## Tuesday, March 27, 2018

### Teach 180: Discovering the Antiderivative (Day 128)

Today in Calculus students were introduced to the idea of the antiderivative.  I wanted them to discover how to take the antiderivative of xn where n ≠ -1.  In addition, I hoped they would discover the reason why +C is necessary when finding an indefinite integral.  I began by having students fold a piece of paper into thirds.  On the top third, they wrote a polynomial function of their choice.  On the middle third, they wrote the derivative.  Then, I had them fold the paper in such a way that the original function was inside.

Next, students traded folded papers with their neighbors and were instructed to find the original function - the one that was now folded inside the paper.  After they had written down what they thought f(x) was, they were allowed to open the paper and see if their function matched the original function.  Here, you can see a sample paper.

This led to the discussion of three main ideas.  First, did you get an exact match to the original function?  If not, why not.  If the original polynomial function had a non-zero constant term, then it was very likely that there was not an exact match.  This led us to adding +C to show that there could be constant added to the end of the original function.  Tomorrow we will consider this as a family of functions.

Second, how did you actually figure out the exponent and coefficient in front of the exponent when you were trying to undo the derivative?  What procedure did you use?  Most groups could articulate adding 1 to the exponent.  But only two groups were able to describe dividing by the new power.

Finally, we looked at why n ≠ -1 when we wrote our general rule for taking the antiderivative  of xn. What would happen if we added 1 to the power and divided by the new power?  Raising a number to the zero power isn't a problem, but dividing by zero IS a problem.  It was at that point that a student recalled that the derivative of ln(x) is 1/x.  So, the antiderivative of 1/x would be ln(x).