Calculus BC — Study Notes & Quiz

Q1 Limits ★ Easy

L'Hôpital → only when you get 0/0 or ∞/∞ !

📎 Quick Example:

\(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\) — memorize this!

What is \(\displaystyle\lim_{x \to 0} \dfrac{1 - \cos x}{x^2}\) ?

📝 Explanation

Apply L'Hôpital's Rule (0/0 form): differentiate top and bottom.
\(\lim_{x\to 0}\dfrac{\sin x}{2x} = \dfrac{1}{2}\lim_{x\to 0}\dfrac{\sin x}{x} = \dfrac{1}{2}(1) = \boxed{\dfrac{1}{2}}\)
Trick: Also works via Taylor series — \(1-\cos x \approx \frac{x^2}{2}\) near 0.

Q2 Continuity ★★ Tricky

Continuity = Limit EXISTS + equals f(c) + f(c) defined

For \(f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2} & x \neq 2 \\ k & x = 2 \end{cases}\), what value of \(k\) makes \(f\) continuous at \(x = 2\)?

📝 Explanation

Factor: \(\dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2\)
So \(\lim_{x\to 2}(x+2) = 4\). For continuity, \(k = 4\). ✓

Q3 Chain Rule ★ Easy

Chain Rule → "Outer' × Inner'"

📎 Quick Example:

\(\dfrac{d}{dx}[\sin(x^2)] = \cos(x^2)\cdot 2x\)

Find \(\dfrac{d}{dx}\left[e^{\sin x}\right]\).

📝 Explanation

Outer function: \(e^u\), derivative = \(e^u\). Inner function: \(u = \sin x\), \(u'=\cos x\).
Answer: \(e^{\sin x} \cdot \cos x\). ✓

Q4 Implicit Differentiation ★★ Tricky

Implicit → treat y as function of x, slap dy/dx after differentiating y

If \(x^2 + y^2 = 25\), find \(\dfrac{dy}{dx}\).

📝 Explanation

Differentiate both sides: \(2x + 2y\dfrac{dy}{dx} = 0\)
Solve: \(\dfrac{dy}{dx} = -\dfrac{2x}{2y} = -\dfrac{x}{y}\) ✓

Q5 Related Rates ★★ Medium

Related Rates → differentiate equation with respect to TIME (t)

📎 Setup Tip:

Draw a diagram. Label what's changing. Write equation. Differentiate w.r.t. \(t\).

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?

📝 Explanation

\(x^2+y^2=100\). At \(x=6\): \(y=8\).
Differentiate: \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\)
\(2(6)(2)+2(8)\dfrac{dy}{dt}=0 \Rightarrow \dfrac{dy}{dt}=-\dfrac{3}{2}\) ft/sec ✓

Q6 MVT ★★ Tricky

MVT → "Secant slope = Tangent slope somewhere inside"

\(f(x) = x^3\) on \([0, 3]\). By the Mean Value Theorem, at what value of \(c\) does \(f'(c)\) equal the average rate of change?

📝 Explanation

Average rate = \(\dfrac{f(3)-f(0)}{3-0}=\dfrac{27}{3}=9\)
\(f'(c)=3c^2=9 \Rightarrow c^2=3 \Rightarrow c=\sqrt{3}\) ✓ (only positive value in (0,3))

Q7 Concavity ★★ Medium

f'' > 0 → Concave UP (cup ☕) | f'' < 0 → Concave DOWN (frown 🙁)

If \(f(x) = x^4 - 4x^2\), on which interval is \(f\) concave up?

📝 Explanation

\(f'=4x^3-8x\), \(f''=12x^2-8\)
\(f''>0\) when \(x^2>\dfrac{2}{3}\), i.e. \(|x|>\dfrac{\sqrt{6}}{3}\)
So concave up on \(\left(-\infty,-\dfrac{\sqrt{6}}{3}\right)\cup\left(\dfrac{\sqrt{6}}{3},\infty\right)\) ✓

Q8 U-Substitution ★ Easy

u-sub → "Let u = inside, find du, swap everything"

📎 Quick Example:

\(\int 2x\cos(x^2)\,dx\): let \(u=x^2\), \(du=2x\,dx\) → \(\int\cos u\,du = \sin u + C\)

Evaluate \(\displaystyle\int x\sqrt{x^2+1}\,dx\).

📝 Explanation

Let \(u = x^2+1\), \(du=2x\,dx\), so \(x\,dx=\dfrac{du}{2}\)
\(\int\sqrt{u}\cdot\dfrac{du}{2} = \dfrac{1}{2}\cdot\dfrac{2}{3}u^{3/2}+C = \dfrac{1}{3}(x^2+1)^{3/2}+C\) ✓

Q9 Integration by Parts ★★ Medium

IBP → LIATE (Log, Inverse trig, Algebraic, Trig, Exponential) — pick u first!

Evaluate \(\displaystyle\int x e^x\,dx\).

📝 Explanation

Let \(u=x\), \(dv=e^x dx\). Then \(du=dx\), \(v=e^x\).
\(\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1)+C\) ✓

Q10 FTC ★★ Tricky

FTC Part 1 → d/dx ∫f(t)dt = f(x) · (chain rule if upper limit is not just x!)

If \(F(x) = \displaystyle\int_1^{x^2} \sin(t)\,dt\), find \(F'(x)\).

