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← Calculus II: Integrals & Series levels

Calculus II: Integrals & Series / LEVEL 1 · DIFFICULTY 4/5

Calculus II: Integrals & Series

Read the prerequisite, study a worked example and justify each step in these selected university topics.

3 stages · 24 practice problems · two 6-question assessment forms

STAGE UC2 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Integration Techniques & Improper Integrals

Useful preparation: Integrals & the Fundamental Theorem

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Integration methods reverse derivative patterns: substitution follows composition, parts rearranges a product rule, and partial fractions simplify rational expressions. For improper integrals, also ask whether finite-cutoff values approach a finite limit.

Substitution
Replacing an inner expression by a new variable while transforming its differential and, for a definite integral, its endpoints consistently.
Integration by parts
Trading an integral of one product for another using the product rule. A useful choice makes the remaining integral easier to evaluate.
Improper integral
An integral defined through one-sided limits because an interval is unbounded or the integrand is unbounded near an endpoint or interior point.

Conditions before calculation

  • Use substitution on intervals where the chain-rule identity applies, and change every occurrence of the variable and the bounds consistently.
  • For the standard integration-by-parts formula, take both selected functions to be continuously differentiable on each finite interval under consideration.
  • At each singularity or infinite endpoint, require its own finite limit. Separate divergent pieces cannot be canceled to claim ordinary convergence.

Why the method works

1. Match structure before manipulating

An inner expression beside its derivative suggests substitution; a product with a simplifying derivative suggests parts. For rational functions, divide if necessary, then factor.

∫f(g(x))g′(x) dx=∫f(u) du,u=g(x)\int f(g(x))g'(x)\,dx=\int f(u)\,du,\qquad u=g(x)

2. Derive the product exchange

Integrate the product rule and rearrange. A useful choice simplifies the remaining integral; applying the formula mechanically may not help.

∫abu(x)v′(x) dx=[u(x)v(x)]ab−∫abu′(x)v(x) dx\int_a^b u(x)v'(x)\,dx=[u(x)v(x)]_a^b-\int_a^b u'(x)v(x)\,dx

3. Evaluate finite pieces, then take limits

Infinity is not an antiderivative input. Use a finite cutoff; at an interior singularity, split and test both one-sided limits separately.

∫a∞f(x) dx=lim⁡R→∞∫aRf(x) dx\int_a^\infty f(x)\,dx=\lim_{R\to\infty}\int_a^R f(x)\,dx

Worked examples

EXAMPLE 1

Evaluate the integral using substitution and parts.

  1. The factor x matches the derivative of the inner quadratic up to a constant. Change the endpoints as well as the differential.

    I=∫01xln⁡(1+x2) dx,u=1+x2,du=2x dx,I=12∫12ln⁡u duI=\int_0^1x\ln(1+x^2)\,dx,\quad u=1+x^2,\quad du=2x\,dx,\quad I=\frac12\int_1^2\ln u\,du

  2. Treat the logarithm as a product with 1. Differentiating the logarithm and integrating 1 leaves a constant integrand.

    ∫ln⁡u du=uln⁡u−∫1 du=uln⁡u−u+C\int\ln u\,du=u\ln u-\int1\,du=u\ln u-u+C

  3. Evaluate in the new variable. Differentiating the resulting antiderivative independently recovers the original integrand.

    I=12[uln⁡u−u]12=ln⁡2−12,ddx((1+x2)ln⁡(1+x2)−(1+x2)2)=xln⁡(1+x2)I=\frac12[u\ln u-u]_1^2=\ln2-\frac12,\qquad \frac{d}{dx}\left(\frac{(1+x^2)\ln(1+x^2)-(1+x^2)}2\right)=x\ln(1+x^2)

Conclusion: The exact value is the natural logarithm of 2 minus one half.

EXAMPLE 2

Determine whether the positive logarithmic tail has finite total accumulation.

