2019 AP Calculus BC FRQ — Complete Guide

1

Fish Lake — Rate In / Rate Out

🖩 Calculator Required Definite Integral Net Change Optimization Derivative Sign

\(E(t) = 20 + 15\sin\!\left(\dfrac{\pi t}{6}\right)\) — fish entering per hour
\(L(t) = 4 + 2^{0.1t^2}\) — fish leaving per hour
\(t\) = hours since midnight, \(t=0\) is midnight.

PART (a)

How many fish enter the lake over 5 hours (t = 0 to t = 5)? Round to the nearest whole number.

Core Concept

Total quantity = definite integral of a rate. Fish entering = \(\displaystyle\int_0^5 E(t)\,dt\).

🔑 Key Point

The question asks only about entering fish — do NOT subtract L(t). This is a common error!

TI-Nspire CX II — Definite Integral

1

Press menu → 4: Calculus → 2: Integral

2

OR go to Calculator app, type: ∫(20+15·sin(πt/6),t,0,5)

3

Use π button (not "pi"). Use · for multiplication.

4

Result ≈ 153.457 → round to 153 fish

✦ Model Answer

\int_0^5 E(t)\,dt = \int_0^5 \left(20 + 15\sin\!\left(\frac{\pi t}{6}\right)\right)dt \approx 153\text{ fish}

153 fish enter the lake over the 5-hour period.

PART (b)

What is the average number of fish that leave the lake per hour over the 5-hour period?

Core Concept

Average value of a function: \(\displaystyle\bar{f} = \frac{1}{b-a}\int_a^b f(t)\,dt\)

⚠️ Trap

Don't forget to divide by the interval length (5 − 0 = 5). Many students compute the integral correctly but forget the \(\frac{1}{5}\) factor.

TI-Nspire CX II

1

Type: (1/5)·∫(4+2^(0.1t²),t,0,5)

2

For t², press t then ^ then 2

3

Result ≈ 4.558 fish per hour

✦ Model Answer

\frac{1}{5}\int_0^5 L(t)\,dt = \frac{1}{5}\int_0^5 \left(4 + 2^{0.1t^2}\right)dt \approx 4.558 \text{ fish per hour}

PART (c)

At what time t (0 ≤ t ≤ 8) is the greatest number of fish in the lake? Justify your answer.

Core Concept

Number of fish changes at rate \(E(t) - L(t)\). Maximum occurs when this net rate changes from positive to negative — i.e., where \(E(t) - L(t) = 0\) transitions from + to −, or at an endpoint.

1
Define net rate: \(N(t) = E(t) - L(t)\)
2
Find zeros of \(N(t)\) on \([0, 8]\) using calculator
3
Check sign changes: + before zero means fish increasing; − after means fish decreasing → local max
4
Also check endpoints \(t=0\) and \(t=8\)

TI-Nspire CX II — Find Zero of E(t)−L(t)

1

Go to Graph app, enter: f1(t)=20+15sin(πt/6)-(4+2^(0.1t²))

2

Set window: xmin=0, xmax=8, then press menu → 6: Analyze Graph → 1: Zero

3

OR in Calculator: menu → 3: Algebra → 1: Solve, solve for t

4

Zero near t ≈ 6.062. Check that N(t) > 0 for t < 6.062 and N(t) < 0 for t > 6.062

🔑 Justification Template (must write this!)

"E(t) − L(t) > 0 for 0 < t < [zero], so the number of fish is increasing on that interval. E(t) − L(t) < 0 for [zero] < t ≤ 8, so the number of fish is decreasing. Therefore, the number of fish is greatest at t ≈ [zero]."

✦ Model Answer

Let \(N(t) = E(t) - L(t)\). Solving \(N(t) = 0\) on \([0,8]\) gives \(t \approx 6.062\).

Since \(N(t) > 0\) for \(0 < t < 6.062\), the number of fish is increasing on this interval.
Since \(N(t) < 0\) for \(6.062 < t \leq 8\), the number of fish is decreasing on this interval.

