### Video Transcript

Calculate the left endpoint estimate of the definite integral between zero and four of π₯ squared plus two with respect to π₯ with π 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 π₯-axis in terms of rectangles. These are called Riemann sums. And here weβre looking to find the left Riemann sum. When working with left Riemann sums, we take values of π from zero to π minus one. And an approximation to our definite integral is the sum of Ξπ₯ times π of π₯π for values of π between zero and π minus one, where Ξπ₯ is π minus π over π. Remember, π is the number of subintervals and π₯π is equal to π plus π lots of Ξπ₯.

If we looked carefully at our integral, we see that the lower limit is zero and the upper limit is four. So weβre going to let π be equal to zero and π be equal to four. And also, weβre told that π is equal to two. The first step with any Riemann sum is usually to work out the value of Ξπ₯. π is zero, π is four. So Ξπ₯ is four minus zero divided by π, which is two. And that tells us that Ξπ₯ is equal to two. Once weβve worked out Ξπ₯, we can then usually work out an expression for π₯π. Itβs π which we know to be zero plus Ξπ₯, which is two lots of π. Well, that simplifies to two π.

And, of course, for the summation formula, we actually need to work out π of π₯π. Well, it follows that π of π₯π must be equal to π of two π. So letβs substitute two π into our function. Thatβs two π squared plus two. Thatβs four π squared plus two. And we now have everything we need to approximate our integral. We take values of π in our sum from zero to π minus one. Well, two minus one is one. Then, itβs Ξπ₯ times π of π₯π, which we just found to be four π squared plus two.

Itβs worth knowing that we can take out any constant factors independent of π. Well, four is independent of π. So we can write this as four times the sum of two π squared plus one for values of π between zero and one. Letβs now substitute π equals zero and π equals one into this expression and find its sum. Thatβs 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 π₯ squared plus two is an increasing or decreasing function over our interval. To do this, weβre going to differentiate π₯ squared plus two. And we get two π₯. Now, of course, this is greater than zero for values of π₯ greater than zero. So this tells us the function π₯ squared plus two must be increasing for values of π₯ greater than zero. What this says for our curve is that itβs sloping upwards for values of π₯ 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.