  1. Replace the infinite endpoint by a finite R greater than 1; only the resulting limit decides convergence.

    J=∫1∞ln⁡xx2 dx,JR=∫1Rln⁡xx2 dxJ=\int_1^\infty\frac{\ln x}{x^2}\,dx,\qquad J_R=\int_1^R\frac{\ln x}{x^2}\,dx

  2. Choose the logarithm for differentiation. Integration by parts reduces the tail to an elementary reciprocal power.

    u=ln⁡x,dv=x−2dx,JR=[−ln⁡xx]1R+∫1Rx−2 dx=1−ln⁡R+1Ru=\ln x,\quad dv=x^{-2}dx,\qquad J_R=\left[-\frac{\ln x}{x}\right]_1^R+\int_1^R x^{-2}\,dx=1-\frac{\ln R+1}{R}

  3. The logarithm grows more slowly than the linear denominator. For example, L′\displaystyle L'Hopital's rule verifies the needed logarithmic quotient limit.

    lim⁡R→∞ln⁡RR=lim⁡R→∞1/R1=0,J=1\lim_{R\to\infty}\frac{\ln R}{R}=\lim_{R\to\infty}\frac{1/R}{1}=0,\qquad J=1

Conclusion: The improper integral converges to 1, although the integration interval is infinite.

Pitfalls & counterexamples

Tempting claim: Symmetric cancellation always makes an improper integral converge.

Correction: Ordinary improper integration requires separate convergence on both sides of a singularity; a symmetric principal value is a different notion.

Test the claim: The reciprocal function on the interval from negative one to one has symmetric cutoff value zero, but its two one-sided integrals diverge.

Check your understanding

Is decay of a positive integrand to zero sufficient for its integral over an infinite interval to converge?

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Compare your reasoning

No.

The reciprocal function tends to zero, but its integral from one to a growing upper bound is the logarithm of that bound, which is unbounded.

Connect this idea

Improper integrals motivate comparison tests for series. Substitution becomes change of variables with a Jacobian in multiple integrals.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    Use integration by parts to find ∫\displaystyle \int from 1 to e of 2 ln⁡ x dx\displaystyle 2\, \operatorname{ln}\, x\, dx.

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  2. Problem 2

    Decompose 1x (x+2)=Ax+Bx+2\displaystyle \frac{1}{x\, \left(x + 2\right)} = \frac{A}{x} + \frac{B}{x + 2}. Find A.

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  3. Problem 3

    Evaluate the convergent improper integral ∫\displaystyle \int from 1 to infinity of 1x2 dx\displaystyle \frac{1}{x^{2}}\, dx.

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  4. Problem 4

    In a trigonometric substitution, let x=2 sin⁡ θ\displaystyle x = 2\, \operatorname{sin}\, \theta with −π2≤θ≤π2\displaystyle \frac{-\pi}{2} \le \theta \le \frac{\pi}{2}. Write 4−x2=A cos⁡ θ\displaystyle \sqrt{4 - x^{2}} = A\, \operatorname{cos}\, \theta. Find A.

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  5. Problem 5New context

    A weighted logarithmic accumulation requires ∫\displaystyle \int from 1 to e of 3 ln⁡ x dx\displaystyle 3\, \operatorname{ln}\, x\, dx.

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  6. Problem 6New context

    To integrate a rational model, decompose 1x (x+3)=Ax+Bx+3\displaystyle \frac{1}{x\, \left(x + 3\right)} = \frac{A}{x} + \frac{B}{x + 3}. Find A.

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  7. Problem 7New context

    A decaying tail has total accumulation ∫\displaystyle \int from 1 to infinity of 1x3 dx\displaystyle \frac{1}{x^{3}}\, dx.

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  8. Problem 8New context

    To simplify a circular cross-section integral, let x=3 sin⁡ θ\displaystyle x = 3\, \operatorname{sin}\, \theta with −π2≤θ≤π2\displaystyle \frac{-\pi}{2} \le \theta \le \frac{\pi}{2}. Write 9−x2=A cos⁡ θ\displaystyle \sqrt{9 - x^{2}} = A\, \operatorname{cos}\, \theta. Find A.

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Open stage UC2 1.1 in the student workspace →

STAGE UC2 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Sequences, Convergence & Taylor Series

Useful preparation: Integration Techniques & Improper Integrals

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For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Sequences track terms; series track running totals. Tiny terms may still accumulate without bound. Taylor series use derivatives at one center, but coefficients, convergence, and equality with the intended function require separate justification.

Sequence and partial sums
A sequence lists terms indexed by positive or nonnegative integers. The partial-sum sequence adds the first finitely many terms of a series.
Absolute and conditional convergence
Absolute convergence means the series of term magnitudes converges. Conditional convergence means the original series converges while that magnitude series diverges.
Taylor polynomial and remainder
The polynomial matches a function's derivatives through a chosen order at a center. The remainder is the actual function value minus this polynomial.

