Question Video: Integrating Functions Numerically - Trapezoidal Rule Mathematics

Video Transcript

Calculate the trapezoidal rule estimate of the definite integral of 𝑥 squared plus two with respect to 𝑥 evaluated between zero and four with 𝑛 equals two subintervals. Is the result an overestimate or underestimate of the actual value?

Remember, the trapezoidal rule says that we can estimate the value of the definite integral of some function 𝑓 of 𝑥 between 𝑎 and 𝑏 by splitting it into 𝑛 subintervals. And, this approximation is given by Δ𝑥 over two of 𝑓 of 𝑥 nought plus 𝑓 of 𝑥 𝑛 plus two times 𝑓 of 𝑥 one plus 𝑓 of 𝑥 two all the way up to 𝑓 of 𝑥 𝑛 minus one. Here, Δ𝑥 is equal to 𝑏 minus 𝑎 over 𝑛. And 𝑥 subscript 𝑖 is equal to 𝑎 plus 𝑖 times Δ𝑥. In other words, to find our values of 𝑥𝑖, we add 𝑖 lots of Δ𝑥 to the lower limit for our integral 𝑎.

Let’s break this down and just begin by working out the value of Δ𝑥. 𝑛 is equal to two. And the lower limit for our integral is zero, so 𝑎 is equal to zero. And the upper limit is four, so 𝑏 must be equal to four in this example. Δ𝑥 is therefore equal to four minus zero over two, which is equal to two. The values for 𝑓 of 𝑥 nought, 𝑓 of 𝑥 one, and so on require little more work. But we could make this process as simple as possible by adding a table. In this table, the 𝑥-values run from 𝑎 to 𝑏, that’s from zero to four. And, the ones in between are found by repeatedly adding Δ𝑥, here that’s two, to 𝑎 which is zero. So this value here is two. And, we now see that we have our two strips with a width of two units. We’re now going to substitute zero, two, and four into our function to find the 𝑓 of 𝑥 values.

Our function is 𝑥 squared plus two. So, 𝑓 of zero is zero squared plus two, which is two. 𝑓 of two is two squared plus two, which is six. And 𝑓 of four is four squared plus two. Here, that’s 18. So, we have everything we need to perform the trapezoidal rule estimate of the definite integral of 𝑥 squared plus two with respect to 𝑥 between the values of zero and four. It’s Δ𝑥 over two, which is two over two, times our first 𝑓 of 𝑥 value and our last 𝑓 of 𝑥 value, that’s two plus 18, plus two lots of essentially everything else. And here, that’s just two lots of six. That gives us one times 32, which is clearly 32. And so, the approximation for the results of the definite integral of 𝑥 squared plus two evaluated between zero and four is 32.

And all that’s left is for us to decide whether this is an under or an overestimate. And, there are two ways we can perform this. We could actually perform the integral of 𝑥 squared plus two between the limits of zero and four. Or alternatively, we can consider the shape of the curve. We know the graph looks a little something like this. We’re interested in the area between the curve and the 𝑥-axis between zero and four. That’s the two trapeziums shown. Notice how these trapeziums overlap the area we’re actually trying to find. And this means our approximation for the value of the integral of 𝑥 squared plus two between zero and four must be an overestimate of the actual value. The solution here is 32, which is an overestimate.