Key Concepts & Formulas to Memorize
AB/BC
Limit Definition of Derivative
\[f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\]
Also: \(f'(a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}\). Memorize both forms.
AB/BC
Chain Rule
\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]
Apply outermost function first, multiply by derivative of inner.
AB/BC
Fundamental Theorem of Calculus
\[\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)\]
FTC Part 1: differentiation undoes integration (with chain rule).
AB/BC
L'Hôpital's Rule
\[\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}\]
Only when form is \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
AB/BC
Integration by Parts
\[\int u\,dv = uv - \int v\,du\]
LIATE order: Logarithmic, Inverse trig, Algebraic, Trig, Exponential.
BC
Taylor/Maclaurin Series
\[f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\]
Key: \(e^x=\sum\frac{x^n}{n!}\), \(\sin x=\sum\frac{(-1)^n x^{2n+1}}{(2n+1)!}\)
BC
Ratio Test for Series
\[L = \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|\]
Converges if L < 1; diverges if L > 1; inconclusive if L = 1.
BC
Polar Area & Arc Length
\[A=\tfrac{1}{2}\int_\alpha^\beta r^2\,d\theta\]
Arc length: \(L=\int\sqrt{r^2+(dr/d\theta)^2}\,d\theta\)
Exam Instructions
- 20 multiple choice questions — one correct answer each
- No partial credit — select the best answer
- No graphing calculator for questions marked ★
- Immediate feedback shown after each question
- Detailed solutions revealed upon submission
- AP scoring: 5 = 70–100%, 4 = 55–69%, 3 = 40–54%
Practice Questions
01
What is \(\displaystyle\lim_{x \to 0} \dfrac{\sin(3x)}{x}\)?
Solution
Use the standard limit: \(\lim_{x\to 0}\frac{\sin(u)}{u}=1\). Rewrite:
\[\frac{\sin(3x)}{x} = 3\cdot\frac{\sin(3x)}{3x} \to 3\cdot 1 = \boxed{3}\]
\[\frac{\sin(3x)}{x} = 3\cdot\frac{\sin(3x)}{3x} \to 3\cdot 1 = \boxed{3}\]
02
\(\displaystyle\lim_{x \to \infty} \frac{5x^3 - 2x + 1}{3x^3 + 7}\) equals
Solution
Divide numerator and denominator by \(x^3\) (the highest power):
\[\frac{5 - 2/x^2 + 1/x^3}{3 + 7/x^3} \to \frac{5-0+0}{3+0} = \boxed{\frac{5}{3}}\] The rule: for equal-degree polynomials, the limit equals the ratio of leading coefficients.
\[\frac{5 - 2/x^2 + 1/x^3}{3 + 7/x^3} \to \frac{5-0+0}{3+0} = \boxed{\frac{5}{3}}\] The rule: for equal-degree polynomials, the limit equals the ratio of leading coefficients.
03
If \(f(x) = x^2 e^x\), then \(f''(x)\) equals
Solution
First derivative (product rule: \((uv)'=u'v+uv'\)):
\[f'(x) = 2x\cdot e^x + x^2\cdot e^x = e^x(x^2+2x)\] Second derivative (product rule again on \(e^x(x^2+2x)\)):
\[f''(x) = e^x(x^2+2x) + e^x(2x+2) = e^x(x^2+4x+2) \] Answer: \(\boxed{e^x(x^2+4x+2)}\)
\[f'(x) = 2x\cdot e^x + x^2\cdot e^x = e^x(x^2+2x)\] Second derivative (product rule again on \(e^x(x^2+2x)\)):
\[f''(x) = e^x(x^2+2x) + e^x(2x+2) = e^x(x^2+4x+2) \] Answer: \(\boxed{e^x(x^2+4x+2)}\)
04
Given \(x^2 + y^2 = 25\), what is \(\dfrac{dy}{dx}\) at the point \((3, 4)\)?
Solution
Differentiate both sides with respect to \(x\):
\[2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}\] At \((3,4)\): \(\dfrac{dy}{dx} = -\dfrac{3}{4}\). This is the slope of the tangent to the circle at that point.
\[2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}\] At \((3,4)\): \(\dfrac{dy}{dx} = -\dfrac{3}{4}\). This is the slope of the tangent to the circle at that point.
05
Let \(f(x)=x^3-x\) on \([0,2]\). What value of \(c \in (0,2)\) satisfies the conclusion of the Mean Value Theorem?
Solution
MVT requires \(f'(c) = \dfrac{f(2)-f(0)}{2-0}\).
\(f(0)=0,\; f(2)=8-2=6\). So \(f'(c) = \dfrac{6}{2}=3\).
\(f'(x)=3x^2-1\). Set \(3c^2-1=3 \implies c^2=\dfrac{4}{3} \implies c = \dfrac{2}{\sqrt{3}}\approx 1.155 \in (0,2)\). ✓
\(f(0)=0,\; f(2)=8-2=6\). So \(f'(c) = \dfrac{6}{2}=3\).