Conditions before calculation

  • Series convergence forces its terms to tend to zero, but that condition alone cannot establish convergence.
  • The alternating-series test requires term magnitudes to tend to zero and be nonincreasing eventually; its next-term error estimate applies after monotonicity begins.
  • A Taylor series represents its function only where the remainders tend to zero. For a power series, test boundary points separately from the open convergence interval.

Why the method works

1. Analyze running totals

Series convergence means its partial sums approach a finite number. This separates term behavior from accumulated behavior.

SN=∑n=1Nan,∑n=1∞an=S  ⟺  lim⁡N→∞SN=SS_N=\sum_{n=1}^{N}a_n,\qquad \sum_{n=1}^{\infty}a_n=S\iff\lim_{N\to\infty}S_N=S

2. Choose a test that reveals structure

Use geometric structure, comparison, or ratios as appropriate. When ratios are eventually defined, a ratio limit of one remains inconclusive.

L=lim⁡n→∞∣an+1an∣:L<1⇒absolute convergence,L>1⇒divergenceL=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|:\quad L<1\Rightarrow\text{absolute convergence},\quad L>1\Rightarrow\text{divergence}

3. Control approximation error

If the next derivative exists and is bounded by M throughout the interval between a and x, Taylor's theorem bounds the error as shown.

PN(x)=∑n=0Nf(n)(a)n!(x−a)n,∣f(x)−PN(x)∣≤M∣x−a∣N+1(N+1)!P_N(x)=\sum_{n=0}^{N}\frac{f^{(n)}(a)}{n!}(x-a)^n,\qquad |f(x)-P_N(x)|\le\frac{M|x-a|^{N+1}}{(N+1)!}

Worked examples

EXAMPLE 1

Sum the geometric series and find its partial-sum error.

  1. The constant ratio has magnitude less than one, so its powers decay.

    SN=∑n=0N(34)n,r=34S_N=\sum_{n=0}^{N}\left(\frac34\right)^n,\qquad r=\frac34

  2. Multiply by the ratio and subtract. All intermediate terms cancel, leaving a finite formula before any infinite operation.

    SN−34SN=1−(34)N+1,SN=4(1−(34)N+1)S_N-\frac34S_N=1-\left(\frac34\right)^{N+1},\qquad S_N=4\left(1-\left(\frac34\right)^{N+1}\right)

  3. Take the limit and subtract the finite sum from it. The error is a remaining tail, not just the next term.

    S=4,S−SN=4(34)N+1S=4,\qquad S-S_N=4\left(\frac34\right)^{N+1}

Conclusion: The sum is 4; every finite cutoff has an explicit positive error.

EXAMPLE 2

Approximate the natural logarithm of 1.2 with three Maclaurin terms and justify accuracy.

  1. Derivatives at zero give the logarithmic coefficients. The expansion is valid inside the stated interval; endpoints require independent tests.

    ln⁡(1+x)=∑n=1∞(−1)n+1xnn,∣x∣<1\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n},\qquad |x|<1

  2. At one fifth, the terms alternate and their magnitudes decrease to zero. Evaluate the first three terms exactly.

    P3(15)=15−150+1375=137750≈0.1826667P_3\left(\frac15\right)=\frac15-\frac1{50}+\frac1{375}=\frac{137}{750}\approx0.1826667

  3. The first omitted term is negative, so this approximation overestimates the value. The alternating-tail estimate bounds the error.

    0<P3(15)−ln⁡(65)<(1/5)44=12500=0.00040<P_3\left(\frac15\right)-\ln\left(\frac65\right)<\frac{(1/5)^4}{4}=\frac1{2500}=0.0004

Conclusion: The three-term estimate has error below four ten-thousandths, with a known direction of error.

Pitfalls & counterexamples

Tempting claim: Terms tending to zero guarantee a convergent series.

Correction: Small terms can still accumulate indefinitely; examine partial sums or apply a justified convergence test.

Test the claim: Reciprocal positive-integer terms tend to zero, but successive doubling blocks in the harmonic series each contribute at least one half.

Check your understanding

If the ratio test gives a limit of one, has the series been shown to diverge?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

The test is inconclusive there: both the harmonic series and the convergent series of reciprocal squares have ratio limit one.

Connect this idea

Taylor polynomials extend tangent-line approximation. Error estimates make infinite-series ideas usable in numerical calculation rather than merely formal expansions.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    Find the sum 2+22+24+28\displaystyle 2 + \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + … .

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  2. Problem 2

    A positive-term series has limiting ratio an+1an=23\displaystyle \frac{a_{n + 1}}{a_{n}} = \frac{2}{3}. Does the ratio test prove convergence? Enter 1 for yes or 0 for no.

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  3. Problem 3

    Find the x2\displaystyle x^{2} coefficient in the Maclaurin series of e2 x\displaystyle e^{2\, x}.

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  4. Problem 4

    For the geometric power series ∑\displaystyle \sum from n=0\displaystyle n = 0 to infinity of (2 x)n\displaystyle \left(2\, x\right)^{n}, find the radius of convergence.

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  5. Problem 5New context

    A repeating process accumulates the series 3+32+34+38\displaystyle 3 + \frac{3}{2} + \frac{3}{4} + \frac{3}{8} + … .

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  6. Problem 6New context

    A series used in a model has limiting ratio an+1an=34\displaystyle \frac{a_{n + 1}}{a_{n}} = \frac{3}{4}. Does the ratio test prove convergence? Enter 1 for yes or 0 for no.

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  7. Problem 7New context

    A second-order approximation needs the x2\displaystyle x^{2} coefficient in the Maclaurin series of e3 x\displaystyle e^{3\, x}.

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  8. Problem 8New context

    A model expands using ∑\displaystyle \sum from n=0\displaystyle n = 0 to infinity of (3 x)n\displaystyle \left(3\, x\right)^{n}, find the radius of convergence.

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STAGE UC2 1.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Parametric & Polar Calculus

Useful preparation: Sequences, Convergence & Taylor Series

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Parametric equations describe both coordinates through a parameter; polar coordinates use angle and signed radius. Calculus follows that geometry: slopes compare coordinate rates, lengths accumulate speed, and polar areas accumulate sectors.

Parametric curve
Both coordinates depend on a parameter. Its interval determines the portion traced, direction, and repeated traversal.
Regular point
A parameter value with nonzero velocity. Vertical tangents are possible even when the slope quotient is unavailable.
Polar coordinates
An angle and signed radial coordinate describing a point. Negative radius places the point opposite the angle's forward ray, and representations are not unique.

Conditions before calculation

  • The quotient formula for slope requires a nonzero derivative of the horizontal coordinate. If both coordinate derivatives vanish, analyze limits or the curve itself.
  • Use continuously differentiable or piecewise smooth parameterizations for the length formulas; repeated traversal contributes repeatedly to the computed length.
  • For ordinary enclosed polar area, choose intervals that cover the intended region once. Signed radii and repeated points require checking the actual tracing.

Why the method works

1. Compare rates using the parameter

Divide the coordinate rates to obtain slope. For the second derivative, differentiate the slope in t, then divide again by the horizontal rate.

dydx=y′(t)x′(t),d2ydx2=ddt(y′(t)/x′(t))x′(t)\frac{dy}{dx}=\frac{y'(t)}{x'(t)},\qquad \frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\left(y'(t)/x'(t)\right)}{x'(t)}

2. Accumulate speed for length

Pythagoras combines coordinate changes into displacement length. Integrating the velocity magnitude therefore accumulates nonnegative length.

L=∫abx′(t)2+y′(t)2 dtL=\int_a^b\sqrt{x'(t)^2+y'(t)^2}\,dt

3. Convert polar geometry carefully

Coordinate conversion gives polar speed. A narrow sector contributes approximately half its squared radius times angle width, explaining the area formula.

x=rcos⁡θ,y=rsin⁡θ;A=12∫αβr2 dθ,L=∫αβr2+(drdθ)2 dθx=r\cos\theta,\quad y=r\sin\theta;\qquad A=\frac12\int_\alpha^\beta r^2\,d\theta,\qquad L=\int_\alpha^\beta\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta

Worked examples

EXAMPLE 1

Find the quarter-circle slope at the stated parameter and its arc length.