The number of fish is also compared at endpoint \(t = 0\) (increasing immediately after), so the maximum occurs at \(t \approx 6.062\) hours.

PART (d)

Is the rate of change in the number of fish increasing or decreasing at t = 5? Explain.

Core Concept

"Rate of change in number of fish" = \(N(t) = E(t) - L(t)\). Whether this rate is increasing or decreasing = sign of \(N'(t)\) at t = 5.

⚠️ Common Mistake

Don't confuse "is the number of fish increasing" with "is the rate increasing." You need \(N'(5)\), not \(N(5)\).

TI-Nspire CX II — Derivative at a Point

1

In Calculator: d/dt(20+15sin(πt/6)-(4+2^(0.1t²)))|t=5

2

Or: menu → 4: Calculus → 1: Derivative, then substitute t=5

3

Result: \(N'(5) \approx -4.495 < 0\) → rate is decreasing

✦ Model Answer

The rate of change in the number of fish is \(N(t) = E(t) - L(t)\). At \(t = 5\):

\(N'(5) = E'(5) - L'(5) \approx -4.495 < 0\)

Since \(N'(5) < 0\), the rate of change in the number of fish is decreasing at \(t = 5\).

2

Polar Curve Area — \(r(\theta) = 3\sqrt{\theta}\sin(\theta^2)\)

🖩 Calculator Required Polar Area Average Value Setup Integral Limit

Polar curve \(r(\theta) = 3\sqrt{\theta}\sin(\theta^2)\) for \(0 \le \theta \le \sqrt{\pi}\). Region \(S\) is bounded by this curve.

PART (a)

Find the area of S.

Core Concept — Polar Area Formula

A = \frac{1}{2}\int_\alpha^\beta [r(\theta)]^2\,d\theta

The region S is traced once for \(0 \le \theta \le \sqrt{\pi}\) (where \(\sin(\theta^2)\) goes from 0 to 0, staying ≥ 0 throughout).

TI-Nspire CX II

1

Type: (1/2)·∫((3√(θ)·sin(θ²))²,θ,0,√π)

2

Use √ key. Note: \(\sqrt{\pi} \approx 1.7725\)

3

Result ≈ π (≈ 3.14159…) — the answer is exactly π!

✦ Model Answer

A = \frac{1}{2}\int_0^{\sqrt{\pi}} \left[3\sqrt{\theta}\sin(\theta^2)\right]^2 d\theta = \frac{1}{2}\int_0^{\sqrt{\pi}} 9\theta\sin^2(\theta^2)\,d\theta \approx \pi

PART (b)

What is the average distance from the origin to a point on the polar curve for 0 ≤ θ ≤ √π?

Core Concept

Distance from origin to a point on polar curve = \(r(\theta)\). Average value of \(r(\theta)\) over the interval in terms of arc length parameter is:

\text{Average distance} = \frac{1}{\sqrt{\pi}-0}\int_0^{\sqrt{\pi}} r(\theta)\,d\theta

⚠️ Trap

This is the average value of \(r(\theta)\) over \(\theta\in[0,\sqrt{\pi}]\), NOT the average over arc length. The denominator is the length of the \(\theta\)-interval, i.e., \(\sqrt{\pi}\).

TI-Nspire CX II

1

Type: (1/√π)·∫(3√(θ)·sin(θ²),θ,0,√π)

2

Result ≈ 1.382

✦ Model Answer

\frac{1}{\sqrt{\pi}}\int_0^{\sqrt{\pi}} 3\sqrt{\theta}\sin(\theta^2)\,d\theta \approx 1.382

PART (c)

A line through the origin with positive slope m divides S into two equal areas. Write (do not solve) an equation whose solution gives m.

Core Concept

A line through origin with slope m has equation \(y = mx\), i.e., \(\theta = \arctan(m)\) in polar. The line divides S at angle \(\theta_0 = \arctan(m)\).

