡 EXPLANATION

Factor: \(\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2\). Cancel (x-2), then plug in: 2+2 = 4. Never substitute before simplifying!

2 Which condition is NOT required for f(x) to be continuous at x = a?

💡 EXPLANATION

Continuity requires: ① limit exists ② f(a) exists ③ limit = f(a). The VALUE of f(a) doesn't need to be zero — it just needs to EXIST and MATCH the limit. f(a) = 0 is NOT a requirement.

3 Find: \(\displaystyle\lim_{x \to \infty} \frac{3x^2 + 5x}{7x^2 - 2}\) ⚠️ TRAP: Degree matters!

💡 EXPLANATION

Divide by highest power (x²). Top: 3 + 5/x → 3. Bottom: 7 - 2/x² → 7. Answer = 3/7. Tip: Same degree → ratio of leading coefficients!

UNIT 2 Derivatives — Differentiation Rules

Chain Rule =
"Outside × Inside'"
f(g(x))' = f'(g(x))·g'(x)

POWER: "Bring it down, reduce by 1"

\(\dfrac{d}{dx}[x^n] = n \cdot x^{n-1}\) — Multiply by n, subtract 1 from exponent.

4 If \(f(x) = x^3 - 4x^2 + 7\), what is \(f'(x)\)?

💡 EXPLANATION

Power rule term by term: \(\frac{d}{dx}x^3 = 3x^2\), \(\frac{d}{dx}(-4x^2) = -8x\), \(\frac{d}{dx}(7) = 0\) (constant → disappears!). Result: \(3x^2 - 8x\).

5 Find the derivative of \(g(x) = \sin(3x^2)\) using the Chain Rule. ⚠️ Don't forget the inner derivative!

💡 EXPLANATION

Chain Rule: outer = sin → cos. Inner = 3x² → 6x. So: \(\cos(3x^2) \cdot 6x = 6x\cos(3x^2)\). The inner derivative (6x) is the part most students miss!

UNIT 2 Product & Quotient Rules

PRODUCT: "Left-Right Rule" — uv' + vu'

"Left × (Right)' + Right × (Left)'" — Keep one, derive the other, then SWITCH.

QUOTIENT: "LO-D-HI minus HI-D-LO over LO²"

\(\left(\dfrac{u}{v}\right)' = \dfrac{vu' - uv'}{v^2}\) — Bottom times top' minus top times bottom', all over bottom squared.

6 Differentiate: \(h(x) = x^2 \sin x\)

💡 EXPLANATION

Product Rule: u = x², v = sin x. \(h'(x) = (x^2)'\sin x + x^2(\sin x)' = 2x\sin x + x^2\cos x\). Always apply to BOTH terms!

7 Find the derivative of \(f(x) = \dfrac{x^2 + 1}{x - 3}\) ⚠️ Quotient Rule — order matters!

💡 EXPLANATION

Quotient Rule: \(\frac{(bottom)(top)' - (top)(bottom)'}{(bottom)^2} = \frac{(x-3)(2x) - (x^2+1)(1)}{(x-3)^2}\). TOP minus BOTTOM order (LO·D·HI - HI·D·LO).

UNIT 3 Applications of Derivatives

INCREASING / DECREASING: "Sign of f'"

f'(x) > 0 → INCREASING ↗ | f'(x) < 0 → DECREASING ↘ | f'(x) = 0 → Critical Point ●

✏️ RELATED RATES — EXAMPLE

A ladder 10 ft long leans on a wall.
If the bottom slides away at 2 ft/s, how fast is the top sliding down when bottom is 6 ft from wall?
\(x^2 + y^2 = 100\) → differentiate: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)

8 If \(f'(x) = (x-2)(x+1)\), on which interval is f(x) decreasing?

💡 EXPLANATION

Critical points: x = -1 and x = 2. Test a value in (-1, 2): x=0 → f'(0) = (-2)(1) = -2 < 0. Negative f' means DECREASING! So f decreases on (-1, 2).

9 The position of a particle is \(s(t) = t^3 - 6t^2 + 9t\). When is the particle at rest? ⚠️ At rest means velocity = 0!

💡 EXPLANATION

At rest → v(t) = s'(t) = 0. \(s'(t) = 3t^2 - 12t + 9 = 3(t^2-4t+3) = 3(t-1)(t-3) = 0\). So t = 1 or t = 3.

10 The second derivative \(f''(x)\) tells us about the concavity. If \(f''(x) > 0\), the graph is:

💡 EXPLANATION

f'' > 0 → Concave UP (∪) — Smile shape! Think: "positive = happy face ☺".
f'' < 0 → Concave DOWN (∩) — Sad face ☹.

