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Calculus 2 extends integration techniques, introduces sequences and series, and explores parametric curves and polar coordinates. Convergence tests for infinite series are a common stumbling block — practice exams help you recognize which test to apply at a glance. Expect both computational and proof-oriented questions.

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Unlike generic question banks, every Crameleon exam reflects the specific framing, terminology, and difficulty level of your course. You'll practise answering questions that look and feel like the real thing — which is exactly what the research on retrieval practice says you should be doing. Students who test themselves with practice exams consistently outperform students who re-read notes, regardless of study time. The generation takes under 2 minutes and your first exam is completely free.

Topics typically tested

  • Integration by parts & partial fractions
  • Improper integrals
  • Sequences & series convergence
  • Power series & Taylor series
  • Parametric equations & arc length
  • Polar coordinates

Sample Calculus 2 exam questions

Three original questions in the style of a real Calculus 2 exam, each with a worked solution. Your own practice exam can be built from your course's notes and syllabus.

Question 1Multiple choiceSequences & series convergence

Consider the series

∑n=1∞(−1)n+1 nn2+1.\sum_{n=1}^{\infty} (-1)^{n+1}\,\frac{n}{n^2 + 1}.

Which statement is correct?

  1. A.

    The series converges absolutely.

  2. B.

    The series converges conditionally.

  3. C.

    The series diverges by the limit comparison test with ∑1n\sum \frac{1}{n}.

  4. D.

    The series diverges because its terms do not approach 0.

Worked solution
Answer:B

Write bn=nn2+1>0b_n = \dfrac{n}{n^2+1} > 0. Apply the alternating series test, which needs bn→0b_n \to 0 and bnb_n decreasing.

Limit: bn=1/n1+1/n2→0b_n = \dfrac{1/n}{1 + 1/n^2} \to 0, so the terms do go to zero and option D is ruled out.

Decreasing: for f(x)=xx2+1f(x) = \dfrac{x}{x^2+1}, the quotient rule gives

f′(x)=(x2+1)−x⋅2x(x2+1)2=1−x2(x2+1)2≤0for x≥1,f'(x) = \frac{(x^2+1) - x\cdot 2x}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2} \le 0 \quad \text{for } x \ge 1,

so bn+1≤bnb_{n+1} \le b_n for all n≥1n \ge 1. Both conditions hold, so the alternating series converges.

Absolute convergence: the series of absolute values is ∑nn2+1\sum \dfrac{n}{n^2+1}. Limit comparison with the harmonic series ∑1n\sum \dfrac{1}{n}:

lim⁡n→∞n/(n2+1)1/n=lim⁡n→∞n2n2+1=1,\lim_{n\to\infty} \frac{n/(n^2+1)}{1/n} = \lim_{n\to\infty} \frac{n^2}{n^2+1} = 1,

a finite positive number, and ∑1n\sum \dfrac{1}{n} diverges, so ∑nn2+1\sum \dfrac{n}{n^2+1} diverges. Hence the series does not converge absolutely.

A series that converges while its series of absolute values diverges is, by definition, conditionally convergent.

The tempting wrong answer is C: the limit comparison test is carried out correctly, but it only tells us about ∑∣an∣\sum |a_n|. Divergence of the absolute series never implies divergence of the original alternating series; that is exactly the situation the word "conditionally" describes.

The series converges conditionally (option B).

Question 2Short answerIntegration by parts & partial fractions

Evaluate

∫01x+5x2+3x+2 dx.\int_0^1 \frac{x + 5}{x^2 + 3x + 2}\,dx.

Give the exact value.

Worked solution
Answer:

7ln⁡2−3ln⁡3=ln⁡12827≈1.5567\ln 2 - 3\ln 3 = \ln\dfrac{128}{27} \approx 1.556

The denominator factors: x2+3x+2=(x+1)(x+2)x^2 + 3x + 2 = (x+1)(x+2), and the numerator has lower degree, so use partial fractions:

x+5(x+1)(x+2)=Ax+1+Bx+2⟹x+5=A(x+2)+B(x+1).\frac{x+5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \quad\Longrightarrow\quad x + 5 = A(x+2) + B(x+1).

Choosing convenient values of xx isolates each constant: x=−1x = -1 gives A=4A = 4, and x=−2x = -2 gives −B=3-B = 3, so B=−3B = -3. (Verification: 4(x+2)−3(x+1)=x+54(x+2) - 3(x+1) = x + 5.)