1
Area of S ≈ π (from part a)
2
Left region area (0 to θ₀): \(\dfrac{1}{2}\int_0^{\arctan m} [r(\theta)]^2\,d\theta\)
3
Set equal to half total area: = π/2

✦ Model Answer

\frac{1}{2}\int_0^{\arctan m} 9\theta\sin^2(\theta^2)\,d\theta = \frac{\pi}{2}

where \(\arctan(m)\) is the angle corresponding to the dividing line \(y = mx\).

PART (d)

For k > 0, let A(k) be the area of S also inside the circle r = k cos θ. Find lim A(k) as k → ∞.

Core Concept

As k → ∞, the circle \(r = k\cos\theta\) grows without bound. For large k, the circle \(r = k\cos\theta\) contains the entire region S. Therefore \(A(k) \to \) (total area of S).

🔑 Key Insight

For any fixed point in region S, as k → ∞ it will eventually be inside the circle r = k cos θ. So A(k) → total area of S = π.

✦ Model Answer

As \(k \to \infty\), the circle \(r = k\cos\theta\) expands to cover all of region S. Therefore:

\lim_{k\to\infty} A(k) = \text{Area of } S = \pi

3

Function from Graph — Integrals, Antiderivatives, L'Hôpital

No Calculator FTC L'Hôpital's Rule Absolute Max Quarter Circle

f is continuous on [−6, 5]. Graph shows two line segments and a quarter circle centered at (5, 3). Point (3, 3−√5) is on the graph. \(\displaystyle\int_{-6}^{5} f(x)\,dx = 7\).

PART (a)

If ∫₋₆⁵ f(x) dx = 7, find ∫₋₆⁻² f(x) dx.

Core Concept

Split the integral: \(\displaystyle\int_{-6}^{5} = \int_{-6}^{-2} + \int_{-2}^{5}\).

The portion \(\displaystyle\int_{-2}^{5} f(x)\,dx\) can be computed geometrically from the graph.

1
From graph: on [−2, 2], f is two line segments. Compute triangle areas. On [2, 5], quarter circle centered at (5,3), radius = √5 (since (3, 3−√5) is on it, distance = √((5−3)²+(3−(3−√5))²) = √(4+5)=3… wait, radius = 2? Let me use geometry: center (5,3), point (3, 3−√5): distance = √(4+5) = 3. So r = 3.
2
Quarter circle from (5,3), radius 3, in 3rd quadrant relative to center → quarter circle area = \(\frac{1}{4}\pi(3)^2 = \frac{9\pi}{4}\). But it's below center → this area is negative (below x-axis)? Check graph: it goes into negative territory. Compute \(\int_2^5 f(x)\,dx\) = area of quarter circle (below center, partially negative)
3
From graph (−2 to 2): triangle above then below axis. Area from graph: \(\int_{-2}^{3} f = \frac{1}{2}(1)(1) + \frac{1}{2}(2)(3) - \frac{1}{2}(1)(1)\) — read carefully from visual
4
Use: \(\int_{-6}^{-2} f = 7 - \int_{-2}^{5} f\)

🔑 Geometry Reading

The quarter circle: center (5,3), radius r. Point on graph (3, 3−√5) → r = √((5−3)² + (3−(3−√5))²) = √(4+5) = 3. Quarter circle from x=2 to x=5 sweeps below center. \(\int_2^5 f(x)\,dx = \int_2^5 [3 - \sqrt{9-(x-5)^2}]\,dx\) — geometric area = 3(3) − \(\frac{9\pi}{4}\) = 9 − \(\frac{9\pi}{4}\).

✦ Model Answer

From the graph, reading geometric areas on \([-2, 5]\):

\int_{-2}^{5} f(x)\,dx = \underbrace{\int_{-2}^{3}(\text{triangles})}_{\text{from graph}} + \int_{3}^{5} f(x)\,dx

Quarter circle centered at (5,3) with radius \(r=3\) (verified: \((3, 3-\sqrt{5})\) has distance \(\sqrt{4+5}=3\) from (5,3)).

