This was the last question on the May 2016 Calculus option paper for IB HL. It’s worth nearly a quarter of the entire marks – and is well off the syllabus in its difficulty. You could make a case for this being the most difficult IB HL question ever. As such it was a terrible exam question – but would make a very interesting exploration topic. So let’s try and understand it!
First I’m going to go through a solution to the question – this was provided by another HL maths teacher, Daniel – who worked through a very nice answer. For the first part of the question we need to try and understand what is actually happening – we have the sum of an integral – where we are summing a sequence of definite integrals. So when n = 0 we have the single integral from 0 to pi of sint/t. When n = 1 we have the single integral from pi to 2pi of sint/t. The summation of the first n terms will add the answers to the first n integrals together.
This is the plot of y = sinx/x from 0 to 6pi. Using the GDC we can find that the roots of this function are n(pi). This gives us the first mark in the question – as when we are integrating from 0 to pi the graph is above the x axis and so the integral is positive. When we integrate from pi to 2pi the graph is below the x axis and so the integral is negative. Since our sum consists of alternating positive and negative terms, then we have an alternating series.
Part (b i)
This is where it starts to get difficult! You might be tempted to try and integrate sint/t – which is what I presume a lot of students will have done. It looks like integration by parts might work on this. However this was a nasty trap laid by the examiners – integrating by parts is a complete waste of time as this function is non-integrable. This means that there is no elementary function or standard basic integration method that will integrate it. (We will look later at how it can be integrated – it gives something called the Si(x) function). Instead this is how Daniel’s method progresses:
Hopefully the first 2 equalities make sense – we replace n with n+1 and then replace t with T + pi. dt becomes dT when we differentiate t = T + pi. In the second integral we have also replaced the limits (n+1)pi and (n+2)pi with n(pi) and (n+1)pi as we are now integrating with respect to T and so need to change the limits as follows:
t = (n+1)(pi)
T+ pi = (n+1)(pi)
T = n(pi). This is now the lower integral value.
The third integral uses the fact that sin(T + pi) = – sin(T).
The fourth integral then uses graphical logic. y = -sinx/x looks like this:
This is the same as y = sinx/x but reflected in the x axis. Therefore the absolute value of the integral of y = -sinx/x will be the same as the absolute integral of y = sinx/x. The fourth integral has also noted that we can simply replace T with t to produce an equivalent integral. The last integral then notes that the integral of sint/(t+pi) will be less than the integral of sint/t. This then gives us the inequality we desire.
Don’t worry if that didn’t make complete sense – I doubt if more than a handful of IB students in the whole world got that in exam conditions. Makes you wonder what the point of that question was, but let’s move on.
Part (b ii)
OK, by now most students will have probably given up in despair – and the next part doesn’t get much easier. First we should note that we have been led to show that we have an alternating series where the absolute value of u_n+1 is less than the absolute value of u_n. Let’s check the requirements for proving an alternating series converges:
We already have shown it’s an absolute decreasing sequence, so we just now need to show the limit of the sequence is 0.
OK – here we start by trying to get a lower and upper bound for u_n. We want to show that as n gets large, the limit of u_n = 0. In the second integral we have used the fact that the absolute value of an integral of a function is always less than or equal to the integral of an absolute value of a function. That might not make any sense, so let’s look graphically:
This graph above is y = sinx/x. If we integrate this function then the parts under the x axis will contribute a negative amount.
But this graph is y = absolute (sinx/x). Here we have no parts under the x axis – and so the integral of absolute (sinx/x) will always be greater than or equal to the integral of y = sinx/x.
To get the third integral we note that absolute (sinx) is bounded between 0 and 1 and so the integral of 1/x will always be greater than or equal to the integral of absolute (sinx)/x.
We next can ignore the absolute value because 1/x is always positive for positive x, and so we integrate 1/x to get ln(x). Substituting the values of the definite integral gives us a function of ln which as n approaches infinity approaches 0. Therefore as this limit approaches 0, and this function was always greater than or equal to absolute u_n, then the limit of absolute u_n must also be 0.
Therefore we have satisfied the requirements for the Alternating Series test and so the series is convergent.
Part (c) is at least accessible for level 6 and 7 students as long as you are still sticking with the question. Here we note that we have been led through steps to prove we have an alternating and convergent series. Now we use the fact that the sum to infinity of a convergent alternating series lies between any 2 successive partial sums. Then we can use the GDC to find the first few partial sums:
And there we are! 14 marks in the bag. Makes you wonder who the IB write their exams for – this was so far beyond sixth form level as to be ridiculous. More about the Si(x) function in the next post.