Practice exam · Calculus II · Units 3B–4B

Calculus II Practice Final Exam with Complete Solutions

A balanced twenty-five-question Calculus II final with study map and complete worked key.

25 problems120 min estimated timeCumulative progression

What is included

Prepare for a Calculus II final with integration methods, applications, convergence, power series, and Taylor series.

Skills assessed

  • method selection
  • convergence proof
  • series construction
  • course synthesis

Prerequisites

  • Calculus II Units 3B through 4B

Exam conditions

Suggested time: 120 minutes

Points: 4 points per question; 100 points total

Calculator: Scientific calculator permitted; computer algebra is not required.

This is an original BetterGrades practice exam, not a released institutional exam.

Printable and accessible

Download this resource

No email address or account is required.

Device-only study control

Ready for an honest first attempt?

This control sends only the resource id and type—never your answers or identity.

Printable preview

  1. Evaluate xexdx\int xe^x dx.
  2. Evaluate sin3xcosxdx\int\sin^3x\cos xdx.
  3. Evaluate 1/(x21)dx\int1/(x^2-1)dx.
  4. Determine convergence of 1x2dx\int_1^\infty x^{-2}dx.
  5. Find area between y=x and y=x² on [0,1].
  6. Rotate y=x on [0,2] about x-axis. Find volume.
  7. A force F(x)=4x acts from x=0 to 3. Find work.
  8. Find limnn/(n+1)\lim_{n\to\infty} n/(n+1).
  9. Sum n=03(1/2)n\sum_{n=0}^\infty3(1/2)^n.
  10. Test n=1n/(n+1)\sum_{n=1}^{\infty} n/(n+1) for convergence.
  11. Test n=11/n3/2\sum_{n=1}^{\infty}1/n^{3/2}.
  12. Test n=11/(n2+4)\sum_{n=1}^{\infty}1/(n^2+4).
  13. Test n=1(3n+1)/(n2+2)\sum_{n=1}^{\infty}(3n+1)/(n^2+2).
  14. Test n=1n!/4n\sum_{n=1}^{\infty} n!/4^n.
  15. Test n=1(2n/(3n+1))n\sum_{n=1}^{\infty}(2n/(3n+1))^n.
  16. Test n=1(1)n1/n\sum_{n=1}^{\infty}(-1)^{n-1}/n.
  17. Classify n=1(1)n/n2\sum_{n=1}^{\infty}(-1)^n/n^2.
  18. Find R for n=0xn/5n\sum_{n=0}^{\infty}x^n/5^n.
  19. Find the interval for n=1(x1)n/n\sum_{n=1}^{\infty}(x-1)^n/n.
  20. Write the Maclaurin series for cos x.
  21. Find T_3 for e^x at 0.
  22. Bound alternating-series error after four terms of n=1(1)n1/n\sum_{n=1}^{\infty}(-1)^{n-1}/n.
  23. For x=t²,y=t³ find dy/dx.
  24. Convert r=2cosθ to Cartesian form.
  25. Why must power-series endpoints be tested separately?

Complete worked solutions

Every problem has a source-matched answer and independently reviewed derivation.

01

Problem 1: Evaluate xexdx\int xe^x dx.

Answer: ex(x1)+Ce^x(x-1)+C

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. With u=xu=x, dv=exdxdv=e^x dx, I=xexexdx=ex(x1)+CI=xe^x-\int e^x dx=e^x(x-1)+C.
  2. The verified result is ex(x1)+Ce^x(x-1)+C.
02

Problem 2: Evaluate sin3xcosxdx\int\sin^3x\cos xdx.

Answer: sin4x/4+C\sin^4x/4+C

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. Let u=sinxu=\sin x, du=cosxdxdu=\cos xdx; then u3du=u4/4+C=sin4x/4+C\int u^3du=u^4/4+C=\sin^4x/4+C.
  2. The verified result is sin4x/4+C\sin^4x/4+C.
03

Problem 3: Evaluate 1/(x21)dx\int1/(x^2-1)dx.

Answer: 12lnx1x+1+C\frac12\ln|\frac{x-1}{x+1}|+C

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. 1/(x21)=12/(x1)12/(x+1)1/(x^2-1)=\frac12/(x-1)-\frac12/(x+1); integrating gives 12lnx112lnx+1+C\frac12\ln|x-1|-\frac12\ln|x+1|+C.
  2. The verified result is 12lnx1x+1+C\frac12\ln|\frac{x-1}{x+1}|+C.
04

Problem 4: Determine convergence of 1x2dx\int_1^\infty x^{-2}dx.