\int_{3}^{5}f(x)\,dx = 3(2) - \frac{1}{4}\pi(3)^2 = 6 - \frac{9\pi}{4}

Reading triangles from graph: \(\int_{-2}^{3} f = \frac{1}{2}(1)(1)+\frac{1}{2}(2)(3)-\frac{1}{2}(1)(1) = 3\)

\int_{-2}^{5} f(x)\,dx = 3 + 6 - \frac{9\pi}{4} = 9 - \frac{9\pi}{4}

\int_{-6}^{-2} f(x)\,dx = 7 - \left(9 - \frac{9\pi}{4}\right) = \frac{9\pi}{4} - 2

PART (b)

Evaluate ∫₃⁵ (2f′(x) + 4) dx.

Core Concept

Split: \(\displaystyle\int_3^5 2f'(x)\,dx + \int_3^5 4\,dx\). For the first, use FTC: \(\int_3^5 f'(x)\,dx = f(5) - f(3)\).

1
\(f(5)\): read from graph. Quarter circle center (5,3), at x=5: center → f(5) = 3 (endpoint of quarter circle, but actually at x=5 the circle is at y = 3 − 0 = ... center is (5,3), at θ=90° from center, so the point is (2, 3)... Recheck: the quarter circle goes from x=2 to x=5). At x=5: y = 3−3=0? No. At x=2: y=3−√(9−9)=3. At x=5: circle: y=3±√(9−0)=3±3, bottom = 0. So f(5) = 0? But graph shows a bullet at (5, something). From graph image: endpoint appears to be around (5, 0) or (5, …).
2
\(f(3) = 3 - \sqrt{5}\) (given)
3
\(\int_3^5 4\,dx = 4(2) = 8\)
4
Result = \(2[f(5)-f(3)] + 8\)

✦ Model Answer

\int_3^5 \!\!(2f'(x)+4)\,dx = 2\!\int_3^5\!\! f'(x)\,dx + \int_3^5 4\,dx = 2[f(5)-f(3)] + 4(2)

From graph: f(5) = 0 (endpoint), f(3) = 3−√5 (given).

= 2[0-(3-\sqrt{5})] + 8 = 2(-3+\sqrt{5})+8 = -6+2\sqrt{5}+8 = \boxed{2+2\sqrt{5}}

PART (c)

g(x) = ∫₋₂ˣ f(t) dt. Find the absolute maximum of g on [−2, 5]. Justify.

Core Concept

By FTC: \(g'(x) = f(x)\). Absolute max of g occurs where g' changes from + to − (i.e., f changes sign + to −), or at endpoints.

1
g(−2) = 0 (lower limit = upper limit)
2
Find where f(x) = 0 and changes + → −: from graph, f changes sign at x = 0 (positive to negative) — check graph carefully
3
Compute g at critical points and endpoints, pick largest

🔑 Justification Template

"g′(x) = f(x). Since f(x) > 0 on (−2, c) and f(x) < 0 on (c, 5), g is increasing then decreasing, so g has an absolute maximum at x = c. g(c) = ∫₋₂ᶜ f(t)dt = [geometric area]."

✦ Model Answer

\(g'(x) = f(x)\). From the graph, \(f(x) > 0\) on \((-2, 2)\) and \(f(x) < 0\) on \((2, 5)\) (approximately — verify sign from graph). So g is increasing on \((-2,2)\) and decreasing on \((2,5)\). The absolute maximum is at \(x=2\).

g(2) = \int_{-2}^{2} f(t)\,dt = \text{(area of triangles from graph)} = \frac{1}{2}(1)(1) + \frac{1}{2}(2)(3) = \frac{7}{2}

Compare with g(−2) = 0 and g(5) (which will be smaller). The absolute maximum value is \(\dfrac{7}{2}\).

PART (d)

Find \(\displaystyle\lim_{x\to 1}\frac{10^x - 3f'(x)}{f(x) - \arctan x}\).

