AP Unit 6 Solutions
Worked solutions to all thirty Unit 6 problems. Each solution gives the mathematical steps and the reason for using them. Compare your setup and justification as well as the final answer.
Problem 1
Worked-example retrieval. Reconstruct the lesson reasoning before checking the solution.
- Positive signed area is ½·2·3=3 m; negative signed area is −½·2·2=−2 m. Reason: Rate times time gives meters, with sign from the graph.
- Add 3+(−2)=1 for displacement and 3+2=5 for distance. Reason: Displacement is net change; distance integrates speed.
Answer: Displacement = 1 m; distance = 5 m.
Final answer
Displacement = 1 m; distance = 5 m.
Review the teaching: Topic 6.1
Problem 2
A tank contains 40 L at t=0. Its net filling rate is 3 L/min for 0≤t≤4 and −2 L/min for 4<t≤7. How much water is in the tank at t=7?
- Net change is 3·4−2·3=6 L. Reason: Constant-rate intervals form signed rectangles.
- Add 40+6=46 L. Reason: The integral gives change, so include the initial amount.
Answer: 46 L.
Final answer
46 L.
Review the teaching: Topic 6.1
Problem 3
| x | 0 | 2 | 5 |
|---|---|---|---|
| f(x) | 1 | 5 | 9 |
- Widths are 2 and 3. Reason: The partition is nonuniform.
- Left=1·2+5·3=17; right=5·2+9·3=37. Reason: Use the matching left or right value on each interval.
- Trapezoidal=(1+5)·2/2+(5+9)·3/2=27. Reason: Each trapezoid uses average endpoint height weighted by its width; the plain mean would give 25, a different procedure.
Answer: Left = 17, right = 37, trapezoidal = 27. A plain mean weights the samples equally rather than using interval widths.
Final answer
Left = 17, right = 37, trapezoidal = 27. A plain mean weights the samples equally rather than using interval widths.
Review the teaching: Topic 6.2
Problem 4
For f(x)=x² on [0,2], use two equal subintervals to find the midpoint approximation. Is it an underestimate or overestimate?
- Width=1 and midpoints are 1/2 and 3/2. Reason: The midpoint rule samples the center of each interval.
- M=1·[(1/2)²+(3/2)²]=5/2. Reason: Add height times width.
- Since f″=2>0, M underestimates. Reason: For concave-up graphs, midpoint rectangles underestimate the integral.
Answer: Midpoint = 5/2; an underestimate.
Final answer
Midpoint = 5/2; an underestimate.
Review the teaching: Topic 6.2
Problem 5
A continuous f is strictly decreasing on [1,5]. Compare left and right Riemann sums using the same partition with ∫₁⁵ f.
- On each interval, f(right)<f(x)<f(left) for interior x. Reason: Strict decrease determines endpoint heights.
- Multiply by positive widths and add. Reason: Integration preserves these comparisons; the claim does not require f to be positive.
Answer: Right sum < integral < left sum.
Final answer
Right sum < integral < left sum.
Review the teaching: Topic 6.2
Problem 6
Worked-example retrieval. Reconstruct the lesson reasoning before checking the solution.
Write limₙ→∞ Σᵢ₌₁ⁿ (2/n)(1+2i/n)² as a definite integral and evaluate it.
- Identify Δx=2/n and right sample xᵢ=1+2i/n. Reason: These correspond to n equal pieces of [1,3].
- Evaluate [x³/3]₁³=(27−1)/3=26/3. Reason: Continuity gives the Riemann-sum limit and the Fundamental Theorem evaluates it.
Answer: ∫₁³ x² dx = 26/3.
Final answer
∫₁³ x² dx = 26/3.
Review the teaching: Topic 6.3
Problem 7
Write ∫₂⁵ cos x dx as a limit of right Riemann sums with n equal subintervals.
- Set Δx=(5−2)/n=3/n and xᵢ=2+3i/n. Reason: These are the uniform widths and right endpoints.
- Sum cos(xᵢ)Δx and take the limit. Reason: The continuous integral is the limiting Riemann sum.
Answer: limₙ→∞ Σᵢ₌₁ⁿ cos(2+3i/n)(3/n).
Final answer
limₙ→∞ Σᵢ₌₁ⁿ cos(2+3i/n)(3/n).
Review the teaching: Topic 6.3
Problem 8
For A(x)=∫₁^(sin x) √(1+t²)dt, find A′(x). Do you need to evaluate the original integral?
- Evaluate the continuous integrand at sin x, giving √(1+sin²x). Reason: The Fundamental Theorem differentiates with respect to the moving endpoint.
- Multiply by cos x. Reason: The upper endpoint sin x requires the chain rule; the theorem works without evaluating the original integral.
