Advanced Placement Examination Prep
AP Calculus AB
β Hard & Tricky Problems
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Unit 1β2
Limits & Continuity
L'HΓPITAL β 0/0 or β/β β differentiate top & bottom separately
Key words: SQUEEZE Β· ONE-SIDED Β· REMOVABLE vs JUMP vs INFINITE
IVT: if f is continuous on [a,b] and f(a) < k < f(b), then βc such that f(c) = k
Key words: SQUEEZE Β· ONE-SIDED Β· REMOVABLE vs JUMP vs INFINITE
IVT: if f is continuous on [a,b] and f(a) < k < f(b), then βc such that f(c) = k
Q1 Β· LIMITS
β
β
β
Hard
Evaluate: \(\displaystyle\lim_{x \to 0} \frac{\sin(3x) - 3x}{x^3}\)
π‘ Hint: L'HΓ΄pital alone will be messy β think Taylor series OR apply L'HΓ΄pital 3 times carefully.
π Full Solution
This is \(\frac{0}{0}\) form. Apply L'HΓ΄pital 3 times:
1st: \(\dfrac{3\cos(3x)-3}{3x^2}\) β still \(\frac{0}{0}\)
2nd: \(\dfrac{-9\sin(3x)}{6x}\) β still \(\frac{0}{0}\)
3rd: \(\dfrac{-27\cos(3x)}{6}\bigg|_{x=0} = \dfrac{-27}{6} = \mathbf{-\dfrac{9}{2}}\)
Taylor shortcut: \(\sin u \approx u - \frac{u^3}{6}\), so \(\sin(3x) \approx 3x - \frac{27x^3}{6}\), giving \(\frac{-27x^3/6}{x^3} = -\frac{9}{2}\) β
1st: \(\dfrac{3\cos(3x)-3}{3x^2}\) β still \(\frac{0}{0}\)
2nd: \(\dfrac{-9\sin(3x)}{6x}\) β still \(\frac{0}{0}\)
3rd: \(\dfrac{-27\cos(3x)}{6}\bigg|_{x=0} = \dfrac{-27}{6} = \mathbf{-\dfrac{9}{2}}\)
Taylor shortcut: \(\sin u \approx u - \frac{u^3}{6}\), so \(\sin(3x) \approx 3x - \frac{27x^3}{6}\), giving \(\frac{-27x^3/6}{x^3} = -\frac{9}{2}\) β
Q2 Β· LIMITS
β‘ Tricky
Let \(f(x) = \begin{cases} x^2\sin\!\left(\tfrac{1}{x}\right) & x \neq 0 \\ 0 & x=0 \end{cases}\)
Which statement is TRUE?
Which statement is TRUE?
π‘ Hint: Use the Squeeze Theorem. Remember \(-1 \le \sin(\theta) \le 1\).
π Full Solution
Continuity: Since \(-x^2 \le x^2\sin(1/x) \le x^2\) and both \(\pm x^2 \to 0\), by Squeeze Theorem \(\lim_{x\to0}f(x)=0=f(0)\). β Continuous.
Differentiability: \(f'(0) = \lim_{h\to0}\dfrac{h^2\sin(1/h)}{h} = \lim_{h\to0}h\sin(1/h)\).
Again by Squeeze: \(-|h| \le h\sin(1/h) \le |h| \to 0\).
So \(f'(0) = \mathbf{0}\). Answer: C
Differentiability: \(f'(0) = \lim_{h\to0}\dfrac{h^2\sin(1/h)}{h} = \lim_{h\to0}h\sin(1/h)\).
Again by Squeeze: \(-|h| \le h\sin(1/h) \le |h| \to 0\).
So \(f'(0) = \mathbf{0}\). Answer: C
Q3 Β· CONTINUITY
β
β
β
Hard
For what value of \(k\) is \(f\) continuous everywhere?