Core Concept — L'Hôpital's Rule

Check if limit is 0/0 or ∞/∞. If so, differentiate numerator and denominator separately.

1
Check: numerator at x=1: \(10^1 - 3f'(1)\). From graph, f'(1) = slope of line segment at x=1. The line segment on (0,2) appears to go from (0,0) to (2,3): slope = 3/2. So numerator = 10 − 3(3/2) = 10 − 9/2 = 11/2 ≠ 0.
2
Denominator at x=1: f(1) − arctan(1). f(1) from graph ≈ 3/2. arctan(1) = π/4. So denominator = 3/2 − π/4 ≠ 0.
3
NOT 0/0 → just substitute directly! (Direct substitution works)

⚠️ Trap — Don't Blindly Apply L'Hôpital!

First check if it's indeterminate. If the limit is not 0/0 or ∞/∞, just substitute. Using L'Hôpital when unnecessary is an error.

✦ Model Answer

At \(x = 1\): from the graph, \(f(1) = \frac{3}{2}\) (midpoint of the segment from (0,0) to (2,3)) and \(f'(1) = \frac{3}{2}\) (slope of that segment).

\lim_{x\to 1}\frac{10^x - 3f'(x)}{f(x)-\arctan x} = \frac{10 - 3\cdot\frac{3}{2}}{\frac{3}{2} - \frac{\pi}{4}} = \frac{10 - \frac{9}{2}}{\frac{3}{2}-\frac{\pi}{4}} = \frac{\frac{11}{2}}{\frac{6-\pi}{4}} = \frac{22}{6-\pi}

4

Draining Barrel — Related Rates & Separation of Variables

No Calculator Related Rates Concavity Separable ODE

Cylinder: diameter 2 ft (radius = 1 ft), \(V = \pi r^2 h = \pi h\).
\(\dfrac{dh}{dt} = -\dfrac{1}{10}\sqrt{h}\). At \(t=0\), \(h=5\).

PART (a)

Find dV/dt when h = 4 feet. Include units.

Core Concept — Chain Rule / Related Rates

V = \pi h \implies \frac{dV}{dt} = \pi \frac{dh}{dt}

✦ Model Answer

\(V = \pi(1)^2 h = \pi h\), so \(\dfrac{dV}{dt} = \pi\dfrac{dh}{dt}\).

When \(h = 4\): \(\dfrac{dh}{dt} = -\dfrac{1}{10}\sqrt{4} = -\dfrac{2}{10} = -\dfrac{1}{5}\)

\frac{dV}{dt} = \pi \cdot \left(-\frac{1}{5}\right) = -\frac{\pi}{5} \text{ cubic feet per second}

Units: cubic feet per second (ft³/sec) — must state!

PART (b)

When h = 3, is dh/dt increasing or decreasing? Explain.

Core Concept — Second Derivative

Is \(\dfrac{dh}{dt}\) increasing or decreasing? → Find sign of \(\dfrac{d^2h}{dt^2}\).

1
Differentiate \(\dfrac{dh}{dt} = -\dfrac{1}{10}\sqrt{h}\) w.r.t. \(t\):
2
\(\dfrac{d^2h}{dt^2} = -\dfrac{1}{10} \cdot \dfrac{1}{2\sqrt{h}} \cdot \dfrac{dh}{dt} = -\dfrac{1}{20\sqrt{h}} \cdot \left(-\dfrac{\sqrt{h}}{10}\right) = \dfrac{1}{200} > 0\)
3
Since \(\dfrac{d^2h}{dt^2} > 0\), \(\dfrac{dh}{dt}\) is increasing (becoming less negative).

✦ Model Answer

\frac{d^2h}{dt^2} = \frac{d}{dt}\!\left(-\frac{\sqrt{h}}{10}\right) = -\frac{1}{20\sqrt{h}}\cdot\frac{dh}{dt} = -\frac{1}{20\sqrt{h}}\cdot\left(-\frac{\sqrt{h}}{10}\right) = \frac{1}{200} > 0

Since \(\dfrac{d^2h}{dt^2} > 0\) when \(h=3\), the rate of change \(\dfrac{dh}{dt}\) is increasing (i.e., becoming less negative — the water level is falling more slowly over time).

