FRBook a Lesson
Vanier · Science · 201-SN3-RE

201-SN3-RE Integral Calculus tutoring

Get one-to-one help with definite and indefinite integrals for Vanier 201-SN3-RE. Try the short application below, or bring a question, even if you do not know how to start.

Ask about a lessonTry the worked example
Exam review with explained answers

Integral calculus at Vanier

Vanier’s Integral Calculus course introduces definite and indefinite integrals and their applications. A useful starting point is deciding what an integral measures. The example below uses a changing velocity to connect signed area, displacement and distance travelled.

Skills to work on

  • Finding antiderivatives and checking them by differentiation
  • Evaluating definite integrals using an antiderivative and both bounds
  • Interpreting an integral as accumulated change
  • Using signs and units to explain an application’s result

Before trying the example

Review the power rule for antiderivatives, factoring a quadratic and deciding where a function is positive or negative. Recall that integrating a rate over a time interval gives an accumulated change: metres per second multiplied by seconds gives metres.

Two steps worth checking

  • Deciding whether the question asks for net change or a total that counts both directions positively
  • Splitting the interval where the rate changes sign before calculating that total

Choose the question you need

Integral calculus: method choice and a volume setup

Use integration by parts for a logarithm, then build a volume integral from disk areas. Write the method or radius before calculating.

Before you start: Allow 20–25 minutes to try both questions, then 15 minutes to check and retry any missed steps.

1. Turn a logarithm into a simpler integral

Evaluate ∫1eln⁡(x) dx\int_1^e \ln(x)\,dx using integration by parts. State uu and dvdv, and explain how the resulting integral is simpler.

Show the explained solution
  1. Choose u=ln⁡xu=\ln x and dv=dxdv=dx. Then du=dx/xdu=dx/x and v=xv=x; the product v duv\,du simplifies to dxdx.
  2. An antiderivative is xln⁡x−xx\ln x-x. Differentiating gives ln⁡x+1−1=ln⁡x\ln x+1-1=\ln x.
  3. ∫1eln⁡(x) dx=0−(−1)=1.\int_1^e\ln(x)\,dx=0-(-1)=1. The positive result is consistent with ln⁡x>0\ln x>0 for 1<x≤e1<x\le e.

Choose an integration method

2. The radius is a function

Rotate the region under y=xy=\sqrt{x}, above the horizontal axis, for 0≤x≤40\le x\le4, about the horizontal axis. Find the volume using disks.

Show the explained solution
  1. A cross-section perpendicular to the axis has radius r=xr=\sqrt{x}, so its area is πr2=πx\pi r^2=\pi x.
  2. V=π∫04x dx=8π.V=\pi\int_0^4 x\,dx=8\pi.
  3. The answer is in cubic units. Integrating πx\pi\sqrt{x} would miss the square in the disk-area formula.

Integration and volume practice

Use your answers to plan your revision
  1. Method selection

    Explain what gets simpler after your choice of method. Check a proposed antiderivative by differentiating.

  2. Geometric setup

    Sketch one representative cross-section; label its radius and thickness before writing the volume integral.

Bring your section’s assigned integration methods and a diagram you found difficult. We can focus a lesson on method selection or setting up applications.

Request 201-SN3-RE exam preparation help

Original Nablio practice, not an official examination.

Course topic sources

Worked example: returning to the starting point

An object moves along a straight line, with the positive direction to the right. Let tt be time in seconds, v(t)v(t) its signed velocity in metres per second, and s(t)s(t) its position in metres. Using these units, v(t)=t2−2tv(t)=t^2-2t for 0≤t≤30\le t\le3, and s(0)=5s(0)=5. Find its displacement, final position and total distance travelled over these three seconds.

Show a hint

Velocity is the rate of change of position. Its integral gives displacement. Find where v(t)v(t) changes sign before adding the distances travelled in each direction.

Show the reasoning

  1. 1. Choose what to integrate

    Displacement is final position minus initial position. Since s′(t)=v(t)s^{\prime}(t)=v(t), integrate velocity over [0,3][0,3]. To find distance, integrate speed, ∣v(t)∣|v(t)|, so travel in either direction contributes positively. The polynomial can be integrated term by term with the power rule.

  2. 2. Locate the change of direction

    Factor v(t)=t(t−2)v(t)=t(t-2). The velocity is negative for 0<t<20<t<2 and positive for 2<t<32<t<3. The object moves left, stops at t=2t=2, then moves right. Split the distance calculation at t=2t=2. The zero at t=0t=0 is already an endpoint.

  3. 3. Calculate the signed contributions

    Choose the antiderivative F(t)=t33−t2F(t)=\tfrac{t^3}{3}-t^2. Then F(0)=0F(0)=0, F(2)=−43F(2)=-\tfrac43 and F(3)=0F(3)=0. The changes in position are

    ∫02v(t) dt=F(2)−F(0)=−43,∫23v(t) dt=F(3)−F(2)=43.\begin{aligned}\int_0^2 v(t)\,dt&=F(2)-F(0)\\&=-\frac43,\\\int_2^3 v(t)\,dt&=F(3)-F(2)\\&=\frac43.\end{aligned}

    These values are in metres. The negative contribution describes motion to the left; the positive contribution describes motion to the right.