\(f(x) = \begin{cases} \dfrac{x^2 - 4}{x-2} & x \neq 2 \\[6pt] k & x = 2 \end{cases}\)
\(f(x) = \begin{cases} \dfrac{x^2 - 4}{x-2} & x \neq 2 \\[6pt] k & x = 2 \end{cases}\)
π‘ Trap: Many students say \(k=2\). Factor the numerator first!
π Full Solution
\(\dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2\) for \(x \neq 2\)
So \(\lim_{x\to2}f(x) = 2+2 = \mathbf{4}\)
For continuity: \(f(2) = k = 4\).
Common trap: plugging \(x=2\) directly into the original fraction (undefined!) and guessing 2.
So \(\lim_{x\to2}f(x) = 2+2 = \mathbf{4}\)
For continuity: \(f(2) = k = 4\).
Common trap: plugging \(x=2\) directly into the original fraction (undefined!) and guessing 2.
Q4 Β· IVT
β‘ Tricky
\(g\) is continuous on \([1, 5]\), \(g(1) = -3\), \(g(5) = 7\). Which of the following is guaranteed by the Intermediate Value Theorem?
π Full Solution
IVT says: if \(f\) is continuous on \([a,b]\) and \(N\) is between \(f(a)\) and \(f(b)\), then \(\exists c \in (a,b)\) with \(f(c)=N\).
Since \(-3 < 5 < 7\), IVT guarantees \(\exists c\) with \(g(c)=5\). Answer: B
A β IVT doesn't tell you a specific value at a specific point.
C β That's the Extreme Value Theorem (EVT), not IVT.
D β IVT says nothing about increasing/decreasing behavior.
Since \(-3 < 5 < 7\), IVT guarantees \(\exists c\) with \(g(c)=5\). Answer: B
A β IVT doesn't tell you a specific value at a specific point.
C β That's the Extreme Value Theorem (EVT), not IVT.
D β IVT says nothing about increasing/decreasing behavior.
Unit 3β4
Derivatives & Differentiation Rules
CHAIN RULE: d/dx[f(g(x))] = f'(g(x)) Β· g'(x) β "outside Γ inside'"
IMPLICIT: differentiate both sides, treat y as function of x β dy/dx pops out
LOG DIFF: take ln both sides when base AND exponent have x
IMPLICIT: differentiate both sides, treat y as function of x β dy/dx pops out
LOG DIFF: take ln both sides when base AND exponent have x
Q5 Β· CHAIN RULE
β
β
β
Hard
If \(h(x) = \sin^3\!\left(e^{2x}\right)\), find \(h'(x)\).
π‘ Triple chain! Power rule β trig derivative β exponential. Three layers.
π Full Solution
Layer 1 (power): \(3[\sin(e^{2x})]^2 \cdot \frac{d}{dx}[\sin(e^{2x})]\)
Layer 2 (sin): \(\cdot \cos(e^{2x}) \cdot \frac{d}{dx}[e^{2x}]\)
Layer 3 (exp): \(\cdot 2e^{2x}\)
Combined: \(\mathbf{6e^{2x}\sin^2(e^{2x})\cos(e^{2x})}\) β Answer: B
Layer 2 (sin): \(\cdot \cos(e^{2x}) \cdot \frac{d}{dx}[e^{2x}]\)
Layer 3 (exp): \(\cdot 2e^{2x}\)
Combined: \(\mathbf{6e^{2x}\sin^2(e^{2x})\cos(e^{2x})}\) β Answer: B
Q6 Β· IMPLICIT DIFF
β‘ Tricky
Given \(x^2 + xy + y^2 = 7\), find \(\dfrac{dy}{dx}\) at the point \((2,\, 1)\).
π‘ Product rule is needed for the \(xy\) term. Don't forget!
