 In this video, we're going to look at the solution of to question five from the midterm practice exam for calculus to math 1220 here. So here we have a table of values for some increasing function f. So our function f is given by this table of values. Using this table, we want to approximate the integral from 10 to 30 of f of x dx using the trapezoidal rule t five. That is, we're going to use the trapezoidal rule with five subdivisions. So some things to note is that 10 is the boundary on the left. So this is our easy or a value 30 is the boundary on the right, which is our B value. And so to approximate using the trapezoidal rule t five, what we're going to do is we're going to take delta x over two times this by f of x zero plus two times f of x one plus two times f of x two. We're going to set in all those parentheses there. We're going to get two times f, oops, I did it again, f of x three plus two times f of x four plus plus one times f of x five like so. So some things to note here is that the a value this is exactly what we mean by x zero. So we already know that this one right here is x zero, and B is going to be our x five right here. At some point we need to identify what is delta x right delta x. It's going to be the last value B minus minus the first value a over five in this case. Well, I mean, in general, it would be over n. But if we use the specifics of this question, we're going to get 30 minus 10 over five 10 take, take away from 30 is 2020 divided by five gives me four. And so we're going to take an increment of four as we go through these. And so if you look at that, you can see that 10 plus four is 14. This is my x one plus four is 18. That's going to be x two plus four is 22 that's going to be your x three plus four is going to be 26 which is x four and then plus four again gives you 30. Now in this situation, the the six values they gave you are exactly the values you need it for this one don't automatically assume that on the test, it actually might give you extra information that you don't need for this problem here. All right, what do we do next? Will we start plugging these values in here. So we're going to get delta x, which was four over two, we'll come back to that one, we're going to take f of x zero. That's this number right here we get a negative 12, like so, we're going to get two times f of x one, which is negative six, we're going to get two times f of x two, which is a negative two. We're going to get two times f of x three, that is f of 22, which is one, then we get two times f of x four, that is f of 26, which is three. And then lastly, we're going to get one times f of x five. x five was 30. So we look at f of five, which is eight, like so. So let's try to simplify these things for goes into two, two times. So we get two right there. We can negative 12. We're going to get two times negative six, which is another negative 12. We're going to get two times negative two, which is a negative four. Two times one, which is a two, two times three, which is a six, and then just an eight to follow up there. And so try to add some of these things together. Notice if we take 12 and 12, that's a 24 minus another negative four, that gives me 20. What did I say before? Sorry, 28, negative 28, 24 and four there. If we add the positives together, two and eight are 10 plus six would be 16, like so. So now this is the hard part of the problem. It's called two digit arithmetic. Yauzer, I might embarrass myself here. But if we do this, 28 take away 16 is going to give us 12, that should be a negative 12, like so times that by two, we're looking for a negative 24, which would give us as our final answer, right here. And so on this question, we did the function for T five, the trapezoidal rule, but on the test, you could be asked to do this with the midpoint rule or the Simpson rule. So using the same table, I would actually recommend computing the midpoint rule or the Simpson's rule and see what you get there. You'll notice that some of those answers are listed as distractors on this one.