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Use the Trapezoidal Rule with $ n = 10 $ to approximate $ \displaystyle \int_0^{20} \cos (\pi x)\ dx $. Compare your result to the actual value. Can you explain the discrepancy?

$$T_{10}=20, \quad \text { actual }=0$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 7

Approximate Integration

Integration Techniques

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Okay. This question wants us to approximate this integral. Using the trap is lead role and compare that to the actual result. So it wants us to use the trap is owed a room with an equal to 10 which says that area is approximately Delta X over too times f off the starting point plus two times the middle values plus f of the other end point. Because again, we're double counting the areas in the middle during this approximation. So doing this, we see the area is approximately Well, what is Delta X? So let's do a little aside here. Delta X. Cols 20 minus zero over 10 which is equal to so it's equal to two over two. So we're not even gonna write anything half of zero plus f of two plus Sorry to effort to plus two of four plus all the way up to yes of 20. So if we do this, we see that our area equals one plus two plus two plus two plus plus one based on this formula. So we have nine twos in the middle, plus two from the end. So we get that area is approximately equal to 20 by the Trap Ysidro. But now let's see what the actual result is. Integral from 0 to 20 co sign of pie Axe D X is well, the anti derivative of co sign is signed. And then we have to divide by the chin loo factor of Tai. We're evaluating that from 0 to 20 and that's just one over pi times sign of 20 pie minus side of zero. And both of these air zero. So it says the rial area zero. So why is there a discrepancy here? Well, here's what's happening. If we look at our graph so co sign of pie axe is going to look something like this. Actually, I'll draw bigger period her smaller period so we can really see what's going on here. So with our trap resides, basically, we are. We're only hitting the positive portions of this graph because our sub intervals are too big. So we're only hitting these positive peaks soar, missing all this negative portion. So, for example, the area of CO sign of pie axe over these periods, if we just want the absolute value of co sign X, it should be 40 because we're only retrieving half the area here. We're just not take into account the negatives. So since Delta X is too large, we, Ms Oh, of the negative area, So that's a danger of approximation. You have to make sure your Delta X is small enough to catch sensitivities like that in the graph.

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