UNIT 3 Optimization & Related Rates

OPTIMIZATION: "Find → Test → Conclude"

① Find critical points (f'=0) ② Test with 2nd derivative or sign chart ③ Check endpoints if on closed interval

11 A rectangle has perimeter 40. What dimensions maximize its area? ⚠️ Classic trap: It's always a SQUARE!

💡 EXPLANATION

Perimeter: 2l + 2w = 40 → l + w = 20. Area: A = l·w = l(20-l). A' = 20-2l = 0 → l = 10, w = 10. A rectangle with fixed perimeter has MAX area when it's a SQUARE!

12 A spherical balloon is being inflated. If the radius increases at 2 cm/s, how fast is the volume increasing when r = 3 cm? (Volume of sphere: \(V = \frac{4}{3}\pi r^3\))

💡 EXPLANATION

Differentiate implicitly: \(\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\). Plug in r=3, dr/dt=2: \(= 4\pi(9)(2) = 72\pi\) cm³/s.

UNIT 4 Integration — Antiderivatives

ANTI-POWER: "Add 1, Divide, Add C"

\(\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C\) — Raise exponent by 1, divide by new exponent, always add +C!

\(\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C \quad (n \neq -1)\) \(\int e^x\,dx = e^x + C\) \(\int \cos x\,dx = \sin x + C\)

13 Evaluate: \(\displaystyle\int (3x^2 - 2x + 5)\,dx\)

💡 EXPLANATION

Term by term: \(\int 3x^2\,dx = x^3\), \(\int -2x\,dx = -x^2\), \(\int 5\,dx = 5x\). Result: \(x^3 - x^2 + 5x + C\). Don't forget +C for indefinite integrals!

14 Evaluate: \(\displaystyle\int_0^2 (x^2 + 1)\,dx\) ⚠️ Definite integral — no +C!

💡 EXPLANATION

\(\int_0^2(x^2+1)\,dx = \left[\frac{x^3}{3}+x\right]_0^2 = \left(\frac{8}{3}+2\right) - (0) = \frac{8}{3}+\frac{6}{3} = \frac{14}{3}\).

UNIT 4 Fundamental Theorem of Calculus

FTC: "PART 1 = Derive the Integral"

\(\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)\) — Derivative cancels integral! Just plug in the upper limit.

FTC PART 2: "Evaluate = F(b) - F(a)"

\(\int_a^b f(x)\,dx = F(b) - F(a)\) — Top minus bottom, always!

15 Using the Fundamental Theorem of Calculus (Part 1):
\(\dfrac{d}{dx}\displaystyle\int_1^x (t^2 + \cos t)\,dt = \) ?

💡 EXPLANATION

FTC Part 1: \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\). Just replace t with x! Answer: \(x^2 + \cos x\). The integral disappears — derivative and integral are inverse operations!

16 What does \(\displaystyle\int_a^b f(x)\,dx\) represent geometrically? ⚠️ Area can be NEGATIVE — careful!

💡 EXPLANATION

The definite integral gives NET signed area — areas below x-axis count as NEGATIVE. If you want total area, use \(\int|f(x)|\,dx\). Option D is a common trap!

17 Evaluate: \(\displaystyle\int \frac{1}{x}\,dx\) ⚠️ Power rule FAILS here (n = -1)!

💡 EXPLANATION

Power rule gives \(\frac{x^0}{0}\) — division by zero! Exception: \(\int \frac{1}{x}\,dx = \ln|x| + C\). The absolute value is needed because ln is only defined for positive numbers.

18 Use u-substitution to evaluate: \(\displaystyle\int 2x\,(x^2+1)^4\,dx\)

💡 EXPLANATION

Let u = x²+1, du = 2x dx. Then \(\int u^4\,du = \frac{u^5}{5} + C = \frac{(x^2+1)^5}{5} + C\). The 2x and du cancel perfectly — that's the sign of a good u-sub!

19 The Mean Value Theorem states: if f is continuous on [a,b] and differentiable on (a,b), there exists c such that:

💡 EXPLANATION

MVT: "Somewhere in the middle, the instantaneous rate of change equals the average rate of change." Formula = slope of secant line = \(\frac{f(b)-f(a)}{b-a}\). Option D is Rolle's Theorem (special case when f(a)=f(b))!

20 🏆 FINAL BOSS: If \(f(x) = e^{x^2}\), find \(f'(x)\). ⚠️ Chain Rule on e^(something)!

💡 EXPLANATION

Chain Rule: outer function is \(e^u\) (derivative stays \(e^u\)), inner is \(u = x^2\) (derivative = 2x). Multiply: \(e^{x^2} \cdot 2x = 2xe^{x^2}\). Remember: \(\frac{d}{dx}[e^u] = e^u \cdot u'\)!

🎯 FINAL SCORE

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