Answer: A′(x)=cos x · √(1+sin²x). No antiderivative calculation is needed.
Final answer
A′(x)=cos x · √(1+sin²x). No antiderivative calculation is needed.
Review the teaching: Topic 6.4
Problem 9
For B(x)=∫ₓ³ e^(−t²)dt, find B′(x).
- Rewrite B(x)=−∫₃ˣ e^(−t²)dt. Reason: Reversing the bounds changes the sign.
- Differentiate to get −e^(−x²). Reason: The Fundamental Theorem applies to the continuous integrand.
Answer: −e^(−x²).
Final answer
−e^(−x²).
Review the teaching: Topic 6.4
Problem 10
For H(x)=∫ₓˣ² (t+1)dt, find H′(x).
- The upper contribution is (x²+1)·2x. Reason: Apply the Fundamental Theorem and chain rule at the upper bound.
- Subtract the lower contribution (x+1)·1. Reason: Increasing the lower endpoint removes accumulated area.
Answer: 2x(x²+1)−(x+1).
Final answer
2x(x²+1)−(x+1).
Review the teaching: Topic 6.4
Problem 11
Let G(x)=∫₀ˣ (t−1)(t−3)dt on [0,4]. Find its absolute maximum value and every point where it occurs.
- G′=(x−1)(x−3), so the critical points are 1 and 3; G increases, decreases, then increases. Reason: The Fundamental Theorem and factor signs identify the candidate behavior.
- An accumulation formula is G(x)=x³/3−2x²+3x. Reason: Integrate from 0 to x with the starting value zero.
- Compare G(0)=0, G(1)=4/3, G(3)=0, G(4)=4/3. Reason: Absolute extrema on a closed interval require endpoint and critical-point values, and ties can occur.
Answer: Maximum value 4/3 at x=1 and x=4.
Final answer
Maximum value 4/3 at x=1 and x=4.
Review the teaching: Topic 6.5
Problem 12
On an interval, a continuous differentiable f is negative and strictly increasing. Describe G(x)=∫ₐˣ f(t)dt there.
- G′=f<0, so G decreases. Reason: The sign of the derivative controls monotonicity.
- G′=f is increasing, so G is concave up. Reason: Increasing slopes characterize concavity; no positive height is needed.
Answer: G is decreasing and concave up.
Final answer
G is decreasing and concave up.
Review the teaching: Topic 6.5
Problem 13
Given ∫₀² f=3 and ∫₂⁵ f=−1, define G(x)=∫₂ˣ f(t)dt. Find G(0) and G(5).
- G(0)=∫₂⁰ f=−∫₀² f=−3. Reason: Reverse the bounds.
- G(5)=∫₂⁵ f=−1. Reason: Use the supplied signed integral directly.
Answer: G(0)=−3; G(5)=−1.
Final answer
G(0)=−3; G(5)=−1.
Problem 14
Given ∫₀² f=3 and ∫₂⁵ f=−1, evaluate ∫₅⁰ [2f(x)+1]dx.
- ∫₀⁵ f=3−1=2. Reason: Add integrals over adjacent intervals.
- ∫₀⁵ (2f+1)=2·2+5=9. Reason: Use linearity and the integral of the constant 1.
- Reverse the bounds to obtain −9. Reason: Bound reversal changes the sign.
Answer: −9.
Final answer
−9.
Review the teaching: Topic 6.6
Problem 15
Evaluate ∫₋₂² √(4−x²)dx using geometry.
- The graph is the upper semicircle of radius 2. Reason: x²+y²=4 with y≥0 describes that semicircle.
- Its area is ½π·2²=2π. Reason: The integrand is nonnegative, so the integral equals geometric area.
Answer: 2π.
Final answer
2π.
Review the teaching: Topic 6.6
Problem 16
Define f(x)=1 for 0≤x<1, f(1)=100, and f(x)=3 for 1<x≤2. Find ∫₀² f(x)dx.
- Integrate the constant pieces: 1·1+3·1=4. Reason: The bounded function has only a finite jump and is integrable.
- The value 100 at one point adds no area. Reason: A single point has zero width.
Answer: 4.
Final answer
4.
Review the teaching: Topic 6.6
Problem 17
Worked-example retrieval. Reconstruct the lesson reasoning before checking the solution.
Evaluate ∫₀² (3x²−4x)dx.
- Use antiderivative F(x)=x³−2x². Reason: F′ matches the continuous integrand.
- Compute F(2)−F(0)=8−8=0. Reason: The Fundamental Theorem evaluates signed accumulation.
Answer: 0.
Final answer
0.
Review the teaching: Topic 6.7
Problem 18
Evaluate ∫₀^(π/2) cos x dx.