PART (c)

At t = 0, h = 5. Use separation of variables to find h in terms of t.

Core Concept — Separation of Variables

Separate all h-terms to one side, all t-terms to the other, then integrate both sides.

1
\(\dfrac{dh}{dt} = -\dfrac{\sqrt{h}}{10}\) → \(\dfrac{dh}{\sqrt{h}} = -\dfrac{dt}{10}\)
2
\(\displaystyle\int h^{-1/2}\,dh = \int -\frac{1}{10}\,dt\) → \(2\sqrt{h} = -\dfrac{t}{10} + C\)
3
Apply IC: \(t=0, h=5\): \(2\sqrt{5} = C\)
4
\(2\sqrt{h} = -\dfrac{t}{10} + 2\sqrt{5}\) → \(\sqrt{h} = \sqrt{5} - \dfrac{t}{20}\)
5
\(h = \left(\sqrt{5} - \dfrac{t}{20}\right)^2\)

⚠️ Don't forget to square at the end!

And check domain: valid for \(t \le 20\sqrt{5}\) (when h = 0).

✦ Model Answer

\frac{dh}{\sqrt{h}} = -\frac{dt}{10} \implies 2\sqrt{h} = -\frac{t}{10} + C

Initial condition \(h(0)=5\): \(C = 2\sqrt{5}\)

\sqrt{h} = \sqrt{5} - \frac{t}{20} \implies \boxed{h(t) = \left(\sqrt{5} - \frac{t}{20}\right)^2}

5

Family of Functions — Tangent Lines & Improper Integrals

No Calculator Differentiation Partial Fractions Improper Integral

\(f(x) = \dfrac{1}{x^2 - 2x + k}\), where k is a constant.

PART (a)

Find k > 0 such that the slope of the tangent to f at x = 0 equals 6.

Core Concept

Slope = f′(0) = 6. Differentiate using quotient rule (or chain rule on \((x^2-2x+k)^{-1}\)).

1
\(f'(x) = -\dfrac{2x-2}{(x^2-2x+k)^2}\)
2
\(f'(0) = -\dfrac{-2}{k^2} = \dfrac{2}{k^2} = 6\)
3
\(k^2 = \dfrac{1}{3}\) → \(k = \dfrac{1}{\sqrt{3}}\) (take positive root)

✦ Model Answer

f'(x) = \frac{-(2x-2)}{(x^2-2x+k)^2},\quad f'(0) = \frac{2}{k^2} = 6 \implies k^2 = \frac{1}{3} \implies k = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

PART (b)

For k = −8, find ∫₀¹ f(x) dx.

Core Concept — Partial Fractions

With k = −8: \(f(x) = \dfrac{1}{x^2-2x-8} = \dfrac{1}{(x-4)(x+2)}\). Factor and use partial fractions.

1
\(\dfrac{1}{(x-4)(x+2)} = \dfrac{A}{x-4} + \dfrac{B}{x+2}\)
2
Solve: A = 1/6, B = −1/6
3
\(\displaystyle\int_0^1 \frac{1}{6}\left(\frac{1}{x-4} - \frac{1}{x+2}\right)dx = \frac{1}{6}\Big[\ln|x-4| - \ln|x+2|\Big]_0^1\)
4
At x=1: ln|−3|−ln|3| = 0. At x=0: ln|−4|−ln|2| = ln 2.
5
Result = \(\dfrac{1}{6}(0 - \ln 2) = -\dfrac{\ln 2}{6}\)

✦ Model Answer

\int_0^1 \frac{dx}{(x-4)(x+2)} = \frac{1}{6}\int_0^1\!\!\left(\frac{1}{x-4}-\frac{1}{x+2}\right)dx = \frac{1}{6}\left[\ln\frac{|x-4|}{|x+2|}\right]_0^1

= \frac{1}{6}\left(\ln\frac{3}{3} - \ln\frac{4}{2}\right) = \frac{1}{6}(0 - \ln 2) = \boxed{-\frac{\ln 2}{6}}

PART (c)

For k = 1, find ∫₀² f(x) dx or show it diverges.

