Area of a rectangle is width length so try
a = 20
(a) By reading values from the given graph of f, use five rectangles to find a lower estimate and an upper estimate for the area under the given graph of f from x = 0 to x = 40.
A (lower estimate)
A (upper estimate)
(b) Find new estimates using ten rectangles in each case.
A (lower estimate)
A (upper estimate)
for (a) i got the width (40-0)/5 which is 8...but it says estimate using 5 rectangles...how would i get the area of 5 rectangles if the width is 8....
same problem with (b)
Each of those small rectangles on the grid has width 4 so 8 is just 2 of those rectangles.
It looks to me like that, for the first two rectangles, the graph is always above 4 and below 12 so the 'lower integral" would use a rectangle of width 8 and height 4, having area 4(8)= 32 while the "upper integral" would use a rectangle of width 8 and height 12, having area 12(8)= 96.
Now, what about the next two integrals, between 8 and 16?