- An antiderivative is sin x. Reason: The derivative of sine is cosine.
- sin(π/2)−sin(0)=1. Reason: Evaluate upper minus lower.
Answer: 1.
Final answer
1.
Review the teaching: Topic 6.7
Problem 19
Find the antiderivative F on x>0 satisfying F′(x)=3/x+cos x and F(1)=2.
- First write F(x)=3ln x+sin x+C. Reason: Linearity and reverse logarithm/cosine rules give the antiderivative family.
- Use F(1)=sin 1+C=2, so C=2−sin 1. Reason: The specified value chooses one member of the family.
- Differentiate to recover 3/x+cos x. Reason: This verifies the antiderivative on its stated positive domain.
Answer: F(x)=3ln x+sin x+2−sin 1, for x>0.
Final answer
F(x)=3ln x+sin x+2−sin 1, for x>0.
Review the teaching: Topic 6.8
Problem 20
Find an antiderivative of sec²(3x) and verify it.
- Reverse the tangent derivative with factor 1/3. Reason: The inside derivative is 3.
- Differentiate: (1/3)sec²(3x)·3=sec²(3x). Reason: The chain rule verifies the coefficient and the domain excludes poles.
Answer: (1/3)tan(3x)+C on an interval where cos(3x)≠0.
Final answer
(1/3)tan(3x)+C on an interval where cos(3x)≠0.
Problem 21
Worked-example retrieval. Reconstruct the lesson reasoning before checking the solution.
Evaluate ∫₀¹ 2x e^(x²)dx.
- Set u=x², du=2x dx; bounds become 0 and 1. Reason: The derivative of the inside matches the factor outside.
- Integrate ∫₀¹ eᵘdu=e−1. Reason: Use the transformed bounds with the u-antiderivative.
Answer: e−1.
Final answer
e−1.
Review the teaching: Topic 6.9
Problem 22
Worked-example retrieval. Reconstruct the lesson reasoning before checking the solution.
Evaluate ∫₀² x/(x²+1)dx.
- Let u=x²+1, du=2x dx; bounds are 1 and 5. Reason: The denominator derivative matches twice the numerator.
- Evaluate ½∫₁⁵ du/u = ½ln 5. Reason: Correct the differential factor and apply the logarithm rule.
Answer: (1/2)ln 5.
Final answer
(1/2)ln 5.
Review the teaching: Topic 6.9
Problem 23
Find ∫sin x cos x dx.
- Let u=sin x, so du=cos x dx. Reason: One trigonometric factor is the derivative of the other.
- Integrate ∫u du=u²/2+C and substitute back. Reason: The power rule applies in u.
Answer: (1/2)sin²x+C.
Final answer
(1/2)sin²x+C.
Review the teaching: Topic 6.9
Problem 24
Evaluate ∫₀² (x²+1)/(x+2)dx. Check that the interval is permitted before integrating.
- The denominator is positive on [0,2]. Divide to obtain x−2+5/(x+2). Reason: Check continuity on the bounds, then use long division to expose basic antiderivatives.
- Use F(x)=x²/2−2x+5ln(x+2). Reason: The rearranged terms use power and logarithm rules.
- F(2)−F(0)=−2+5ln4−5ln2=5ln2−2. Reason: The Fundamental Theorem evaluates the definite integral, and logarithm laws combine the ratio.
Answer: 5ln2−2.
Final answer
5ln2−2.
Review the teaching: Topic 6.10
Problem 25
Worked-example retrieval. Reconstruct the lesson reasoning before checking the solution.
Find ∫dx/(x²+4x+8).
- Complete the square: (x+2)²+4. Reason: This exposes a scaled arctangent form.
- Set u=(x+2)/2, dx=2du, giving ½∫du/(1+u²). Reason: The denominator is 4(1+u²); retain the scale factor.
- Integrate to ½arctan u+C. Reason: Reverse the arctangent derivative.
Answer: (1/2)arctan((x+2)/2)+C.
Final answer
(1/2)arctan((x+2)/2)+C.
Review the teaching: Topic 6.10
Problem 26
The continuous function e^(−x²) has no elementary antiderivative. Does ∫₀¹ e^(−x²)dx exist? Explain how to find a useful approximation.
- Check that e^(−x²) is continuous on [0,1]. Reason: Continuity ensures integrability.
- Choose a partition and a numerical method, or a calculator integral when permitted. Reason: An elementary closed form is not required for a definite integral.
Answer: Yes. Continuity guarantees the integral; a Riemann/trapezoidal sum or numerical integration gives an approximation.
Final answer
Yes. Continuity guarantees the integral; a Riemann/trapezoidal sum or numerical integration gives an approximation.