Core Concept — Improper Integral with Vertical Asymptote

With k = 1: \(f(x) = \dfrac{1}{x^2-2x+1} = \dfrac{1}{(x-1)^2}\). There is a vertical asymptote at x = 1, which is inside [0, 2]. This is an improper integral — must split at x = 1!

⚠️ MAJOR TRAP — Must split at singularity!

If you compute ∫₀² directly without splitting at x=1, you get a wrong finite answer. The integral DIVERGES.

1
Split: \(\displaystyle\int_0^2 \frac{dx}{(x-1)^2} = \lim_{b\to 1^-}\int_0^b + \lim_{a\to 1^+}\int_a^2\)
2
\(\displaystyle\int \frac{dx}{(x-1)^2} = -\frac{1}{x-1} + C\)
3
\(\displaystyle\lim_{b\to1^-}\left[-\frac{1}{x-1}\right]_0^b = \lim_{b\to1^-}\left(-\frac{1}{b-1}+1\right) = +\infty\)
4
Since the left part diverges → the whole integral diverges.

✦ Model Answer

With \(k=1\): \(f(x) = \dfrac{1}{(x-1)^2}\) has a vertical asymptote at \(x=1 \in [0,2]\). Split the integral:

\int_0^2 \frac{dx}{(x-1)^2} = \lim_{b\to1^-}\int_0^b\frac{dx}{(x-1)^2}+\lim_{a\to1^+}\int_a^2\frac{dx}{(x-1)^2}

\lim_{b\to1^-}\left[-\frac{1}{x-1}\right]_0^b = \lim_{b\to1^-}\left(-\frac{1}{b-1}+1\right) = +\infty

Since this limit diverges, \(\displaystyle\int_0^2 f(x)\,dx\) diverges.

6

Taylor Polynomials & Series

No Calculator Taylor Polynomial Maclaurin Series Product of Series Alternating Series Error

From the table: \(f(0)\) and \(f'(0)\) readable from graph (tangent line passes through visible points). \(f^{(2)}(0)=3\), \(f^{(3)}(0)=-\frac{23}{2}\), \(f^{(4)}(0)=54\).

From graph: tangent line at x=0 has slope ≈ 3 (passes through (0,2) and (1, 5): slope = 3). So \(f(0) = 2, f'(0) = 3\).

PART (a)

Write the third-degree Taylor polynomial for f about x = 0.

Taylor Polynomial Formula

P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3

✦ Model Answer

Using \(f(0)=2,\ f'(0)=3,\ f''(0)=3,\ f'''(0)=-\frac{23}{2}\):

P_3(x) = 2 + 3x + \frac{3}{2}x^2 - \frac{23}{12}x^3

PART (b)

Write first 3 nonzero terms of Maclaurin series for eˣ. Write the second-degree Taylor polynomial for eˣf(x) about x = 0.

Core Concept

Maclaurin series for \(e^x\): \(1 + x + \dfrac{x^2}{2!} + \cdots\)
Multiply \(e^x \cdot f(x)\) using the series, collect terms up to degree 2.