\(f'(x)=3x^2-1\). Set \(3c^2-1=3 \implies c^2=\dfrac{4}{3} \implies c = \dfrac{2}{\sqrt{3}}\approx 1.155 \in (0,2)\). ✓
06
A ladder 10 ft long rests against a vertical wall. If the bottom slides away at 2 ft/sec, how fast is the top of the ladder sliding down when the bottom is 6 ft from the wall?
Solution
Let \(x\) = distance from wall, \(y\) = height. Then \(x^2+y^2=100\).
Differentiate: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\).
When \(x=6\): \(y=\sqrt{100-36}=8\). Given \(\frac{dx}{dt}=2\):
\[2(6)(2)+2(8)\frac{dy}{dt}=0 \implies \frac{dy}{dt}=-\frac{3}{2}\text{ ft/sec}\] The top slides down at \(\frac{3}{2}\) ft/sec.
Differentiate: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\).
When \(x=6\): \(y=\sqrt{100-36}=8\). Given \(\frac{dx}{dt}=2\):
\[2(6)(2)+2(8)\frac{dy}{dt}=0 \implies \frac{dy}{dt}=-\frac{3}{2}\text{ ft/sec}\] The top slides down at \(\frac{3}{2}\) ft/sec.
07
\(\displaystyle\int_0^{\pi} \sin x\,dx\) equals
Solution
\(\displaystyle\int_0^{\pi}\sin x\,dx = \Big[-\cos x\Big]_0^{\pi} = (-\cos\pi)-(-\cos 0) = (-(-1))-(-1)=1+1=\boxed{2}\)
08
If \(g(x) = \displaystyle\int_1^{x^2} \ln(t)\,dt\), then \(g'(x) =\)
Solution
By FTC Part 1 with Chain Rule: \(\dfrac{d}{dx}\int_a^{u(x)}f(t)\,dt = f(u(x))\cdot u'(x)\).
Here \(u(x)=x^2\), \(u'(x)=2x\), \(f(t)=\ln t\).
\[g'(x) = \ln(x^2)\cdot 2x = \boxed{2x\ln(x^2)}\] Note: this can be simplified to \(4x\ln x\), but \(2x\ln(x^2)\) is the direct form matching choice B.
Here \(u(x)=x^2\), \(u'(x)=2x\), \(f(t)=\ln t\).
\[g'(x) = \ln(x^2)\cdot 2x = \boxed{2x\ln(x^2)}\] Note: this can be simplified to \(4x\ln x\), but \(2x\ln(x^2)\) is the direct form matching choice B.
09
\(\displaystyle\int_0^1 x\,e^{x^2}\,dx =\)
Solution
Let \(u=x^2\), \(du=2x\,dx\), so \(x\,dx = \frac{du}{2}\). New limits: \(u(0)=0\), \(u(1)=1\).
\[\int_0^1 xe^{x^2}\,dx = \int_0^1 e^u\frac{du}{2} = \frac{1}{2}\Big[e^u\Big]_0^1 = \frac{e-1}{2}\]
\[\int_0^1 xe^{x^2}\,dx = \int_0^1 e^u\frac{du}{2} = \frac{1}{2}\Big[e^u\Big]_0^1 = \frac{e-1}{2}\]
10
The area of the region enclosed by \(y = x^2\) and \(y = x\) is
Solution
Intersections: \(x^2=x \Rightarrow x(x-1)=0\), so \(x=0,1\). On \([0,1]\), \(x \ge x^2\).
\[A = \int_0^1(x-x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \frac{1}{2}-\frac{1}{3} = \boxed{\frac{1}{6}}\]
\[A = \int_0^1(x-x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \frac{1}{2}-\frac{1}{3} = \boxed{\frac{1}{6}}\]
11
The solution to \(\dfrac{dy}{dx} = ky\) with \(y(0)=y_0\) is
Solution
Separate variables: \(\dfrac{dy}{y}=k\,dx\). Integrate both sides:
\(\ln|y|=kx+C \implies y=Ae^{kx}\).
Apply \(y(0)=y_0\): \(A=y_0\). Thus \(y=\boxed{y_0 e^{kx}}\). This is exponential growth/decay.
\(\ln|y|=kx+C \implies y=Ae^{kx}\).
Apply \(y(0)=y_0\): \(A=y_0\). Thus \(y=\boxed{y_0 e^{kx}}\). This is exponential growth/decay.
12
A slope field for the differential equation \(\dfrac{dy}{dx} = \dfrac{x}{y}\) shows slopes of 0 along which line?
Solution
Slope \(= \frac{x}{y} = 0\) when \(x = 0\) and \(y \ne 0\). This is the \(y\)-axis. (When \(y=0\), the slope is undefined, not zero.) So horizontal segments appear along \(x=0\).
