Free practice final
Calculus 2 practice final (free, with answers)
A typical Calculus 2 final covers techniques of integration, improper integrals, volumes of revolution, and infinite series, from convergence tests through power series and Taylor series. Work each problem on paper before opening the answer, and notice which method or test each one calls for, since choosing it is often the hardest step.
Topics
- Integration by Parts
- Trigonometric Integrals and Substitution
- Partial Fractions
- Improper Integrals
- Volumes of Revolution
- Sequences and Series
- Convergence Tests
- Power Series
- Taylor Series
15 questions
Try each one before you open its answer.
1. Evaluate the indefinite integral ∫ x² eˣ dx.
- A eˣ(x² + 2x + 2) + C
- B (x³/3)eˣ + C
- C eˣ(x² − 2x + 2) + C
- D eˣ(x² − 2x) + C
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Answer: C. eˣ(x² − 2x + 2) + C
Integrate by parts twice, each time differentiating the power of x: ∫ x²eˣ dx = x²eˣ − 2∫ xeˣ dx, and ∫ xeˣ dx = xeˣ − eˣ. Putting these together gives x²eˣ − 2xeˣ + 2eˣ + C = eˣ(x² − 2x + 2) + C, which you can check by differentiating.
2. Evaluate ∫₁^e ln x dx.
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Answer: 1
Take u = ln x and dv = dx, so du = dx/x and v = x, which gives ∫ ln x dx = x ln x − ∫ 1 dx = x ln x − x. Evaluating from 1 to e gives (e − e) − (0 − 1) = 1.
3. Evaluate the indefinite integral ∫ sin³x cos²x dx.
- A −cos³x/3 + cos⁵x/5 + C
- B cos³x/3 − cos⁵x/5 + C
- C −cos³x/3 − cos⁵x/5 + C
- D (sin⁴x/4)(cos³x/3) + C
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Answer: A. −cos³x/3 + cos⁵x/5 + C
Save one factor of sin x, write sin²x = 1 − cos²x, and let u = cos x, so du = −sin x dx. The integral becomes −∫ (1 − u²)u² du = −u³/3 + u⁵/5 + C = −cos³x/3 + cos⁵x/5 + C.
4. Evaluate ∫₀² dx / (4 + x²)^(3/2).
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Answer: √2/8, about 0.177
Let x = 2 tan θ, so dx = 2 sec²θ dθ and (4 + x²)^(3/2) = 8 sec³θ, and the integral becomes (1/4)∫ cos θ dθ = (1/4) sin θ = x / (4√(4 + x²)). From 0 to 2 this is 2/(4√8) − 0 = √2/8 ≈ 0.177.
5. Evaluate the indefinite integral ∫ (x + 5) / ((x − 1)(x + 2)) dx.
- A 2 ln|x − 1| + ln|x + 2| + C
- B −ln|x − 1| + 2 ln|x + 2| + C
- C ln|x² + x − 2| + C
- D 2 ln|x − 1| − ln|x + 2| + C
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Answer: D. 2 ln|x − 1| − ln|x + 2| + C
Write (x + 5)/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2), so x + 5 = A(x + 2) + B(x − 1). Setting x = 1 gives A = 2 and setting x = −2 gives B = −1, so the integral is 2 ln|x − 1| − ln|x + 2| + C.
6. For which values of p does the improper integral ∫₀¹ 1/xᵖ dx converge?
- A p > 1
- B p < 1
- C p ≤ 1
- D p ≥ 1
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Answer: B. p < 1
For p ≠ 1 the antiderivative is x^(1 − p)/(1 − p), which has a finite limit as x → 0⁺ only when 1 − p > 0, and for p = 1 the antiderivative ln x goes to −∞. So the integral converges exactly when p < 1, the opposite of ∫₁^∞ 1/xᵖ dx, which converges only when p > 1.
7. Evaluate ∫₁^∞ 1 / (x(x + 1)) dx.
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Answer: ln 2, about 0.693
Partial fractions give 1/(x(x + 1)) = 1/x − 1/(x + 1), so an antiderivative is ln x − ln(x + 1) = ln(x/(x + 1)). As b → ∞, ln(b/(b + 1)) → ln 1 = 0, so the integral is 0 − ln(1/2) = ln 2.