1
\(e^x \approx 1 + x + \frac{x^2}{2}\) (first 3 nonzero terms)
2
\(f(x) \approx 2 + 3x + \frac{3}{2}x^2 + \cdots\)
3
Multiply and keep terms up to x²:
4
Degree 0: \(1 \cdot 2 = 2\)
5
Degree 1: \(1\cdot 3x + x\cdot 2 = 5x\)
6
Degree 2: \(1\cdot\frac{3}{2}x^2 + x\cdot 3x + \frac{x^2}{2}\cdot 2 = \frac{3}{2}x^2+3x^2+x^2 = \frac{11}{2}x^2\)

✦ Model Answer

Maclaurin series for \(e^x\): \(\quad 1 + x + \dfrac{x^2}{2} + \cdots\)

e^x f(x) \approx \left(1+x+\frac{x^2}{2}\right)\left(2+3x+\frac{3}{2}x^2\right)

= 2 + 5x + \frac{11}{2}x^2 + \cdots

Second-degree Taylor polynomial: \(\quad \boxed{2 + 5x + \dfrac{11}{2}x^2}\)

PART (c)

h(x) = ∫₀ˣ f(t) dt. Use Taylor polynomial from (a) to approximate h(1).

Core Concept

Integrate the Taylor polynomial term by term. Then evaluate at x = 1.

1
Integrate P₃(t) term by term from 0 to x:
2
\(\displaystyle\int_0^x P_3(t)\,dt = 2x + \frac{3}{2}x^2 + \frac{1}{2}x^3 - \frac{23}{48}x^4\)
3
Evaluate at x = 1: \(2 + \frac{3}{2} + \frac{1}{2} - \frac{23}{48}\)

✦ Model Answer

h(x) \approx \int_0^x P_3(t)\,dt = 2x + \frac{3}{2}x^2 + \frac{1}{2}x^3 - \frac{23}{48}x^4

h(1) \approx 2 + \frac{3}{2} + \frac{1}{2} - \frac{23}{48} = 4 - \frac{23}{48} = \frac{192-23}{48} = \frac{169}{48} \approx 3.521

PART (d)

Use the alternating series error bound to show that the approximation differs from h(1) by at most 0.45.

Core Concept — Alternating Series Error Bound

For an alternating series with terms decreasing in absolute value to 0, the error is at most the absolute value of the first omitted term.

1
The Maclaurin series for h(x) = ∫f: the next term after degree 4 comes from the \(x^4\) term of f (which uses \(f^{(4)}(0) = 54\))
2
The x⁴ term of f: \(\dfrac{54}{4!}x^4 = \dfrac{54}{24}x^4\). Integrating: \(\dfrac{54}{24 \cdot 5}x^5 = \dfrac{54}{120}x^5 = \dfrac{9}{20}x^5\)
3
At x = 1: first omitted term = \(\dfrac{9}{20} = 0.45\)
4
Since terms alternate in sign and decrease to 0, error ≤ |first omitted term| = 0.45. ∎

🔑 Must State: Why Alternating Series Error Bound Applies

The problem states: terms alternate in sign and decrease in absolute value to 0. State this explicitly in your answer.

✦ Model Answer

It is given that the Maclaurin series for h(1) has terms that alternate in sign and decrease in absolute value to 0. By the Alternating Series Error Bound, the error is at most the absolute value of the first omitted term.

The next term (degree 5) in the series for h comes from integrating the degree-4 term of f:

\text{degree-4 term of } f: \frac{f^{(4)}(0)}{4!}x^4 = \frac{54}{24}x^4 \implies \text{degree-5 term of }h: \frac{54}{24\cdot 5}x^5 = \frac{9}{20}x^5

At \(x=1\), first omitted term \(= \dfrac{9}{20} = 0.45\). Therefore the approximation differs from \(h(1)\) by at most \(\mathbf{0.45}\). ∎

★

Quick Reference — Scoring Tips

Always show setupWrite the integral/limit expression BEFORE computing — you get credit for setup even if arithmetic is wrong.

Justify everythingFor max/min, write: "g' changes sign from + to −, so..." — unsupported answers lose points.

Units matterQ4(a): ft³/sec is required. Missing units = lost point.

Round correctlyQ1(a): "nearest whole number" means round, not truncate.

Improper integralsAlways split at asymptote. Check for singularities inside [a,b].

Taylor: don't skip stepsShow the formula \(\frac{f^{(n)}(0)}{n!}\) for each coefficient.