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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. We want to show that as n gets large, the limit of u_n = 0. OK – here we start by trying to get a lower and upper bound for u_n. We already have shown it’s an absolute decreasing sequence, so we just now need to show the limit of the sequence is 0.
#Answer of maths hl ib free series#
Let’s check the requirements for proving an alternating series converges: 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. OK, by now most students will have probably given up in despair – and the next part doesn’t get much easier. Makes you wonder what the point of that question was, but let’s move on. This then gives us the inequality we desire.ĭon’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. The last integral then notes that the integral of sint/(t+pi) will be less than the integral of sint/t. The fourth integral has also noted that we can simply replace T with t to produce an equivalent integral. Therefore the absolute value of the integral of y = -sinx/x will be the same as the absolute integral of y = sinx/x. This is the same as y = sinx/x but reflected in the x axis. The fourth integral then uses graphical logic. The third integral uses the fact that sin(T + pi) = – sin(T). 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: dt becomes dT when we differentiate t = T + pi. Hopefully the first 2 equalities make sense – we replace n with n+1 and then replace t with T + pi. Instead this is how Daniel’s method progresses: (We will look later at how it can be integrated – it gives something called the Si(x) function). This means that there is no elementary function or standard basic integration method that will integrate it. 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. It looks like integration by parts might work on this. 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. Since our sum consists of alternating positive and negative terms, then we have an alternating series. When we integrate from pi to 2pi the graph is below the x axis and so the integral is negative. 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. Using the GDC we can find that the roots of this function are n(pi). This is the plot of y = sinx/x from 0 to 6pi. The summation of the first n terms will add the answers to the first n integrals together. When n = 1 we have the single integral from pi to 2pi of sint/t. So when n = 0 we have the single integral from 0 to pi of sint/t. 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 let’s try and understand it!įirst 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. As such it was a terrible exam question – but would make a very interesting exploration topic. You could make a case for this being the most difficult IB HL question ever.
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It’s worth nearly a quarter of the entire marks – and is well off the syllabus in its difficulty. This was the last question on the May 2016 Calculus option paper for IB HL.
#Answer of maths hl ib free pdf#
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