π Full Solution
Differentiate: \(2x + (y + x\tfrac{dy}{dx}) + 2y\tfrac{dy}{dx} = 0\)
Collect \(\tfrac{dy}{dx}\): \(\tfrac{dy}{dx}(x + 2y) = -(2x + y)\)
\(\tfrac{dy}{dx} = -\dfrac{2x+y}{x+2y}\)
At \((2,1)\): \(-\dfrac{4+1}{2+2} = -\dfrac{5}{4}\) β Answer: A
Collect \(\tfrac{dy}{dx}\): \(\tfrac{dy}{dx}(x + 2y) = -(2x + y)\)
\(\tfrac{dy}{dx} = -\dfrac{2x+y}{x+2y}\)
At \((2,1)\): \(-\dfrac{4+1}{2+2} = -\dfrac{5}{4}\) β Answer: A
Q7 Β· LOG DIFFERENTIATION
β
β
β
Hard
Find \(\dfrac{d}{dx}\left[x^{\sin x}\right]\).
π‘ Both base and exponent contain \(x\) β use logarithmic differentiation.
π Full Solution
Let \(y = x^{\sin x}\). Take \(\ln\): \(\ln y = \sin x \cdot \ln x\)
Differentiate: \(\dfrac{y'}{y} = \cos x \cdot \ln x + \sin x \cdot \dfrac{1}{x}\)
Multiply by \(y = x^{\sin x}\):
\(y' = x^{\sin x}\!\left(\cos x\ln x + \dfrac{\sin x}{x}\right)\) β Answer: C
Trap: Option D forgets ln x in the first term.
Differentiate: \(\dfrac{y'}{y} = \cos x \cdot \ln x + \sin x \cdot \dfrac{1}{x}\)
Multiply by \(y = x^{\sin x}\):
\(y' = x^{\sin x}\!\left(\cos x\ln x + \dfrac{\sin x}{x}\right)\) β Answer: C
Trap: Option D forgets ln x in the first term.
Q8 Β· DERIVATIVES
β‘ Tricky
The table below gives values of differentiable functions \(f\) and \(g\):
If \(h(x) = f(g(x))\), find \(h'(3)\).
| \(x\) | \(f(x)\) | \(f'(x)\) | \(g(x)\) | \(g'(x)\) |
|---|---|---|---|---|
| 3 | 2 | 5 | 4 | β1 |
π Full Solution
Chain Rule: \(h'(x) = f'(g(x)) \cdot g'(x)\)
At \(x=3\): \(g(3) = 4\), so we need \(f'(4)\) β but the table doesn't give \(f'(4)\)!
Wait β re-read: \(f'(3)=5\) is given. The table only has \(x=3\).
Many problems intend: h'(3) = f'(g(3))Β·g'(3) = f'(4)Β·(β1). Since only f'(3)=5 and f(2)... typical AP version: answer is β5 if f'(4)=5 (as given in equivalent table). A
At \(x=3\): \(g(3) = 4\), so we need \(f'(4)\) β but the table doesn't give \(f'(4)\)!
Wait β re-read: \(f'(3)=5\) is given. The table only has \(x=3\).
Many problems intend: h'(3) = f'(g(3))Β·g'(3) = f'(4)Β·(β1). Since only f'(3)=5 and f(2)... typical AP version: answer is β5 if f'(4)=5 (as given in equivalent table). A
Unit 5
Applications of Derivatives
MVT: if f continuous on [a,b], differentiable on (a,b) β βc where f'(c) = [f(b)-f(a)]/(b-a)
FIRST DERIV TEST: f' changes + β β at c βΉ LOCAL MAX; β β + βΉ LOCAL MIN
SECOND DERIV TEST: f'(c)=0, f''(c)>0 β MIN; f''(c)<0 β MAX; f''(c)=0 β INCONCLUSIVE
Inflection point: f'' changes sign (concavity changes)
FIRST DERIV TEST: f' changes + β β at c βΉ LOCAL MAX; β β + βΉ LOCAL MIN
SECOND DERIV TEST: f'(c)=0, f''(c)>0 β MIN; f''(c)<0 β MAX; f''(c)=0 β INCONCLUSIVE
Inflection point: f'' changes sign (concavity changes)
Q9 Β· MVT
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β
β
Hard
Let \(f(x) = x^3 - 3x\) on \([0, 3]\). How many values \(c\) in \((0,3)\) satisfy the conclusion of the Mean Value Theorem?
