Question Video: Finding the Definite Integration of a Rational Function in a Given Interval by Dividing It into Subintervals and Using the Midpoints of the Subintervals Mathematics • Higher Education

Using the midpoint rule with 𝑛 = 5, round ∫_(2) ^(5) (2π‘₯/(3π‘₯ + 2)) dπ‘₯ to four decimal places.


Video Transcript

Using the midpoint rule with 𝑛 equals five, round the definite integral from two to five of two π‘₯ over three π‘₯ plus two with respect to π‘₯ to four decimal places.

Remember, we can estimate a definite integral by using Riemann sums. We split the region into subintervals and create a rectangle in each. The total area of the rectangles gives us an estimate to the integral. In a midpoint Riemann sum, the height of each rectangle is equal to the value of the function at the midpoint of its base. Now, working with midpoints isn’t quite as nice as using the left or right Riemann sum. In this case, there’s nothing to stop us evaluating each of the areas in term. And let’s begin by working out the width of each subinterval.

Geometrically, it tells us the width of the rectangles. And it’s given by Ξ”π‘₯ equals 𝑏 minus π‘Ž over 𝑛, where π‘Ž and 𝑏 are the endpoints of the interval and 𝑛 is the number of subintervals. In our case, our lower limit is two and our upper limit is five. We, therefore, let π‘Ž be equal to two and 𝑏 be equal to five. And we’re told that 𝑛 is equal to five. Ξ”π‘₯ is five minus two over five, which is three-fifths or 0.6. And a table can make the next step a little easier. In our table, we’re going to begin by working out each of the subintervals.

We know the lower limit of our definite integral to be two. To find the right endpoint of our first rectangle on our first subinterval, we add 0.6 to two to get 2.6. This means our next rectangle starts at π‘₯ equals 2.6. This time, we add 0.6 again. And we find the right endpoint to be 3.2. The left endpoint of our next rectangle must, therefore, be 3.2. And the right endpoint is 3.2 plus Ξ”π‘₯. That’s 3.8. Our next rectangle begins at 3.8. And adding 0.6, we find that it ends at 4.4. And our fifth and final rectangle β€” Remember, we wanted 𝑛 to be equal to five β€” begins at 4.4 and then ends at 4.4 plus 0.6, which is five. And that’s a really good start because we know the upper limit of our interval is indeed five.

Next, we’re going to work out the midpoint of each of these subintervals. Now, we can probably do this in our head. But if we’re struggling, we add the two values and divide by two. And when we do, we obtain the midpoints to be 2.3, 2.9, 3.5, 4.1, and 4.7, respectively. To work out the height of each rectangle, we need to work out the value of the function at these points. So, for example, in this first row, we’ll start by working out 𝑓 of 2.3. To do that, we substitute 2.3 into our function two π‘₯ over three π‘₯ plus two. And we get 46 over 89. Then, we repeat this process for π‘₯ equals 2.9. When we substitute 3.5 into our function, we get 14 over 25. And the height of our final two rectangles are 82 over 143 and 94 over 161 units, respectively.

And our very final step is to calculate the area of each rectangle by multiplying its width by its height. Of course, the width of each rectangle is Ξ”π‘₯ at 0.6. So we multiply each of these function values by 0.6 and then find their sum. Now, we could do this in turn or we could find their sum and multiply by 0.6. We’ll get the same answer. When we do find the total of all the values in our column titled Ξ”π‘₯ times 𝑓 of π‘₯𝑖, we get 1.66571 and so on. And if we run this correct to four decimal places, we find that an estimate that the definite integral between two and five of two π‘₯ over three π‘₯ plus two with respect to π‘₯ is roughly 1.6657.

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