  1. The parameter interval traces the upper-right quarter once, moving counterclockwise. Differentiate both coordinates.

    x(t)=3cos⁡t,y(t)=3sin⁡t,0≤t≤π2;x′=−3sin⁡t,y′=3cos⁡tx(t)=3\cos t,\quad y(t)=3\sin t,\quad 0\le t\le\frac\pi2;\qquad x'=-3\sin t,\quad y'=3\cos t

  2. At the chosen parameter the horizontal rate is nonzero, so division is legitimate.

    dydx∣t=π/6=3cos⁡(π/6)−3sin⁡(π/6)=−3\left.\frac{dy}{dx}\right|_{t=\pi/6}=\frac{3\cos(\pi/6)}{-3\sin(\pi/6)}=-\sqrt3

  3. The velocity magnitude is constant even though its components vary. Integrating this speed gives the arc length.

    9sin⁡2t+9cos⁡2t=3,L=∫0π/23 dt=3π2\sqrt{9\sin^2t+9\cos^2t}=3,\qquad L=\int_0^{\pi/2}3\,dt=\frac{3\pi}{2}

Conclusion: The slope is negative square root of 3; arc length is three pi over two.

EXAMPLE 2

Find the area enclosed by the polar curve and explain why a full angle interval would overcount it.

  1. Convert to Cartesian form to identify the geometry. On the chosen interval, the radius is nonnegative and the circle is traced once.

    r=2cos⁡θ,−π2≤θ≤π2;x2+y2=2x  ⟺  (x−1)2+y2=1r=2\cos\theta,\quad -\frac\pi2\le\theta\le\frac\pi2;\qquad x^2+y^2=2x\iff(x-1)^2+y^2=1

  2. Integrate the sector areas on this single traversal. The double-angle identity evaluates the squared cosine.

    A=12∫−π/2π/24cos⁡2θ dθ=[θ+12sin⁡(2θ)]−π/2π/2=πA=\frac12\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta=\left[\theta+\frac12\sin(2\theta)\right]_{-\pi/2}^{\pi/2}=\pi

  3. Increasing the angle by pi reverses both the radius sign and direction, producing the same point. A full turn therefore traces the circle twice.

    r(θ+π)=−r(θ),(x(θ+π),y(θ+π))=(x(θ),y(θ))r(\theta+\pi)=-r(\theta),\qquad (x(\theta+\pi),y(\theta+\pi))=(x(\theta),y(\theta))

Conclusion: The enclosed area is pi; using an interval of length two pi would give twice that area.

Pitfalls & counterexamples

Tempting claim: A zero horizontal rate means the curve has no tangent.

Correction: With nonzero vertical rate it gives a vertical tangent; if both rates vanish, the quotient is inconclusive.

Test the claim: At the starting point of the quarter-circle example, the horizontal rate is zero and the vertical rate is 3, so the tangent is vertical.

Check your understanding

Does tracing the same circle twice double its geometric circumference?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No, but it doubles the parameterized path length.

The geometric set is unchanged, while the length integral counts every traversal.

Connect this idea

Parametric speed becomes the arc-length element in line integrals. Polar area anticipates the radial Jacobian factor in double integrals.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    For the parametric curve x=2 t\displaystyle x = 2\, t, y=2 t2\displaystyle y = 2\, t^{2}. Find dydx\displaystyle \frac{dy}{dx} at t=1\displaystyle t = 1.

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  2. Problem 2

    Find the length x=3 t\displaystyle x = 3\, t, y=4 t\displaystyle y = 4\, t for 0≤t≤2\displaystyle 0 \le t \le 2.

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  3. Problem 3

    A polar circle has r=2\displaystyle r = 2 for 0≤θ≤π\displaystyle 0 \le \theta \le \pi. Its swept area is a π\displaystyle a\, \pi. Find a.

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  4. Problem 4

    For x=t2\displaystyle x = t^{2} and y=3 t\displaystyle y = 3\, t, find dydx\displaystyle \frac{dy}{dx} at t=2\displaystyle t = 2.

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  5. Problem 5New context

    A moving point follows x=2 t\displaystyle x = 2\, t, y=3 t2\displaystyle y = 3\, t^{2}. Find dydx\displaystyle \frac{dy}{dx} at t=1\displaystyle t = 1.

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  6. Problem 6New context

    Find the distance traveled along x=3 t\displaystyle x = 3\, t, y=4 t\displaystyle y = 4\, t for 0≤t≤3\displaystyle 0 \le t \le 3.

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  7. Problem 7New context

    A circular sweep has r=3\displaystyle r = 3 for 0≤θ≤π\displaystyle 0 \le \theta \le \pi. Its swept area is a π\displaystyle a\, \pi. Find a.

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  8. Problem 8New context

    A path is x=t2\displaystyle x = t^{2} and y=4 t\displaystyle y = 4\, t, find dydx\displaystyle \frac{dy}{dx} at t=2\displaystyle t = 2.

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