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Marianopolis · Science · 201-SN3-RE

201-SN3-RE Integral Calculus tutoring

Get one-to-one help with integration methods and applications for Marianopolis Science 201-SN3-RE. Work through the short substitution example below, or bring a question, even if you do not know how to start.

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Exam review with explained answers

Your Marianopolis Science integral calculus course

201-SN3-RE covers integration methods and their applications, and includes an introduction to series and power series. The example below focuses on keeping a definite integral consistent when you change variables.

Course topics

  • Definite and indefinite integrals; the Fundamental Theorem of Calculus
  • Integration techniques
  • Indeterminate forms and improper integrals
  • Applications to area, volume and differential equations
  • An introduction to series and power series

Your 201-SN3-RE revision route

Choose only topics listed for your assessment. The course description supports this subject, but does not establish your section’s exam date or chapter cutoff.

  1. Choose topics. Decide on an integration method before calculating
  2. Follow a worked example. Compare identities with trigonometric substitution
  3. Try without solutions. Choose existing mixed questions from your announced chapters
  4. Get help with a step. Ask about 201-SN3-RE tutoring Include the chapter, assessment date and question; attempted work is useful but not required. Sessions are currently in English.

Official course source Your section outline and instructor announcements take priority. These are Nablio study routes and exercises, not official exams.

Before trying the example

Review the chain rule, the power rule for antiderivatives and evaluating a function at an endpoint. For the independent check, recall that the derivative of the natural logarithm is 1/u1/u for positive uu.

Two steps worth checking

  • Replacing the whole integrand and differential when substituting
  • Using bounds that belong to the variable of integration

Choose the step you need to check

Use your current section outline to choose which of these skills to practise for an assessment.

Integral calculus: choose the method before calculating

Try integration by parts, an area between curves and an improper integral. Explain why each setup fits before doing the calculation.

Add series if they are on your review list.

Before you start: Allow 20 minutes for the three questions, then 10 minutes to check your method choices.

1. A definite integral by parts

Evaluate ∫01xe2x dx\displaystyle\int_0^1 xe^{2x}\,dx. State your choices for integration by parts.

Show the explained solution
  1. Choose u=xu=x and dv=e2xdxdv=e^{2x}dx, so du=dxdu=dx and v=e2x/2v=e^{2x}/2. Differentiating the polynomial simplifies it.
  2. An antiderivative is F(x)=xe2x2−e2x4.F(x)=\frac{xe^{2x}}2-\frac{e^{2x}}4. The factor 1/21/2 must appear when integrating e2xe^{2x}.
  3. Now F(1)=e2/4F(1)=e^2/4 and F(0)=−1/4F(0)=-1/4, so the integral equals (e2+1)/4(e^2+1)/4. Differentiating FF gives xe2xxe^{2x}, checking the antiderivative.

Practise choosing parts that simplify the remaining integral

2. Area requires an order

Find the area enclosed by y=2xy=2x and y=x2y=x^2. Explain which function belongs on top between the intersections.

Show the explained solution
  1. Solve 2x=x22x=x^2: the intersections have x=0x=0 and x=2x=2. For 0<x<20<x<2, 2x−x2=x(2−x)>02x-x^2=x(2-x)>0, so the line lies above the parabola.
  2. The area is A=∫02(2x−x2) dx.A=\int_0^2(2x-x^2)\,dx. An antiderivative is x2−x3/3x^2-x^3/3.
  3. Evaluating gives A=4−8/3=4/3A=4-8/3=4/3 square units. Reversing the subtraction produces a negative signed integral, not the geometric area.

Practise intersections, curve order and area

3. Give infinity a limiting meaning

Decide whether ∫1∞x−3 dx\displaystyle\int_1^\infty x^{-3}\,dx converges, and find its value if it does. Write the limit explicitly.

Show the explained solution
  1. Replace the infinite endpoint by a finite b>1b>1: ∫1bx−3 dx=12−12b2.\int_1^b x^{-3}\,dx=\frac12-\frac{1}{2b^2}.
  2. By definition, ∫1∞x−3 dx=lim⁡b→∞∫1bx−3 dx=lim⁡b→∞(12−12b2)=12.\begin{gathered}\int_1^\infty x^{-3}\,dx\\=\lim_{b\to\infty}\int_1^b x^{-3}\,dx\\=\lim_{b\to\infty}\left(\frac12-\frac{1}{2b^2}\right)\\=\frac12.\end{gathered} Since 1/(2b2)→01/(2b^2)\to0, the limit exists and is finite, so the improper integral converges.
  3. The area beyond a finite bb is 1/(2b2)1/(2b^2). The interval is unbounded, but this remaining area shrinks to zero. Do not treat infinity as a number substituted into the antiderivative.

Review improper integral limits

Use your answers to plan your revision
  1. Write the clue first

    For each assigned integral, name the feature suggesting substitution, parts or another method. Explain why integrating the factors separately would be invalid.

  2. Sketch before setting bounds

    Mark intersections and the upper curve before writing an area integral. Recheck the order if curves cross inside the interval.