8. The region under y = √x from x = 0 to x = 4 is rotated about the x-axis. What is the volume of the resulting solid?
- A 16π/3
- B 8π
- C 16π
- D 128π/5
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Answer: B. 8π
By the disk method, V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π · 16/2 = 8π. Forgetting to square the radius gives 16π/3, and rotating about the y-axis instead gives 128π/5.
9. The region enclosed by y = x and y = x² is rotated about the y-axis. Find the volume of the resulting solid.
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Answer: π/6
The curves meet at x = 0 and x = 1, and y = x is on top between them. Using cylindrical shells, V = 2π∫₀¹ x(x − x²) dx = 2π(1/3 − 1/4) = π/6.
10. Find the sum of the series Σ (n=1 to ∞) 2ⁿ / 3^(n+1).
- A 2/3
- B 1
- C 2
- D The series diverges
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Answer: A. 2/3
Rewrite the terms as (1/3)(2/3)ⁿ, a geometric series with ratio 2/3 whose first term, at n = 1, is (1/3)(2/3) = 2/9. Since the ratio is less than 1 in absolute value, the sum is (2/9)/(1 − 2/3) = 2/3, while starting the count at n = 0 by mistake gives 1.
11. Exactly one of these series converges. Which one? (Each sum runs from n = 1 to ∞.)
- A Σ 1/√n
- B Σ n/(n + 1)
- C Σ n/(n² + 1)
- D Σ n²/3ⁿ
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Answer: D. Σ n²/3ⁿ
For n²/3ⁿ, the ratio test gives ((n + 1)/n)² · (1/3) → 1/3, which is less than 1, so that series converges. The others diverge: Σ 1/√n is a p-series with p = 1/2, n/(n + 1) → 1 fails the divergence test, and n/(n² + 1) behaves like 1/n by limit comparison.
12. Which statement best describes the series Σ (n=1 to ∞) (−1)^(n+1) / √n?
- A It converges absolutely
- B It diverges, because Σ 1/√n diverges
- C It converges conditionally
- D It diverges, because its terms do not approach 0
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Answer: C. It converges conditionally
The terms 1/√n decrease and approach 0, so the series converges by the alternating series test. But the series of absolute values, Σ 1/√n, is a divergent p-series with p = 1/2, so the convergence is conditional, not absolute.
13. What is the interval of convergence of the power series Σ (n=1 to ∞) (x − 2)ⁿ / (n · 3ⁿ)?
- A (−1, 5)
- B [−1, 5)
- C (−1, 5]
- D [−1, 5]
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Answer: B. [−1, 5)
The ratio test gives |x − 2|/3 < 1, so the radius of convergence is 3 and the series converges for −1 < x < 5. At x = 5 the series becomes Σ 1/n, the divergent harmonic series, and at x = −1 it becomes Σ (−1)ⁿ/n, which converges by the alternating series test, so the interval is [−1, 5).
14. Which gives the first three nonzero terms of the Maclaurin series for cos(2x)?
- A 1 − x² + x⁴/12
- B 2 − x² + x⁴/12
- C 2x − (4/3)x³ + (4/15)x⁵
- D 1 − 2x² + (2/3)x⁴
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Answer: D. 1 − 2x² + (2/3)x⁴
Substitute u = 2x into cos u = 1 − u²/2 + u⁴/24 − ⋯ to get 1 − (2x)²/2 + (2x)⁴/24 − ⋯ = 1 − 2x² + (2/3)x⁴ − ⋯. Forgetting to raise the 2 to each power gives the wrong 1 − x² + x⁴/12.
15. Find the second-degree Taylor polynomial of f(x) = √x centered at a = 4.
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Answer: T₂(x) = 2 + (x − 4)/4 − (x − 4)²/64
Here f(4) = 2, f′(x) = 1/(2√x) gives f′(4) = 1/4, and f″(x) = −1/(4x^(3/2)) gives f″(4) = −1/32. So T₂(x) = f(4) + f′(4)(x − 4) + f″(4)(x − 4)²/2 = 2 + (x − 4)/4 − (x − 4)²/64.