13
\(\displaystyle\int x\cos x\,dx =\)
Solution
Let \(u=x\), \(dv=\cos x\,dx\). Then \(du=dx\), \(v=\sin x\).
\[\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x -(-\cos x)+C = \boxed{x\sin x+\cos x+C}\] Verify: \(\frac{d}{dx}[x\sin x+\cos x] = \sin x + x\cos x - \sin x = x\cos x\) ✓
\[\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x -(-\cos x)+C = \boxed{x\sin x+\cos x+C}\] Verify: \(\frac{d}{dx}[x\sin x+\cos x] = \sin x + x\cos x - \sin x = x\cos x\) ✓
14
Which of the following series converges?
Solution
By the p-series test: \(\sum \frac{1}{n^p}\) converges if and only if \(p > 1\).
A: \(p=1\) (harmonic series) — diverges.
B: \(p=\frac{1}{2}<1\) — diverges.
C: \(p=2>1\) — converges (Basel problem: \(=\frac{\pi^2}{6}\)).
D: \(\sum n\) diverges (terms \(\to\infty\) by divergence test).
A: \(p=1\) (harmonic series) — diverges.
B: \(p=\frac{1}{2}<1\) — diverges.
C: \(p=2>1\) — converges (Basel problem: \(=\frac{\pi^2}{6}\)).
D: \(\sum n\) diverges (terms \(\to\infty\) by divergence test).
15
The Maclaurin series for \(e^x\) is \(\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}\). Using this, the Maclaurin series for \(e^{-x^2}\) is
Solution
Substitute \(x \to -x^2\) into \(e^x=\sum\frac{x^n}{n!}\):
\[e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!}\] This technique — substituting directly into a known series — is faster and less error-prone than computing derivatives.
\[e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!}\] This technique — substituting directly into a known series — is faster and less error-prone than computing derivatives.
16
If \(x(t) = t^2\) and \(y(t) = t^3\), then \(\dfrac{dy}{dx} =\)
Solution
For parametric curves: \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\).
\(\dfrac{dy}{dt}=3t^2\), \(\dfrac{dx}{dt}=2t\).
\[\frac{dy}{dx} = \frac{3t^2}{2t} = \boxed{\frac{3t}{2}}\]
\(\dfrac{dy}{dt}=3t^2\), \(\dfrac{dx}{dt}=2t\).
\[\frac{dy}{dx} = \frac{3t^2}{2t} = \boxed{\frac{3t}{2}}\]
17
The area enclosed by the polar curve \(r = 2\cos\theta\) is
Solution
\(r=2\cos\theta\) is a circle of radius 1 centered at \((1,0)\). Area \(=\pi r^2=\pi\).
Using the formula: \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2\,d\theta = 2\int_{-\pi/2}^{\pi/2}(1+\cos2\theta)\,d\theta\)
\[= 2\Big[\theta+\tfrac{\sin2\theta}{2}\Big]_{-\pi/2}^{\pi/2} = 2\cdot\pi = \pi\text{ (after simplification)}=\boxed{\pi}\]
Using the formula: \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2\,d\theta = 2\int_{-\pi/2}^{\pi/2}(1+\cos2\theta)\,d\theta\)
\[= 2\Big[\theta+\tfrac{\sin2\theta}{2}\Big]_{-\pi/2}^{\pi/2} = 2\cdot\pi = \pi\text{ (after simplification)}=\boxed{\pi}\]
18
\(\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} =\)
Solution
Form is \(\frac{0}{0}\), so apply L'Hôpital's Rule:
\[\lim_{x\to 0}\frac{e^x-1}{x} = \lim_{x\to 0}\frac{e^x}{1} = e^0 = \boxed{1}\] Alternatively: this is the definition of \(\frac{d}{dx}[e^x]\big|_{x=0}=1\).
\[\lim_{x\to 0}\frac{e^x-1}{x} = \lim_{x\to 0}\frac{e^x}{1} = e^0 = \boxed{1}\] Alternatively: this is the definition of \(\frac{d}{dx}[e^x]\big|_{x=0}=1\).
19
The volume of the solid generated by revolving the region bounded by \(y=\sqrt{x}\), \(y=0\), and \(x=4\) about the \(x\)-axis is
Solution
Disk method: \(V = \pi\displaystyle\int_0^4 [\sqrt{x}]^2\,dx = \pi\int_0^4 x\,dx\)
\[= \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\cdot\frac{16}{2} = \boxed{8\pi}\]
\[= \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\cdot\frac{16}{2} = \boxed{8\pi}\]
20
The alternating series \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\). The error in approximating the sum with the first 3 terms is at most
Solution
By the Alternating Series Estimation Theorem: for a convergent alternating series whose terms decrease in absolute value to 0, the error is at most \(|a_{n+1}|\), the absolute value of the first omitted term.
Using 3 terms means the first omitted term is \(a_4 = \frac{1}{4}\). Error \(\le \boxed{\frac{1}{4}}\).
Using 3 terms means the first omitted term is \(a_4 = \frac{1}{4}\). Error \(\le \boxed{\frac{1}{4}}\).
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