### Video Transcript

Find the lower Riemann sum approximation for π of π₯ equals five minus π₯ squared on the close interval one to two, given that π equals four subintervals.

The lower Riemann sum approximation is going to be the smallest estimate for the area between the curve of π of π₯ and the π₯-axis and bounded by the lines π₯ equals one on π₯ equals two that weβll get. So we recall if π of π₯ is increasing, that is, π prime of π₯ is greater than zero over some interval, then a left Riemann sum will yield an underestimate for the area between the curve on the π₯-axis in that interval. And a right Riemann sum will yield an overestimate. For a decreasing function, the reverse is true. So letβs begin by finding the derivative of our function π of π₯ and working out whether itβs greater than or less than zero over the closed interval one to two.

The first derivative of our function is negative two π₯. Now for values of π₯ that are greater than zero, in other words, positive values of π₯, negative two π₯ is going to be less than zero. This means that our function must be a decreasing function in the closed interval one to two. So we need to evaluate a right Riemann sum. To find a right Riemann sum, we take values of π from one to π. Itβs the sum of Ξπ₯ times π of π₯π for these values of π, where Ξπ₯ is π minus π divided by π. And in our case, π is equal to one and π is equal to two. And of course, π is the number of subintervals. And π₯π is π plus π lots of Ξπ₯.

Letβs begin by working out the value of Ξπ₯. Itβs π minus π, thatβs two minus one, all divided by π, which is four. Two minus one divided by four is a quarter or 0.25. And now, we have enough information to be able to evaluate π₯π. Itβs π which is one plus Ξπ₯ times π; thatβs 0.25π. Remember, to evaluate the summation, weβre going to need to find π of π₯π. Thatβs, of course, π of one plus 0.25π. So letβs swap π₯ in our function for one plus 0.25π. When we do, we obtain π of π₯π to be five minus one plus 0.25π all squared. Now whilst we could distribute these parentheses and simplify, thereβs really no need. We have enough to substitute into the summation formula. Weβre interested in values of π from one to π, so thatβs one to four. Then itβs Ξπ₯ which is 0.25 times five minus one plus 0.25π squared.

Now, itβs useful to know that 0.25 is independent of π. So we can take this out as a common factor. And we need to work out the sum of five minus one plus 0.25π squared between π equals one and four and then multiply that by 0.25. To do this, we simply substitute values of π from one to four into this formula and then add them all together. When π is one, itβs five minus one plus 0.25 squared. When π is two, we get five minus one plus 0.5 squared. When π is three, itβs five minus one plus 0.75 squared. And when π is four, its five minus one plus one all squared. Remember, weβre gonna add these together and then multiply it all by 0.25. When we do, we obtain an answer of 2.28125 which correct to two decimal places is 2.28. The lower Riemann sum approximation for our function is 2.28.