📝 Explanation

FTC Part 1 with chain rule: \(F'(x) = \sin(x^2)\cdot\dfrac{d}{dx}(x^2) = \sin(x^2)\cdot 2x\) ✓
The chain rule multiplier \(2x\) is the most commonly forgotten step!

Q11 Separable ODE ★ Easy

Separable → "y's to one side, x's to the other, then integrate both sides"

Solve \(\dfrac{dy}{dx} = \dfrac{x}{y}\) with \(y(0) = 3\).

📝 Explanation

\(y\,dy = x\,dx \Rightarrow \dfrac{y^2}{2} = \dfrac{x^2}{2}+C\)
\(y(0)=3\): \(\dfrac{9}{2}=C\), so \(y^2 = x^2+9 \Rightarrow y=\sqrt{x^2+9}\) ✓

Q12 Euler's Method ★★ Medium

Euler → y_new = y_old + f(x,y)·Δx (always approximate, never exact!)

Given \(\dfrac{dy}{dx}=x+y\), \(y(0)=1\), use Euler's method with step \(\Delta x = 0.1\) to estimate \(y(0.1)\).

📝 Explanation

At \((0,1)\): slope = \(0+1=1\)
\(y(0.1)\approx 1 + 1\cdot(0.1) = 1.1\) ✓

Q13 Convergence Tests ★★ Medium

Ratio Test → L = lim|a_(n+1)/a_n|: L<1 converge, L>1 diverge, L=1 inconclusive

Does \(\displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{n^n}\) converge or diverge? Use Ratio Test!

📝 Explanation

\(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)!\cdot n^n}{(n+1)^{n+1}\cdot n!}=\dfrac{n^n}{(n+1)^n}=\left(\dfrac{n}{n+1}\right)^n\to\dfrac{1}{e}\)
Since \(L=\dfrac{1}{e}<1\), the series converges ✓

Q14 Taylor Series ★★ Tricky

e^x series → \(\sum \frac{x^n}{n!}\) — memorize this and substitute!

Which of the following is the Maclaurin series for \(e^{-x^2}\)?

📝 Explanation

Start with \(e^u = \sum\dfrac{u^n}{n!}\). Substitute \(u=-x^2\):
\(e^{-x^2}=\sum\dfrac{(-x^2)^n}{n!}=\sum\dfrac{(-1)^n x^{2n}}{n!}\) ✓

Q15 Interval of Convergence ★★ Tricky

IoC → Ratio Test for radius, then CHECK ENDPOINTS separately!

Find the interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\dfrac{(x-2)^n}{n}\).

📝 Explanation

Ratio test gives \(|x-2|<1\), so center=2, radius=1: open interval \((1,3)\).
Check endpoints:
• \(x=3\): \(\sum\dfrac{1}{n}\) → harmonic series → diverges
• \(x=1\): \(\sum\dfrac{(-1)^n}{n}\) → alternating harmonic → converges
Interval: \([1,3)\) ✓

Q16 Parametric Derivatives ★★ Medium

dy/dx parametric → (dy/dt) ÷ (dx/dt)

Given \(x=t^2\), \(y=t^3\), find \(\dfrac{dy}{dx}\) at \(t=2\).

📝 Explanation

\(\dfrac{dx}{dt}=2t\), \(\dfrac{dy}{dt}=3t^2\)
\(\dfrac{dy}{dx}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}\)
At \(t=2\): \(\dfrac{dy}{dx}=\dfrac{3(2)}{2}=3\) ✓

Q17 Arc Length ★★ Tricky

Arc length (param) → \(\int\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt\)

The arc length of the curve \(x=3t\), \(y=4t\) from \(t=0\) to \(t=1\) is:

📝 Explanation

\(\dfrac{dx}{dt}=3\), \(\dfrac{dy}{dt}=4\)
\(L=\int_0^1\sqrt{9+16}\,dt=\int_0^1 5\,dt = 5\) ✓
(This is just a straight line — 3-4-5 triangle!)

Q18 Area Between Curves ★ Easy

Area = ∫(TOP − BOTTOM)dx — always subtract the lower function!

Find the area between \(y = x^2\) and \(y = x\) from \(x=0\) to \(x=1\).

📝 Explanation

On \([0,1]\): \(x \geq x^2\), so area = \(\int_0^1(x-x^2)\,dx\)
\(=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2}-\dfrac{1}{3} = \dfrac{1}{6}\) ✓

Q19 Disk/Washer Method ★★ Medium

Disk → π∫[R(x)]²dx | Washer → π∫([R(x)]²−[r(x)]²)dx

The volume of the solid formed by rotating \(y = \sqrt{x}\) from \(x=0\) to \(x=4\) about the \(x\)-axis is:

📝 Explanation

\(V=\pi\int_0^4(\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=\pi\left[\dfrac{x^2}{2}\right]_0^4=\pi\cdot 8 = 8\pi\) ✓

Q20 Improper Integrals ★★ Tricky

Improper → replace ∞ with limit variable b, then let b→∞

Does \(\displaystyle\int_1^{\infty}\dfrac{1}{x^2}\,dx\) converge? If so, find the value.

📝 Explanation

\(\int_1^b x^{-2}\,dx = \left[-\dfrac{1}{x}\right]_1^b = -\dfrac{1}{b}+1\)
As \(b\to\infty\): \(-\dfrac{1}{b}\to 0\), so the integral = \(1\). Converges! ✓
p-series: \(\int_1^\infty x^{-p}dx\) converges iff \(p>1\)