Review the teaching: Topic 6.14 Topic 6.8
Problem 27
Select and carry out a method for ∫x²/(x+1)dx.
- Use long division: x²/(x+1)=x−1+1/(x+1). Reason: The rational integrand is improper.
- Integrate the three basic pieces. Reason: Simplification exposes power and logarithm rules.
Answer: x²/2−x+ln|x+1|+C, for x≠−1 on an interval.
Final answer
x²/2−x+ln|x+1|+C, for x≠−1 on an interval.
Review the teaching: Topic 6.14 Topic 6.10
Problem 28
Use a calculator’s numerical-integration feature to approximate ∫₀¹ e^(−x²)dx to three decimal places. Write the expression entered.
- Enter the integrand e^(−x²), variable x, lower bound 0, and upper bound 1. Reason: The setup specifies the intended continuous definite integral.
- The numerical value is approximately 0.7468241328, which rounds to 0.747. Reason: Round the final value rather than intermediate data.
Answer: ∫₀¹ e^(−x²)dx ≈ 0.747.
Final answer
∫₀¹ e^(−x²)dx ≈ 0.747.
Review the teaching: Topic 6.2 Topic 6.14
Problem 29
| t (s) | 0 | 2 | 5 |
|---|---|---|---|
| r(t) (m/s) | 1 | 3 | 4 |
- T=½(1+3)·2+½(3+4)·3=4+10.5=14.5 m. Reason: Use both interval widths and average endpoint rates.
- No concavity between samples is supplied. Reason: Trapezoid error direction needs a justified concavity condition, not just endpoint values.
Answer: 14.5 m; the table alone does not establish an error direction.
Final answer
14.5 m; the table alone does not establish an error direction.
Review the teaching: Topic 6.2
Problem 30
Let F(x)=∫₁ˣ (t²+2t)dt. Find F′(x), F(1), and an explicit formula for F(x).
- Differentiate by the Fundamental Theorem to get F′=x²+2x. Reason: The integrand is continuous.
- Evaluate [t³/3+t²]₁ˣ=x³/3+x²−4/3. Reason: The Fundamental Theorem gives a formula for the accumulation function.
- At x=1, the expression is zero. Reason: An accumulation from 1 to itself is zero.
Answer: F′(x)=x²+2x; F(1)=0; F(x)=x³/3+x²−4/3.
Final answer
F′(x)=x²+2x; F(1)=0; F(x)=x³/3+x²−4/3.
Explain and apply: worked solutions
Compare the setup and the reasoning, as well as the final answer. Use the rubric to find your next step; equivalent correct reasoning earns the same credit.
Estimate a supplied curved graph and qualify the result
The supplied graph of f is continuous, strictly decreasing and strictly concave down throughout [0,4]. The accessible readings, rounded to the nearest 0.1, are: x = 0,0.5,1,1.5,2,2.5,3,3.5,4; f(x) ≈ 4.0,4.0,3.9,3.7,3.5,3.2,2.9,2.5,2.0 in that order. Each displayed height has at most 0.05 reading error. No formula for this curve is supplied.
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Diagnose an estimate from a verbal rate description
A continuous flow rate is positive and strictly concave up on the entire time interval [0,2] minutes. Its exact values at t = 0,1,2 are 1,2,4 liters per minute. Use one trapezoid on each of [0,1] and [1,2].
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Build an accumulation function from an integrand graph
The exact graph of the continuous net inflow rate f is made of straight segments joining (0,2), (2,−2), (4,0) and (6,2), in that order. Time is measured in minutes and f in liters per minute. A tank contains 10 liters at t = 0. Define G to mean net change since time 0, and A to mean the amount in the tank.
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The diagram keeps its full size here. Focus the view and use the arrow keys, or scroll across it, to inspect every label.
Inspect diagram labels
The diagram keeps its full size here. Focus the view and use the arrow keys, or scroll across it, to inspect every label.
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Accumulate a nonuniform rate table under a stated model
A continuous rate is sampled at the exact pairs (t,f(t)) = (0,2), (1,4), (3,−2), (4,0), with time in minutes and rate in liters per minute. For parts a–c, explicitly assume linear interpolation between each pair of consecutive rows; call that rate model f_PL. The net change is measured from time 0.
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Transfer accumulation reasoning to a formula and words
For parts a–b, h(t) = t² − 2t on [0,3] and net accumulation is measured from t = 1. For parts c–d, r is continuous, strictly negative and strictly increasing on the entire interval [2,5]; its numerical values are not supplied.
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Diagnose what raw rate samples establish
A continuous rate f has exact samples f(0) = 2, f(1) = 4 and f(2) = 6. No interpolation, monotonicity or concavity assumption is supplied. Let G(x) = ∫₀ˣ f(t) dt.
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