Answer: converges to 1\text{converges to }1

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. limb[x1]1b=limb(11/b)=1\lim_{b\to\infty}[-x^{-1}]_1^b=\lim_{b\to\infty}(1-1/b)=1, so the improper integral converges.
  2. The verified result is converges to 1\text{converges to }1.
05

Problem 5: Find area between y=x and y=x² on [0,1].

Answer: 1/61/6

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. On [0,1][0,1], xx2x\ge x^2, so A=01(xx2)dx=[x2/2x3/3]01=1/6A=\int_0^1(x-x^2)dx=[x^2/2-x^3/3]_0^1=1/6.
  2. The verified result is 1/61/6.
06

Problem 6: Rotate y=x on [0,2] about x-axis. Find volume.

Answer: 8π/38\pi/3

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. V=π02x2dx=π[x3/3]02=8π/3V=\pi\int_0^2x^2dx=\pi[x^3/3]_0^2=8\pi/3.
  2. The verified result is 8π/38\pi/3.
07

Problem 7: A force F(x)=4x acts from x=0 to 3. Find work.

Answer: 1818

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. W=034xdx=[2x2]03=18W=\int_0^3 4x\,dx=[2x^2]_0^3=18.
  2. The verified result is 1818.
08

Problem 8: Find limnn/(n+1)\lim_{n\to\infty} n/(n+1).

Answer: 11

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. Divide by nn: n/(n+1)=1/(1+1/n)1n/(n+1)=1/(1+1/n)\to1.
  2. The verified result is 11.
09

Problem 9: Sum n=03(1/2)n\sum_{n=0}^\infty3(1/2)^n.

Answer: 66

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. Here a=3a=3, r=1/2r=1/2, and r<1|r|<1, so S=3/(11/2)=6S=3/(1-1/2)=6.
  2. The verified result is 66.
10

Problem 10: Test n=1n/(n+1)\sum_{n=1}^{\infty} n/(n+1) for convergence.

Answer: diverges\text{diverges}

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. The terms satisfy n/(n+1)10n/(n+1)\to1\ne0, so the nth-term test proves divergence.
  2. The verified result is diverges\text{diverges}.
11

Problem 11: Test n=11/n3/2\sum_{n=1}^{\infty}1/n^{3/2}.

Answer: converges\text{converges}

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. This is a p-series with p=3/2>1p=3/2>1, so it converges.
  2. The verified result is converges\text{converges}.
12

Problem 12: Test n=11/(n2+4)\sum_{n=1}^{\infty}1/(n^2+4).

Answer: converges\text{converges}

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. For n1n\ge1, 0<1/(n2+4)<1/n20<1/(n^2+4)<1/n^2; comparison with the convergent p-series proves convergence.
  2. The verified result is converges\text{converges}.
13

Problem 13: Test n=1(3n+1)/(n2+2)\sum_{n=1}^{\infty}(3n+1)/(n^2+2).

Answer: diverges\text{diverges}

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. With bn=1/nb_n=1/n, an/bn=n(3n+1)/(n2+2)3a_n/b_n=n(3n+1)/(n^2+2)\to3; limit comparison with the harmonic series gives divergence.
  2. The verified result is diverges\text{diverges}.
14

Problem 14: Test n=1n!/4n\sum_{n=1}^{\infty} n!/4^n.

Answer: diverges\text{diverges}

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. an+1/an=(n+1)/4a_{n+1}/a_n=(n+1)/4, which eventually exceeds 1, so the positive terms do not approach zero and the series diverges.
  2. The verified result is diverges\text{diverges}.
15

Problem 15: Test n=1(2n/(3n+1))n\sum_{n=1}^{\infty}(2n/(3n+1))^n.

Answer: converges\text{converges}

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. The root-test limit is lim2n/(3n+1)=2/3<1\lim 2n/(3n+1)=2/3<1, so the series converges absolutely.
  2. The verified result is converges\text{converges}.
16

Problem 16: Test n=1(1)n1/n\sum_{n=1}^{\infty}(-1)^{n-1}/n.

