Calculate the left endpoint estimate of the definite integral between zero and four of 𝑥 squared plus two with respect to 𝑥 with 𝑛 equals two subintervals. Is the result an overestimate or underestimate of the actual value?
Remember, we can estimate the solution to definite integrals by considering the area between the curve and the 𝑥-axis in terms of rectangles. These are called Riemann sums. And here we’re looking to find the left Riemann sum. When working with left Riemann sums, we take values of 𝑖 from zero to 𝑛 minus one. And an approximation to our definite integral is the sum of Δ𝑥 times 𝑓 of 𝑥𝑖 for values of 𝑖 between zero and 𝑛 minus one, where Δ𝑥 is 𝑏 minus 𝑎 over 𝑛. Remember, 𝑛 is the number of subintervals and 𝑥𝑖 is equal to 𝑎 plus 𝑖 lots of Δ𝑥.
If we looked carefully at our integral, we see that the lower limit is zero and the upper limit is four. So we’re going to let 𝑎 be equal to zero and 𝑏 be equal to four. And also, we’re told that 𝑛 is equal to two. The first step with any Riemann sum is usually to work out the value of Δ𝑥. 𝑎 is zero, 𝑏 is four. So Δ𝑥 is four minus zero divided by 𝑛, which is two. And that tells us that Δ𝑥 is equal to two. Once we’ve worked out Δ𝑥, we can then usually work out an expression for 𝑥𝑖. It’s 𝑎 which we know to be zero plus Δ𝑥, which is two lots of 𝑖. Well, that simplifies to two 𝑖.
And, of course, for the summation formula, we actually need to work out 𝑓 of 𝑥𝑖. Well, it follows that 𝑓 of 𝑥𝑖 must be equal to 𝑓 of two 𝑖. So let’s substitute two 𝑖 into our function. That’s two 𝑖 squared plus two. That’s four 𝑖 squared plus two. And we now have everything we need to approximate our integral. We take values of 𝑖 in our sum from zero to 𝑛 minus one. Well, two minus one is one. Then, it’s Δ𝑥 times 𝑓 of 𝑥𝑖, which we just found to be four 𝑖 squared plus two.
It’s worth knowing that we can take out any constant factors independent of 𝑖. Well, four is independent of 𝑖. So we can write this as four times the sum of two 𝑖 squared plus one for values of 𝑖 between zero and one. Let’s now substitute 𝑖 equals zero and 𝑖 equals one into this expression and find its sum. That’s four lots of two times zero squared plus one plus two times one squared plus one, which is 16. And so, we found an estimation of this definite integral.
Our next step is to work out whether this is an overestimate or underestimate of the exact value. Well, we need to decide whether 𝑥 squared plus two is an increasing or decreasing function over our interval. To do this, we’re going to differentiate 𝑥 squared plus two. And we get two 𝑥. Now, of course, this is greater than zero for values of 𝑥 greater than zero. So this tells us the function 𝑥 squared plus two must be increasing for values of 𝑥 greater than zero. What this says for our curve is that it’s sloping upwards for values of 𝑥 greater than zero.
Well, when working with increasing functions, the left Riemann sum gives an underestimate, whereas the right Riemann sum gives an overestimate for the approximation of the integral. The reverse, of course, is true for decreasing functions. And so, this means the left endpoint estimate of our definite integral is an underestimate.