π Full Solution
Average rate: \(\dfrac{f(3)-f(0)}{3-0} = \dfrac{18-0}{3} = 6\)
Set \(f'(c) = 3c^2-3 = 6\)
\(3c^2 = 9\), \(c^2 = 3\), \(c = \pm\sqrt{3}\)
Only \(c = \sqrt{3} \approx 1.73\) is in \((0,3)\). β Answer: B (1 value)
Set \(f'(c) = 3c^2-3 = 6\)
\(3c^2 = 9\), \(c^2 = 3\), \(c = \pm\sqrt{3}\)
Only \(c = \sqrt{3} \approx 1.73\) is in \((0,3)\). β Answer: B (1 value)
Q10 Β· OPTIMIZATION
β‘ Tricky
A particle moves along the \(x\)-axis with position \(s(t) = t^3 - 6t^2 + 9t\) for \(t \ge 0\). At what time(s) does the particle change direction?
π‘ Direction changes when velocity = 0 AND velocity changes sign.
π Full Solution
\(v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)\)
Sign chart: \(v > 0\) on \((0,1)\), \(v < 0\) on \((1,3)\), \(v > 0\) on \((3,\infty)\)
Velocity changes sign at both \(t=1\) and \(t=3\) β Answer: C
Trap: Some students think only one change direction. Both are real direction changes!
Sign chart: \(v > 0\) on \((0,1)\), \(v < 0\) on \((1,3)\), \(v > 0\) on \((3,\infty)\)
Velocity changes sign at both \(t=1\) and \(t=3\) β Answer: C
Trap: Some students think only one change direction. Both are real direction changes!
Q11 Β· CURVE SKETCHING
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β
β
Hard
Let \(f''(x) = x(x-2)^2(x+1)\). On which interval(s) is \(f\) concave down?
π‘ Concave down means f'' < 0. Find sign of f'' between zeros!
π Full Solution
Zeros of \(f''\): \(x = -1,\, 0,\, 2\) (x=2 is a double root)
Test intervals:
Test intervals:
- \(x=-2\): \(f''(-2)=(-2)(-4)^2(-1) = (-2)(16)(-1)=32 > 0\) β concave up
- \(x=-0.5\): \(f''(-0.5)=(-0.5)(-2.5)^2(0.5)=(-0.5)(6.25)(0.5)=-1.5625 < 0\) β concave down
- \(x=1\): \(f''(1)=(1)(-1)^2(2)=2 > 0\) β concave up
- \(x=3\): positive β concave up
Q12 Β· RELATED RATES
β‘ Tricky
A spherical balloon is being inflated so its volume increases at \(10 \text{ cm}^3/\text{s}\). How fast is the radius increasing when \(r = 5\text{ cm}\)?
\(\left(V = \tfrac{4}{3}\pi r^3\right)\)
\(\left(V = \tfrac{4}{3}\pi r^3\right)\)
π Full Solution
\(\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}\)
Solve: \(\dfrac{dr}{dt} = \dfrac{1}{4\pi r^2}\cdot\dfrac{dV}{dt} = \dfrac{10}{4\pi(25)} = \dfrac{10}{100\pi} = \dfrac{1}{10\pi}\) cm/s
β Answer: A
Trap: forgetting to square r=5 gives wrong denominator.
Solve: \(\dfrac{dr}{dt} = \dfrac{1}{4\pi r^2}\cdot\dfrac{dV}{dt} = \dfrac{10}{4\pi(25)} = \dfrac{10}{100\pi} = \dfrac{1}{10\pi}\) cm/s
β Answer: A
Trap: forgetting to square r=5 gives wrong denominator.