  4. 4. Interpret and check the results

    The displacement is −43+43=0-\tfrac43+\tfrac43=0 metres, so the final position is s(3)=5+0=5s(3)=5+0=5 metres. The total distance, denoted by DD, adds both lengths:

    D=∫03∣v(t)∣ dt=−∫02v(t) dt+∫23v(t) dt=43+43=83 metres.\begin{aligned}D&=\int_0^3|v(t)|\,dt\\&=-\int_0^2v(t)\,dt+\int_2^3v(t)\,dt\\&=\frac43+\frac43=\frac83\text{ metres}.\end{aligned}

    The object returns to its starting position after travelling 8/38/3 metres. On a velocity–time graph, the two regions have equal areas and lie on opposite sides of the time axis. Differentiating FF gives F′(t)=t2−2t=v(t)F^{\prime}(t)=t^2-2t=v(t), which checks the antiderivative.

Your turn: the object finishes somewhere else

For a different straight-line motion, keep the same units and positive direction, but use v(t)=t2−4v(t)=t^2-4 on 0≤t≤30\le t\le3, with initial position s(0)=5s(0)=5. (1) Find when the direction changes. (2) Calculate the displacement and final position. (3) Find the total distance and explain why taking the absolute value of the displacement does not give that distance.

Show a hint

Solve t2−4=0t^2-4=0, keeping only times in the stated interval. Choose an antiderivative, evaluate it at the endpoints and the turning time, then keep the signed contributions separate.

Check the three steps

  1. 1. Find the turning time

    The roots are t=−2t=-2 and t=2t=2; only t=2t=2 lies in [0,3][0,3]. Velocity is negative before 2 seconds and positive afterwards. Split the distance integral at 2 seconds.

  2. 2. Calculate displacement and position

    Let G(t)=t33−4tG(t)=\tfrac{t^3}{3}-4t. This is an antiderivative because G′(t)=t2−4G^{\prime}(t)=t^2-4. Its values are G(0)=0G(0)=0, G(2)=−163G(2)=-\tfrac{16}{3} and G(3)=−3G(3)=-3.

    The displacement is G(3)−G(0)=−3G(3)-G(0)=-3 metres. The object finishes at s(3)=5−3=2s(3)=5-3=2 metres, three metres to the left of its starting position.

  3. 3. Add the two distances

    On [0,2][0,2], the displacement is −16/3-16/3 metres. On [2,3][2,3], it is −3−(−16/3)=7/3-3-(-16/3)=7/3 metres. Therefore

    D=163+73=233 metres.D=\frac{16}{3}+\frac73=\frac{23}{3}\text{ metres}.

    Taking ∣−3∣=3|-3|=3 metres gives the separation between the starting and finishing positions. It leaves out the extra travel caused by reversing direction. For distance, take the magnitude of each signed contribution before adding.

If your distance was zero, revisit the sign on each interval. If a contribution was wrong, check the antiderivative and write the upper-bound value minus the lower-bound value before simplifying.

Work through the difficult step with a tutor

Your tutor can help you identify the quantity being accumulated, label the rate and its units, and sketch where it is positive or negative. Together, you can decide what each bound represents and whether the interval needs to be split before integrating.

For a calculation, your tutor can ask you to explain your method choice, trace a missing factor or endpoint term, and check the antiderivative by differentiation. Then try a changed function or interval yourself and explain how the setup changes.

Plan your review

Choose one integration-method question and one accumulation problem from your section’s review topics. Explain your method choice and check the antiderivative by differentiating. For the application, label the rate and units, decide whether the question asks for net change or a total, and split the interval where needed. Redo a missed setup without the solution. Follow your teacher’s instructions for permitted tools.

Independent practice

Choose your next practice step

35 min

Choose an integration method

Compare substitution, integration by parts and partial fractions when the power rule alone does not apply.

Vanier course information

Before requesting a lesson

Can I request help with 201-SN3-RE?

Yes. Send the topic and your preferred times. You can include a question and any work you have started, but you can also ask for help getting started. Add your course code or outline if helpful. We confirm tutor availability, lesson details and price before payment.

What should I use to prepare for an assessment?

Start with your section’s current topic list, assigned questions and assessment instructions. Bring a question you are allowed to share. Your tutor can help you choose the first step or review a step you cannot explain.

How are lessons delivered?

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

Is Nablio part of Vanier?

Nablio provides independent tutoring and is not affiliated with or endorsed by Vanier College.

Ask for help with your question

Tell us the topic and a few times that work for you. You can include any work you have started. Add the course code or outline if helpful.

Ask about a lesson