Video: Estimating Definite Integrals Using the Trapezoidal Rule

Use the trapezoidal rule to estimate ∫_(0)^(1) √(π‘₯ + 1) dπ‘₯ using four subintervals. Round your answer to three decimal places.

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Video Transcript

Use the trapezoidal rule to estimate the definite integral between the limits of zero and one of the square root of π‘₯ plus one 𝑑π‘₯ using four subintervals. Round your answer to three decimal places.

Remember, the trapezoidal rule says that we can find an estimate for the definite integral evaluated between π‘Ž and 𝑏 of 𝑓 of π‘₯ by splitting the area between the curve and the π‘₯-axis into 𝑛 subintervals. The formula we will require is Ξ”π‘₯ over two times 𝑓 of π‘₯ naught plus 𝑓 of π‘₯ 𝑛 plus two lots of 𝑓 of π‘₯ one all the way up to 𝑓 of π‘₯ 𝑛 minus one. And here Ξ”π‘₯ is given by 𝑏 minus π‘Ž over 𝑛 and π‘₯𝑖 is given by π‘Ž plus 𝑖 times Ξ”π‘₯.

Let’s break this down and just begin by working out the value of Ξ”π‘₯. Contextually, Ξ”π‘₯ is the width of each of our subintervals. In this question, we’re working with four subintervals. So 𝑛 is equal to four. π‘Ž is the lower limit of our integral. It’s zero. And 𝑏 is the upper limit. It’s one. Ξ”π‘₯ is therefore equal to one minus zero over four.

We know that to be equal to a quarter or 0.25. The values for 𝑓 of π‘₯ naught, 𝑓 of π‘₯ one, and so on require little more work. But we can make this as simple as possible by adding a table. The π‘₯-values in our table run from π‘Ž to 𝑏. That’s zero to one. And it’s worth noting that there will always be one more π‘₯ value than the value of 𝑛. So here I’ve included five columns. The other values of π‘₯ are found by repeatedly adding Ξ”π‘₯ β€” that’s 0.25 β€” to π‘Ž, which is zero. So they are 0.25, 0.5, and 0.75. And this gives us our four strips of width 0.25 units.

We’re now going to substitute each of these π‘₯-values into our function. We begin with 𝑓 of naught. That’s the square root of zero plus one, which is simply one. 𝑓 of 0.25 is the square root of 0.25 plus one. This is root five over two. Now we could actually use a decimal value here. But for accuracy, it will be important to include at least six decimal places. We then have 𝑓 of 0.5. That’s the square root of 0.5 plus one, which is root six over two. Repeating this process for 0.75 and one, we end up with root seven over two and root two, respectively.

Let’s substitute everything we have into the formula for the trapezoidal rule. Ξ”π‘₯ over two is 0.25 over two. 𝑓 of π‘₯ naught and 𝑓 of π‘₯ 𝑛 are one and root two. We have two lots of everything else. That’s root five over two, root six over two, and root seven over two. And typing this into our calculator, we get a value of 1.21819 and so on. Correct to three decimal places, that gives us an estimate for the definite integral between zero and one of the square root of π‘₯ plus one as 1.218.

And at this stage, we do have a couple of ways that we can check our solution. We could type the exact problem into our calculator. And when we do, we get a value of 1.21895 and so on. This is, of course, extremely close to the solution we got. Alternatively, we could use integration by substitution to evaluate the exact integral.

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