# Find a lower estimate and an upper estimate

• May 16th 2010, 07:10 PM
rushin25
Find a lower estimate and an upper estimate
a = 20
http://www.webassign.net/scalcet/5-1-001alt.gif
(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 http://www.webassign.net/images/apx.gif (lower estimate)
A http://www.webassign.net/images/apx.gif (upper estimate)

(b) Find new estimates using ten rectangles in each case.

A http://www.webassign.net/images/apx.gif (lower estimate)
A http://www.webassign.net/images/apx.gif (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)
• May 16th 2010, 07:41 PM
pickslides
Area of a rectangle is width $\times$ length so try $8\times f(8)+ 8\times f(16)+\dots$
• May 17th 2010, 04:28 AM
HallsofIvy
Quote:

Originally Posted by rushin25
a = 20
http://www.webassign.net/scalcet/5-1-001alt.gif
(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 http://www.webassign.net/images/apx.gif (lower estimate)
A http://www.webassign.net/images/apx.gif (upper estimate)

(b) Find new estimates using ten rectangles in each case.

A http://www.webassign.net/images/apx.gif (lower estimate)
A http://www.webassign.net/images/apx.gif (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?