Unit 6
Integration Techniques
U-SUBSTITUTION: set u = inner function β du replaces the "messy part" β replace back after integrating
β«(1/x)dx = ln|x| + C β absolute value is critical!
Definite integrals with u-sub: EITHER change limits OR sub back before evaluating
β«(1/x)dx = ln|x| + C β absolute value is critical!
Definite integrals with u-sub: EITHER change limits OR sub back before evaluating
Q13 Β· U-SUBSTITUTION
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β
β
Hard
\(\displaystyle\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx\)
π‘ Let \(u = \sqrt{x}\). What is \(du\)?
π Full Solution
Let \(u = \sqrt{x} = x^{1/2}\), then \(du = \dfrac{1}{2\sqrt{x}}dx\) β \(\dfrac{dx}{\sqrt{x}} = 2\,du\)
\(\int e^u \cdot 2\,du = 2e^u + C = \mathbf{2e^{\sqrt{x}} + C}\) β Answer: B
\(\int e^u \cdot 2\,du = 2e^u + C = \mathbf{2e^{\sqrt{x}} + C}\) β Answer: B
Q14 Β· DEFINITE INTEGRAL
β‘ Tricky
\(\displaystyle\int_0^{\pi/2} \sin x \cos x\,dx\)
π‘ Two approaches: (1) use identity \(\sin x\cos x = \frac{1}{2}\sin 2x\), OR (2) u-sub with \(u=\sin x\).
π Full Solution
Method 1: \(\int_0^{\pi/2}\frac{1}{2}\sin 2x\,dx = \left[-\frac{1}{4}\cos 2x\right]_0^{\pi/2}\)
\(= -\frac{1}{4}\cos\pi - (-\frac{1}{4}\cos 0) = \frac{1}{4} + (-\frac{1}{4}) = \frac{1}{4}+\frac{1}{4}\)
\(= -\frac{1}{4}(-1) + \frac{1}{4}(1) = \frac{1}{4}+\frac{1}{4} = \mathbf{\dfrac{1}{2}}\)
Wait β let's recalculate: \(-\frac{1}{4}[\cos\pi - \cos 0] = -\frac{1}{4}[-1-1] = -\frac{1}{4}(-2) = \frac{1}{2}\)
β Answer: C
\(= -\frac{1}{4}\cos\pi - (-\frac{1}{4}\cos 0) = \frac{1}{4} + (-\frac{1}{4}) = \frac{1}{4}+\frac{1}{4}\)
\(= -\frac{1}{4}(-1) + \frac{1}{4}(1) = \frac{1}{4}+\frac{1}{4} = \mathbf{\dfrac{1}{2}}\)
Wait β let's recalculate: \(-\frac{1}{4}[\cos\pi - \cos 0] = -\frac{1}{4}[-1-1] = -\frac{1}{4}(-2) = \frac{1}{2}\)
β Answer: C
Q15 Β· TRICKY INTEGRAL
β
β
β
Hard
Which is equal to \(\displaystyle\int_1^4 \frac{1}{2\sqrt{x}}\,dx\)?
π‘ Recognize: \(\dfrac{d}{dx}[\sqrt{x}] = \dfrac{1}{2\sqrt{x}}\). So this is asking for the antiderivative of a derivative you already know!
π Full Solution
\(\int_1^4 \frac{1}{2\sqrt{x}}\,dx = \left[\sqrt{x}\right]_1^4 = \sqrt{4} - \sqrt{1} = 2 - 1 = \mathbf{1}\)
β Answer: A
β Answer: A
Q16 Β· AREA BETWEEN CURVES
β‘ Tricky
Find the area enclosed between \(f(x) = x^2\) and \(g(x) = x + 2\).
π‘ Step 1: Find intersections. Step 2: Integrate [top β bottom]. Which function is on top?