Now integrate term by term; each piece is a logarithm:

∫01(4x+1−3x+2)dx=[ 4ln⁡∣x+1∣−3ln⁡∣x+2∣ ]01.\int_0^1 \left(\frac{4}{x+1} - \frac{3}{x+2}\right)dx = \Big[\,4\ln|x+1| - 3\ln|x+2|\,\Big]_0^1.

Evaluate at the limits (x=1x = 1 then x=0x = 0):

(4ln⁡2−3ln⁡3)−(4ln⁡1−3ln⁡2)=4ln⁡2−3ln⁡3+3ln⁡2=7ln⁡2−3ln⁡3.\left(4\ln 2 - 3\ln 3\right) - \left(4\ln 1 - 3\ln 2\right) = 4\ln 2 - 3\ln 3 + 3\ln 2 = 7\ln 2 - 3\ln 3.

Using the laws of logarithms, 7ln⁡2−3ln⁡3=ln⁡2733=ln⁡12827≈1.5567\ln 2 - 3\ln 3 = \ln\dfrac{2^7}{3^3} = \ln\dfrac{128}{27} \approx 1.556.

∫01x+5x2+3x+2 dx=7ln⁡2−3ln⁡3=ln⁡12827≈1.556\displaystyle\int_0^1 \frac{x+5}{x^2+3x+2}\,dx = 7\ln 2 - 3\ln 3 = \ln\frac{128}{27} \approx 1.556

Question 3Worked problemPower series & Taylor series

Let I=∫01/21−cos⁡xx2 dxI = \displaystyle\int_0^{1/2} \frac{1 - \cos x}{x^2}\,dx.

(a) Use the Maclaurin series for cos⁡x\cos x to express II as an infinite series of rational numbers.

(b) Using the alternating series estimation theorem, determine how many terms of that series are needed to approximate II with error less than 10−510^{-5}, and give that approximation to 5 decimal places.

Worked solution
Answer:

I=14−1576+1115200−⋯I = \dfrac14 - \dfrac{1}{576} + \dfrac{1}{115200} - \cdots; two terms suffice, giving I≈0.24826I \approx 0.24826

(a) Start from the Maclaurin series cos⁡x=∑k=0∞(−1)kx2k(2k)!=1−x22!+x44!−⋯\cos x = \displaystyle\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots, valid for all xx. Subtracting from 1 removes the constant term, and dividing by x2x^2 lowers every power by 2:

1−cos⁡xx2=12!−x24!+x46!−⋯=∑k=1∞(−1)k+1x2k−2(2k)!.\frac{1 - \cos x}{x^2} = \frac{1}{2!} - \frac{x^2}{4!} + \frac{x^4}{6!} - \cdots = \sum_{k=1}^{\infty} (-1)^{k+1}\frac{x^{2k-2}}{(2k)!}.

A power series may be integrated term by term inside its interval of convergence (here, everywhere):

I=∑k=1∞(−1)k+11(2k)!∫01/2x2k−2 dx=∑k=1∞(−1)k+1(1/2)2k−1(2k−1) (2k)!.I = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(2k)!}\int_0^{1/2} x^{2k-2}\,dx = \sum_{k=1}^{\infty} (-1)^{k+1}\frac{(1/2)^{2k-1}}{(2k-1)\,(2k)!}.

The first terms are

I=1/21⋅2!−(1/2)33⋅4!+(1/2)55⋅6!−⋯=14−1576+1115200−⋯I = \frac{1/2}{1\cdot 2!} - \frac{(1/2)^3}{3\cdot 4!} + \frac{(1/2)^5}{5\cdot 6!} - \cdots = \frac14 - \frac{1}{576} + \frac{1}{115200} - \cdots

(b) The series alternates, its terms decrease in size, and they tend to 0, so the alternating series estimation theorem applies: the error after NN terms is at most the size of the first omitted term. The third term is 1115200≈8.7×10−6<10−5\dfrac{1}{115200} \approx 8.7\times 10^{-6} < 10^{-5}, so two terms are enough (one term is not, since the second term 1576≈1.7×10−3\dfrac{1}{576} \approx 1.7\times 10^{-3} would be too large an error).

I≈14−1576=143576≈0.24826.I \approx \frac14 - \frac{1}{576} = \frac{143}{576} \approx 0.24826.

(The true value is 0.248272…, within the promised 10−510^{-5}.)

Two terms suffice: I≈143576≈0.24826I \approx \dfrac{143}{576} \approx 0.24826, with error below 1115200<10−5\dfrac{1}{115200} < 10^{-5}.

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