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

The table shows the values of a function obtained from an experiment. Estimate ∫_(1)^(7) 𝑓(π‘₯) dπ‘₯ using three equal subintervals with midpoints.

04:07

Video Transcript

The table shows the values of a function obtained from an experiment. Estimate the integral from one to seven of 𝑓 of π‘₯ with respect to π‘₯ using three equal subintervals with midpoints.

The question is asking us to estimate the value of the integral from one to seven of 𝑓 of π‘₯ with respect to π‘₯. It wants us to do this by using three equal subintervals with midpoints as our sample points. And instead of being given our function 𝑓 of π‘₯, we’re given a table obtained experimentally. It gives us our values of π‘₯ and their respective outputs 𝑓 of π‘₯.

Let’s start by recalling what it means to estimate our integral by using three equal subintervals with midpoints as our sample points. First, let’s sketch an approximation of our function 𝑓 of π‘₯. Our function 𝑓 of π‘₯ might look something like this. Remember, it’s not important that we get this sketch perfectly. We’re just using this to help us estimate our integral by using three subintervals with midpoints as the sample points.

Our integral ranges from one to seven, so our first value of π‘₯ will be one and our last value of π‘₯ will be seven. So, we need to split this interval up into three subintervals of equal width. We call the width of our subintervals Ξ”π‘₯. And we know this is the length of our internal divided by the number of subintervals. This is equal to 𝑏 minus π‘Ž divided by 𝑛. And we can just calculate this. We know that our number of subintervals 𝑛 is three and the length of our line is seven minus one. This gives us seven minus one divided by three, which is, of course, equal to two.

We can then approximate our integral by finding the area of these three rectangles. The width of these rectangles is Ξ”π‘₯. And the height of these rectangles is our function 𝑓 of π‘₯ evaluated at the midpoints of our subintervals. One easy way to collect all of this information is to use a table. We’ll find our sample points, which are the midpoints of each of these three intervals. We’ll then evaluate our function 𝑓 at these three sample points. Finally, we’ll calculate the area of each of these rectangles. It will be positive if our rectangle is above the π‘₯-axis and negative if our rectangle is below the π‘₯-axis.

And this is calculated by the width of the rectangle Ξ”π‘₯ times 𝑓 of π‘₯ 𝑖. Let’s start by finding the endpoints for our first subinterval. We start at one and then we go up Ξ”π‘₯ units. And we know Ξ”π‘₯ is equal to two. So, our first subinterval ranges from one to three. We can then do the same to find our second subinterval. It starts at three and then goes up by Ξ”π‘₯, which is two. So, the second subinterval is from three to five. And we can do the same for our third subinterval. It ranges from five to seven.

We need to find the midpoint of each of these subintervals. We can probably do this by eye, but we could also just add our two endpoints together and divide by two. We have three plus one all divided by two is equal to two. Five plus three all divided by two is equal to four. And seven plus five all divided by two is equal to six.

Next, we need to find our function 𝑓 evaluated at each of our midpoints. We haven’t been given our function 𝑓, so we need to use our table. First, when π‘₯ is equal to two, we see that 𝑓 of π‘₯ is equal to negative 2.1. Next, when π‘₯ is equal to four, we can see that 𝑓 of π‘₯ is equal to negative 0.1. Finally, when π‘₯ is equal to six, we can see that 𝑓 of π‘₯ is equal to 2.1.

Now, we want to calculate the signed area of our rectangles. Remember, our width, Ξ”π‘₯, is equal to two. So, we’re just multiplying 𝑓 of π‘₯ 𝑖 by two in each case. We get negative 2.1 times two is negative 4.2, negative 0.1 times two is negative 0.2, and 2.1 times two is 4.2.

Finally, we just need to add these areas together. We see 4.2 minus 4.2 is equal to zero. So, this sums to give us negative 0.2. And remember, the sum of these three areas is our approximation of the integral.

Therefore, by using a Riemann’s sum with midpoints as our sample points and three equal subintervals, we’ve shown the integral from one to seven of 𝑓 of π‘₯ with respect to π‘₯ is approximately equal to negative 0.2.

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