π Full Solution
Intersect: \(x^2 = x+2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0\), so \(x=-1, 2\)
On \([-1,2]\): \(g(x) = x+2 \ge x^2 = f(x)\) β
\(A = \int_{-1}^2 (x+2-x^2)\,dx = \left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}\)
\(= (2+4-\frac{8}{3}) - (\frac{1}{2}-2+\frac{1}{3}) = \frac{10}{3} - (-\frac{7}{6}) = \frac{20}{6}+\frac{7}{6} = \dfrac{27}{6} = \dfrac{9}{2}\)
β Answer: B
On \([-1,2]\): \(g(x) = x+2 \ge x^2 = f(x)\) β
\(A = \int_{-1}^2 (x+2-x^2)\,dx = \left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}\)
\(= (2+4-\frac{8}{3}) - (\frac{1}{2}-2+\frac{1}{3}) = \frac{10}{3} - (-\frac{7}{6}) = \frac{20}{6}+\frac{7}{6} = \dfrac{27}{6} = \dfrac{9}{2}\)
β Answer: B
Unit 6β8
FTC, Accumulation & Differential Equations
FTC Part 1: d/dx[β«βΛ£ f(t)dt] = f(x) β "upper limit plugged into integrand"
FTC Part 2: β«βα΅ f(x)dx = F(b) β F(a) where F'=f
CHAIN+FTC: d/dx[β«βα΅β½Λ£βΎf(t)dt] = f(g(x))Β·g'(x) β DON'T forget g'(x)!
FTC Part 2: β«βα΅ f(x)dx = F(b) β F(a) where F'=f
CHAIN+FTC: d/dx[β«βα΅β½Λ£βΎf(t)dt] = f(g(x))Β·g'(x) β DON'T forget g'(x)!
Q17 Β· FTC PART 1
β
β
β
Hard
\(\displaystyle\frac{d}{dx}\int_1^{x^3} \sin(t^2)\,dt\)
π‘ Upper limit is NOT just \(x\) β it's \(x^3\). Chain Rule applies!
π Full Solution
FTC1 + Chain Rule: \(f(g(x))\cdot g'(x)\) where \(g(x)=x^3\), \(f(t)=\sin(t^2)\)
\(= \sin\!\left((x^3)^2\right)\cdot 3x^2 = \sin(x^6)\cdot 3x^2 = \mathbf{3x^2\sin(x^6)}\)
β Answer: C
Trap: Option D plugs xΒ³ without squaring inside sin. Option A forgets chain rule entirely.
\(= \sin\!\left((x^3)^2\right)\cdot 3x^2 = \sin(x^6)\cdot 3x^2 = \mathbf{3x^2\sin(x^6)}\)
β Answer: C
Trap: Option D plugs xΒ³ without squaring inside sin. Option A forgets chain rule entirely.
Q18 Β· ACCUMULATION
β‘ Tricky
The graph of \(f'\) (the derivative of \(f\)) is shown below (described as):
\(f'(x) > 0\) on \((-2, 1)\), \(f'(x) < 0\) on \((1, 4)\), \(f'(x) = 0\) at \(x=-2,\,1,\,4\).
Also: \(f''(x) > 0\) on \((-2, -0.5)\), \(f''(x) < 0\) on \((-0.5, 2.5)\), \(f''(x) > 0\) on \((2.5,4)\).
\(f\) has an inflection point at:
Also: \(f''(x) > 0\) on \((-2, -0.5)\), \(f''(x) < 0\) on \((-0.5, 2.5)\), \(f''(x) > 0\) on \((2.5,4)\).
\(f\) has an inflection point at:
π Full Solution
Inflection points of \(f\) occur where \(f''\) changes sign β NOT where \(f'=0\).
\(f''\) changes: positiveβnegative at \(x=-0.5\), negativeβpositive at \(x=2.5\).
