Work with an online tutor on integration methods, applications and introductory series in McGill MATH 141. You can start with an integral you have not been able to begin. Your tutor can help you choose a method, check an antiderivative by differentiating and use a sketch to explain the bounds.
McGill MATH 141 covers definite integrals, integration techniques and applications, and an introduction to sequences and series.
Practise the reasoning between steps
Choose a workshop when the topic is included in your instructor’s announced material. These original exercises are not an exam reproduction or a complete course review.
Find the area enclosed by y=3x−x2 and y=x. Coordinates are dimensionless. Set up the integral before evaluating it.
Show the explained solution
Intersections satisfy 3x−x2=x, so x(2−x)=0, giving x=0,2.
Between those values, (3x−x2)−x=x(2−x)≥0; the parabola is above the line. Thus A=∫02(2x−x2)dx.
A=[x2−3x3]02=34. This is a positive area in square coordinate units.
If you reverse the top and bottom functions, the integral becomes negative. If you insert arbitrary bounds, it no longer describes the enclosed region.
Compare substitution and parts in the linked integration-methods lesson. Then use the worked example on this page that combines both methods, keeping its transformed bounds beside the new variable.
If question 2 had the wrong sign or bounds
Sketch the two curves and mark each intersection. Use a test input between consecutive intersections before deciding which function goes on top.
If series are on your current outline
Use the sequences-and-series practice on this page after the integral review. Distinguish the limit of a sequence from the sum of a series, and write the assumptions for the convergence argument you choose.
Bring the integral you could not set up or the line where your method stopped helping. A focused lesson can compare two possible first moves and give you a fresh problem to solve without the earlier solution.
Review derivative rules, basic antiderivatives and algebra. Distinguish a signed accumulation from geometric area before choosing integration bounds.
Registration requirements
The guidance above is study preparation, not an eligibility decision. Confirm current prerequisites, restrictions and availability with McGill University.
Let x be the integration variable. Evaluate ∫023x2dx.
Show the reasoning
An antiderivative is F(x)=x3, because F′(x)=3x2. The fundamental theorem of calculus gives the upper-end value minus the lower-end value: F(2)−F(0)=23−03=8. A constant of integration would cancel in this difference.
A definite integral that needs two decisions
A useful substitution can simplify an integral without finishing it. This exercise applies two methods in sequence and keeps the bounds consistent throughout.
1. Identify the inside function and its derivative
Let x be the real integration variable and I the value of the definite integral below. The symbol ln denotes the natural logarithm. Evaluate I=∫01xln(1+x2)dx. The derivative of 1+x2 is 2x, so the outside factor xdx is half the needed differential. This makes u=1+x2 a useful first move.
2. Change the differential and both bounds
Since du=2xdx, we have xdx=21du. The lower bound x=0 becomes u=1, and x=1 becomes u=2. Thus I=21∫12lnudu. Every part now uses the same variable.
3. Use integration by parts on the simpler integral
Write lnu as lnu⋅1. Differentiating the logarithm and integrating 1 gives ∫lnudu=ulnu−∫uu1du=ulnu−u+C. This works because the remaining product simplifies to 1. The method follows the product rule in reverse. Here C is an arbitrary constant of integration. It cancels when the antiderivative is evaluated at the two bounds.
4. Evaluate and check the scale
Using the new bounds, I=21[(2ln2−2)−(1ln1−1)]=ln2−21≈0.1931. To check the scale, 0≤x≤1 gives 1≤1+x2≤2. Since the logarithm is increasing, 0≤ln(1+x2)≤ln2. Multiplying by nonnegative x and integrating gives 0≤I≤∫01xln2dx=2ln2≈0.3466. Our result lies within this range. As a separate check of the antiderivative, differentiating 21[(1+x2)ln(1+x2)−(1+x2)] recovers xln(1+x2).
Try it yourself
Use the same reasoning to evaluate J=∫01xln(1+2x2)dx. State the new differential and bounds before integrating.
Show the explained solution
With u=1+2x2, du=4xdx, and bounds 1,3: J=41∫13lnudu=43ln3−2≈0.3240. The changed factor and upper bound both matter.
Before each calculation, give a reason for the next move. If the first substitution leaves a new integral, inspect that new structure instead of forcing the original method again.
Original practice
A sequence term and a series sum answer different questions
Let an=3(1/2)n−1 for integers n≥1. Find the limit of the sequence (an), then find the sum of the series ∑n=1∞an. These are different questions.
Show the reasoning
The terms begin 3,3/2,3/4,3/8,…. Each is half the previous term, so an→0.
To study the series, define its N-term partial sum:SN=3+23+⋯+2N−13.Subtracting half this sum cancels the middle terms:SN−21SN=3−2N3.Therefore SN=6(1−2−N). Since 2−N→0, the partial sums approach 6. The sequence approaches zero, while the series sums to six.
After four terms, S4=45/8=5.625. The remaining infinite tail is 6−S4=3/8=0.375.
Avoid this mistake: A term approaching zero does not, by itself, prove that a series converges. Here convergence follows from the partial-sum formula.
Try it
For bn=2(1/3)n−1, find the term limit and series sum.
Check your reasoning
The term limit is 0; the sum is 2/(1−1/3)=3, because the geometric ratio has absolute value below one.
Why terms tending to zero are not enough
The harmonic series is ∑n=1∞1/n. Its terms tend to zero. Does the series converge?
See why the series diverges
No. Let HN=∑n=1N1/n. Convergence would require these partial sums to approach a finite number. Group the terms after the first one into blocks: 1/2, then 1/3+1/4, then 1/5+⋯+1/8, and so on.
For every integer k≥1, the block from n=2k−1+1 to n=2k has 2k−1 terms. Each term is at least 1/2k, so each block contributes at least 1/2. After m blocks, for every integer m≥1,H2m≥1+2m.
The bound grows without limit as m increases, so the partial sums are unbounded. The harmonic series diverges even though 1/n→0. The argument works for arbitrarily many blocks; checking a few partial sums alone would not prove divergence.
If your current MATH 141 outline includes comparison, ratio and alternating-series tests, continue with the series-convergence test practice. Otherwise, stay with term limits and partial sums, and check your instructor’s current outline before moving to those tests.
How tutoring can help
You can start with an integral you do not know how to approach. Your tutor can explain how to choose a method, work through a change of variable and bounds with you, and show how differentiation checks an antiderivative. For sequences and series, you can first clarify the difference between a term and a partial sum, then work through the reasoning about convergence. Your current course material guides the examples; you can try a similar question with help before working independently.
Exam preparation for MATH 141
Mix methods with geometric applications and the sequence/series material in your outline. Explain why a convergence test applies rather than naming one without conditions.
Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.
Related courses
Useful links for preparation or further study—not transfer equivalencies or a required registration sequence.
Compare substitution, integration by parts, partial fractions and trigonometric substitution, with sign-aware radical choices, worked examples and progressive checks.
Build volume integrals from slices, radii and bounds. Practise disks and washers, then compare washers with cylindrical shells in an optional worked example.