Well maybe a little bit more than that. Okay here's the deal:
We devided into 20 rectangles each of width then the approximation could be done by adding the rectangles of width and length/hight that is whuc will turn out to be very soon to be
The area between the graph of and the x-axis for 1 ≤ x ≤ 4 s estimated using 20 rectangles and x* k is taken to be the right-hand boundary of each interval.
The result is rounded off to three decimal places.
The answer is 4.741
Could someone help explain this? Thanks.
Well if you have reached Rieman sums you should be getting that, but I'm assuming you've never heard about the guy. So, when approximating areas under a curve of , above the x-axis and between 2 vertical lines lets say and we divide the segment into partitions of width each... It's like cutting a piece of bread into slices of equal width (the s don't have to be of equal width though, later to that...). So when you finish dividing the segment into smaller subsegments, and then drawing a recatangle over each subsegment of height = the (average) value of on that segment (the avergae value of on is around 1 because )
Then, you add the areas of these rectangles, each has an area of where is any point in the interval , I would go for the edge points( and so on...) So when you sum up the whole lot you're doing , note that 20 is your choice, it could ran up to 100, a 1000 or infinity.