  3. Finish with a different check

    Differentiate a proposed antiderivative and check that a geometric area is nonnegative. For an improper integral, write its defining limit or limits and check that each exists and is finite. Otherwise, state that the integral diverges. Add series practice only if it belongs to your current review.

Send 201-SN3-RE, your assigned topic list and an integral you want help with, even if you have not started it. Your tutor can help you choose a method, work through the calculation or interpret the result. If you have an attempt, you can review it together.

Request 201-SN3-RE exam preparation help

Original Nablio practice, not an official examination.

Course topic sources

Worked example: change the variable and the bounds

Evaluate ∫016x1+3x2 dx\int_0^1 6x\sqrt{1+3x^2}\,dx by substitution. Write the integral in the new variable before calculating.

Show a hint

Differentiate the expression inside the square root. Which factor in the integral matches that derivative?

Show the reasoning

  1. 1. Choose the new variable

    Let u=1+3x2u=1+3x^2. Its derivative is du/dx=6xdu/dx=6x, so du=6x dxdu=6x\,dx. This replaces both the input of the square root and the remaining factor 6x dx6x\,dx.

  2. 2. Change both bounds

    The old bounds describe x. At x=0x=0, the new variable is u=1u=1; at x=1x=1, it is u=4u=4. Thus ∫016x1+3x2 dx=∫14u du.\begin{gathered}\int_0^1 6x\sqrt{1+3x^2}\,dx\\=\int_1^4\sqrt{u}\,du.\end{gathered} Everything on the right now uses u.

  3. 3. Integrate and evaluate

    The power rule gives the antiderivative 23u3/2\frac23u^{3/2}. Evaluate at the upper bound, then subtract the value at the lower bound: ∫14u1/2 du=23(43/2−13/2)=23(8−1)=143.\begin{gathered}\int_1^4 u^{1/2}\,du\\=\frac23(4^{3/2}-1^{3/2})\\=\frac23(8-1)=\frac{14}{3}.\end{gathered}

  4. 4. Check in the original variable

    An antiderivative in x is F(x)=23(1+3x2)3/2F(x)=\frac23(1+3x^2)^{3/2}. Differentiating with the chain rule gives F′(x)=23⋅32⋅6x⋅(1+3x2)1/2=6x1+3x2.\begin{aligned}F^{\prime}(x)&=\frac23\cdot\frac32\cdot6x\\&\quad{}\cdot(1+3x^2)^{1/2}\\&=6x\sqrt{1+3x^2}.\end{aligned} Then F(1)−F(0)=14/3F(1)-F(0)=14/3, the same answer. Returning to x lets you use the original x-bounds.

Try it yourself: find the incorrect bounds

A student sets u=3+4x2u=3+4x^2, so du=8x dxdu=8x\,dx, and writes ∫018x3+4x2 dx=∫011u du.\begin{gathered}\int_0^1\frac{8x}{3+4x^2}\,dx\\=\int_0^1\frac1u\,du.\end{gathered} Repair the right-hand integral and evaluate it. Does the original integrand have a zero denominator anywhere on [0,1][0,1]? Here ln⁡\ln denotes the natural logarithm.

Check your reasoning

  1. 1. Translate the endpoints

    The substitution is suitable, but the bounds were left in x. At x=0x=0, u=3u=3; at x=1x=1, u=7u=7. The corrected integral is ∫371u du\int_3^7\frac1u\,du.

  2. 2. Evaluate on the correct interval

    Since u is positive from 3 to 7, an antiderivative is ln⁡u\ln u. The result is ln⁡7−ln⁡3=ln⁡(7/3)\ln7-\ln3=\ln(7/3).

  3. 3. Check the apparent problem at zero

    The original denominator satisfies 3+4x2≥33+4x^2\ge3, so it never vanishes. Its integrand is continuous on [0,1][0,1]. The incorrect lower bound u=0u=0 makes 1/u1/u undefined there. This division by zero comes from the wrong bound, not from the original problem.

If choosing u was difficult, review substitution in the integration-methods lesson. If only the endpoints went wrong, practise writing the new value beside each old bound before integrating.

Work through the sticking point with a tutor

Your tutor can help you choose a substitution, write each replacement and explain the new bounds. If you have already started, you can check the calculation together. Then compare the integral with a similar-looking one to decide whether the same method still works.

For applications, your tutor can help you sketch the region and choose the bounds before integrating. For improper integrals or series, your tutor can help you state the limit or convergence question and explain why the chosen method applies. Bring a problem, whether you want help starting it or reviewing a solution.

Plan your review

Use your section’s topic list and instructions. Practise choosing a method before doing the calculation, then check the bounds and differentiate your antiderivative when possible. Redo a missed question without the solution and explain the step you changed.

Independent practice

Choose your next practice step

Course description

Before requesting a lesson

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Use your own section’s assessment instructions to choose what to review. Send your topic and preferred times. You can add a question you are allowed to share, any work you have started, and the course code or outline if helpful.

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This guide is available in English and French. Tutoring sessions are currently in English, online through Zoom or Google Meet.

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