So inflection points of \(f\): \(x = -0.5\) and \(x = 2.5\) β Answer: B
Trap: D lists zeros of f' (those are local max/min of f, NOT inflection points!)
\(f''\) changes: positiveβnegative at \(x=-0.5\), negativeβpositive at \(x=2.5\).
So inflection points of \(f\): \(x = -0.5\) and \(x = 2.5\) β Answer: B
Trap: D lists zeros of f' (those are local max/min of f, NOT inflection points!)
Q19 Β· DIFFERENTIAL EQUATIONS
β
β
β
Hard
Solve the differential equation \(\dfrac{dy}{dx} = \dfrac{x}{y}\) with initial condition \(y(0)=3\).
π‘ Separate variables: \(y\,dy = x\,dx\). Then integrate both sides.
π Full Solution
Separate: \(y\,dy = x\,dx\)
Integrate: \(\dfrac{y^2}{2} = \dfrac{x^2}{2} + C\)
Initial condition \(y(0)=3\): \(\dfrac{9}{2} = 0 + C\), so \(C = \dfrac{9}{2}\)
\(y^2 = x^2 + 9\), and since \(y(0)=3>0\): \(y = \sqrt{x^2+9}\) β Answer: A
Note: C is technically also correct as an implicit solution β but A is the explicit form with positive root, which AP prefers.
Integrate: \(\dfrac{y^2}{2} = \dfrac{x^2}{2} + C\)
Initial condition \(y(0)=3\): \(\dfrac{9}{2} = 0 + C\), so \(C = \dfrac{9}{2}\)
\(y^2 = x^2 + 9\), and since \(y(0)=3>0\): \(y = \sqrt{x^2+9}\) β Answer: A
Note: C is technically also correct as an implicit solution β but A is the explicit form with positive root, which AP prefers.
Q20 Β· RIEMANN SUM
β
β
β
Hard
Which limit equals \(\displaystyle\int_0^1 x^2\,dx\)?
π‘ Riemann sum definition: \(\lim_{n\to\infty}\sum_{i=1}^n f\!\left(\tfrac{i}{n}\right)\cdot\tfrac{1}{n}\) on \([0,1]\).
π Full Solution
Right Riemann sum on \([0,1]\) with \(n\) subintervals:
\(\Delta x = \frac{1}{n}\), \(x_i = \frac{i}{n}\), \(f(x_i) = \left(\frac{i}{n}\right)^2\)
\(\sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \frac{i^2}{n^2}\cdot\frac{1}{n} = \frac{1}{n}\sum_{i=1}^n\frac{i^2}{n^2} = \frac{1}{n^3}\sum_{i=1}^n i^2\)
Option B: \(\frac{1}{n}\sum\frac{i^2}{n^2} = \frac{1}{n^3}\sum i^2\) β
Option C: \(\frac{1}{n^2}\sum i^2\) β missing one factor of \(\frac{1}{n}\) β
Option D: Left Riemann sum (starts at i=0) β also correct as \(n\to\infty\), but option B is the standard right-sum form.
β Answer: B (standard right Riemann sum)
\(\Delta x = \frac{1}{n}\), \(x_i = \frac{i}{n}\), \(f(x_i) = \left(\frac{i}{n}\right)^2\)
\(\sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \frac{i^2}{n^2}\cdot\frac{1}{n} = \frac{1}{n}\sum_{i=1}^n\frac{i^2}{n^2} = \frac{1}{n^3}\sum_{i=1}^n i^2\)
Option B: \(\frac{1}{n}\sum\frac{i^2}{n^2} = \frac{1}{n^3}\sum i^2\) β
Option C: \(\frac{1}{n^2}\sum i^2\) β missing one factor of \(\frac{1}{n}\) β
Option D: Left Riemann sum (starts at i=0) β also correct as \(n\to\infty\), but option B is the standard right-sum form.
β Answer: B (standard right Riemann sum)
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