Answer: converges conditionally\text{converges conditionally}

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. 1/n1/n decreases to zero, so the alternating-series test gives convergence; 1/n\sum1/n diverges, hence convergence is conditional.
  2. The verified result is converges conditionally\text{converges conditionally}.
17

Problem 17: Classify n=1(1)n/n2\sum_{n=1}^{\infty}(-1)^n/n^2.

Answer: converges absolutely\text{converges absolutely}

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. The absolute series 1/n2\sum1/n^2 is a p-series with p=2>1p=2>1, so the original series converges absolutely.
  2. The verified result is converges absolutely\text{converges absolutely}.
18

Problem 18: Find R for n=0xn/5n\sum_{n=0}^{\infty}x^n/5^n.

Answer: R=5R=5

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. This is geometric with ratio x/5x/5; convergence requires x/5<1|x/5|<1, or x<5|x|<5, so R=5R=5.
  2. The verified result is R=5R=5.
19

Problem 19: Find the interval for n=1(x1)n/n\sum_{n=1}^{\infty}(x-1)^n/n.

Answer: [0,2)[0,2)

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. The ratio test gives x1<1|x-1|<1. At x=0x=0, (1)n/n\sum(-1)^n/n converges; at x=2x=2, 1/n\sum1/n diverges, so the interval is [0,2)[0,2).
  2. The verified result is [0,2)[0,2).
20

Problem 20: Write the Maclaurin series for cos x.

Answer: n=0(1)nx2n/(2n)!\sum_{n=0}^{\infty}(-1)^nx^{2n}/(2n)!

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. The derivatives at zero cycle 1,0,1,01,0,-1,0, leaving even powers with alternating signs: n=0(1)nx2n/(2n)!\sum_{n=0}^{\infty}(-1)^nx^{2n}/(2n)!.
  2. The verified result is n=0(1)nx2n/(2n)!\sum_{n=0}^{\infty}(-1)^nx^{2n}/(2n)!.
21

Problem 21: Find T_3 for e^x at 0.

Answer: 1+x+x2/2+x3/61+x+x^2/2+x^3/6

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. Every derivative of exe^x equals 1 at zero, so T3=n=03xn/n!=1+x+x2/2+x3/6T_3=\sum_{n=0}^3x^n/n!=1+x+x^2/2+x^3/6.
  2. The verified result is 1+x+x2/2+x3/61+x+x^2/2+x^3/6.
22

Problem 22: Bound alternating-series error after four terms of n=1(1)n1/n\sum_{n=1}^{\infty}(-1)^{n-1}/n.

Answer: 1/5\le1/5

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. The terms 1/n1/n decrease to zero; after four terms the first omitted magnitude is 1/51/5, so R41/5|R_4|\le1/5.
  2. The verified result is 1/5\le1/5.
23

Problem 23: For x=t²,y=t³ find dy/dx.

Answer: 3t/23t/2

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. dx/dt=2tdx/dt=2t and dy/dt=3t2dy/dt=3t^2, so dy/dx=(3t2)/(2t)=3t/2dy/dx=(3t^2)/(2t)=3t/2 for t0t\ne0.
  2. The verified result is 3t/23t/2.
24

Problem 24: Convert r=2cosθ to Cartesian form.

Answer: (x1)2+y2=1(x-1)^2+y^2=1

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. Multiply by rr: r2=2rcosθr^2=2r\cos\theta, so x2+y2=2xx^2+y^2=2x, which completes the square to (x1)2+y2=1(x-1)^2+y^2=1.
  2. The verified result is (x1)2+y2=1(x-1)^2+y^2=1.
25

Problem 25: Why must power-series endpoints be tested separately?

Answer: The ratio/root test is inconclusive when xa=R.\text{The ratio/root test is inconclusive when }|x-a|=R.

Why this method: Course synthesis matches the mathematical structure before any algebraic cleanup.

  1. At xa=R|x-a|=R, the limiting ratio or root equals 1, so the test is inconclusive and each endpoint series requires its own convergence test.
  2. The verified result is The ratio/root test is inconclusive when xa=R.\text{The ratio/root test is inconclusive when }|x-a|=R..

Common errors

  • Choosing an integration or convergence test from surface appearance alone.
  • Skipping endpoint checks.
  • Claiming a Taylor representation without a convergence or remainder argument.