Question Video: Estimating the Definite Integration of a Function in a Given Interval by Dividing It into Subintervals and Using the Left Endpoint of the Subintervals | Nagwa Question Video: Estimating the Definite Integration of a Function in a Given Interval by Dividing It into Subintervals and Using the Left Endpoint of the Subintervals | Nagwa

Question Video: Estimating the Definite Integration of a Function in a Given Interval by Dividing It into Subintervals and Using the Left Endpoint of the Subintervals Mathematics

The table shows the values of a function obtained from an experiment. Estimate ∫_(5) ^(17) 𝑓(π‘₯) dπ‘₯ using three equal subintervals with left endpoints.

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Video Transcript

The table shows the values of a function obtained from an experiment. Estimate the definite integral between five and 17 of 𝑓 of π‘₯ with respect to π‘₯ using three equal subintervals with left endpoints.

Remember, we can estimate a definite integral by using Riemann sums. In this case, we’re estimating the integral between five and 17 of 𝑓 of π‘₯. Now, it doesn’t really matter that we don’t know what the function is. We have enough information in our table to perform the left Riemann sum. The left Riemann sum involves taking the heights of our rectangles as the function value at the left endpoint of the subinterval. We want to use three equally sized subintervals. So let’s recall the formula that allows us to work out the size of each subinterval, in other words, the widths of the rectangle.

It’s Ξ”π‘₯ equals 𝑏 minus π‘Ž over 𝑛, where π‘Ž and 𝑏 are the endpoints of our interval and 𝑛 is the number of subintervals. In our case, we’re looking to evaluate the definite integral between five and 17. So we let π‘Ž be equal to five and 𝑏 be equal to 17. And we want three equal subintervals. So we’ll let 𝑛 be equal to three. Ξ”π‘₯ is then 17 minus five all divided by three, which is simply four. Then when writing a left Riemann sum, we take values of 𝑖 from zero to 𝑛 minus one. It’s the sum of Ξ”π‘₯ times 𝑓 of π‘₯𝑖 for values of 𝑖 from zero to 𝑛 minus one. π‘₯𝑖 is π‘Ž plus 𝑖 lots of Ξ”π‘₯. In this case, we know that π‘Ž is equal to five, and Ξ” π‘₯ is equal to four. So our π‘₯𝑖 value is given by five plus four 𝑖.

Well, since we’re using the left Riemann sum, we begin by letting 𝑖 be equal to zero. We need to work out π‘₯ zero. It’s five plus four times zero, which is simply five. We can find 𝑓 of π‘₯ nought in our table. It’s negative three. Next, we let 𝑖 be equal to one. And we get π‘₯ one to be five plus four times one, which is nine. We look up the value π‘₯ equals nine in our table. And we see that 𝑓 of nine is negative 0.6. Next, we let 𝑖 be equal to two. And remember, we’re looking for values of 𝑖 up to 𝑛 minus one. Well, three minus one is two. So this is the last value of 𝑖 we’re interested in. This time, that’s five plus four times two which is 13. We look up π‘₯ equals 13 in our table. And we get that 𝑓 of 13 and 𝑓 of π‘₯ two is 1.8.

Then, according to our summation formula, we find the sum of the products of Ξ”π‘₯ and these values of 𝑓 of π‘₯𝑖. And so, an estimate for our definite integral is four times negative three plus four times negative 0.6 plus four times 1.8, which is negative 7.2. An estimate for the definite integral between five and 17 of 𝑓 of π‘₯ with respect to π‘₯ using three equal subintervals is negative 7.2.

Now, we don’t need to worry here that our answer is negative. Remember, when we’re working with Riemann sums, we’re looking at areas. But when the function values are negative, the rectangle sits below the π‘₯-axis. And so, its area is subtracted.

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