Inside Our Calculus Courseware: Integration by Parts

The integration by parts method is not straightforward. It requires some thought, and the student must make two initial choices. Successfully working the exercises demands these choices be wise ones. And it may be necessary to repeat the process. Students often think they’ve failed if one application doesn’t yield the solution.

Sample Problem #1

Below is an example of a question where integration by parts is applied twice. Note that in the directions of the question we point out that it might be used more than once.

7.1-Integration-by-PartsA.png

Solution

In the Practice mode, students have access to learning aids to help them understand how to tackle each problem. For example, they can choose to view the solution to the problem in the Tutor area.

The solution to this problem clearly explains how and why we pick u and dv and shows all the steps that take place to get the final answer.

7.1-Integration-by-PartsBCDE.png

Step-by-Step

If students want to try answering the problem, but they do not know where to start, they have access to Step-by-Step. Step-by-Step provides a step-by-step breakdown of the problem, walking the student through the problem in manageable pieces. While it provides plenty of guidance, the Step-by-Step portion does ask the student to input the results of each step so they are learning as they go.

Below is Step 1, which reminds the student that their choices for u and dv should be made with the goal of producing a simpler integral.

7.1-Integration-by-PartsFG.png

Once the choices for u and dv are made, in Step 2 the student needs to find du and v.

7.1-Integration-by-PartsH.png

In Step 3 the given integral is rewritten based on the method of integration by parts, and the student is being prompted and guided that integration by parts needs to be applied again. Therefore, the student needs to determine u and dv for the new integral resulting from the first application of the integration by parts method.

7.1-Integration-by-PartsIJ.png

Once the choices for u and dv for the new integral to be evaluated by integration by parts were made, in Step 4 the student needs to find du and v.

7.1-Integration-by-PartsK.png

In Step 5 the intermediate integral is rewritten based on the method of integration by parts and the student is prompted to evaluate it.

7.1-Integration-by-PartsL.png

In the last step, the student puts together all the pieces found in the previous steps to find the result of the given integral.

7.1-Integration-by-PartsM.png

Explain Error

Another helpful learning aid provided in Hawkes’ courseware is Explain Error, which gives students precise feedback from the system’s artificial intelligence. It anticipates and
diagnoses specific errors, stopping students in their tracks and showing them not only that their answer is incorrect, but why it is incorrect.

Let’s say the student forgets to use C for the constant of integration.

7.1-Integration-by-PartsN.png

When the student selects Explain Error, they receive this detailed feedback:

7.1-Integration-by-PartsO.png

After the student reads the explanation, they can go back into Practice to add the constant of integration C:

7.1-Integration-by-PartsP.png

Now, when applying the integration by parts the second time, let’s say the student makes a mistake in the sign of the antiderivative of sint. So, instead of having v = – cos t, the student writes v = cos t.

7.1-Integration-by-PartsQ.png

This sign mistake leads to the following incorrect answer and the corresponding explanation.

7.1-Integration-by-PartsR.png

7.1-Integration-by-PartsS.png

After the student reads the Explain Error explanation, they can go back into Practice to modify their answer.

7.1-Integration-by-PartsT.png


Sample Problem #2

Below is a new question, which can be solved by different methods: integration by parts or u-substitution.

7.1-Integration-by-PartsU.png

If the student were to choose u = x and dv = ln(2x2)dx, then v is very difficult to find and the integral becomes more complicated. Therefore, the best choices in this case are u = ln(2x2) and dv = xdv.

7.1-Integration-by-PartsV.png

Solution

Below is the thorough solution. Note that this question also can be solved by starting with u-substitution. Our solution first shows the method of integration by parts, then it shows the u-substitution method.

7.1-Integration-by-PartsWXYZ.png


Calc Book and Computer

Interested in seeing more of the calculus question bank? Contact us today at info@hawkeslearning.com or 1-800-426-9538 to get free access to the student courseware!

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