OpenStudy (anonymous):

Help please? use upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width).

3 years ago
OpenStudy (anonymous):

3 years ago
OpenStudy (anonymous):

calculator exercise

3 years ago
OpenStudy (anonymous):

?

3 years ago
OpenStudy (anonymous):

How do I do it?

3 years ago
OpenStudy (anonymous):

there are 8 pieces so each are of length \(\frac{1}{4}=.25\)

3 years ago
OpenStudy (anonymous):

evaluate at \(0,.25,.5,.75,1,1.25,1.5,1.75\) for the lower limit, and multiply it all by \(.25\)

3 years ago
OpenStudy (anonymous):

so here you should look up Riemann summation....must say that is quit annoying question when there is an easier way of finding the area....just google it surely you'll find it

3 years ago
OpenStudy (anonymous):

do I evaluate each at x=2? as in .25 * 2 + .5 *2?

3 years ago
OpenStudy (anonymous):

@satellite73

3 years ago
OpenStudy (anonymous):

or is the width one?

3 years ago
OpenStudy (turingtest):

the width of each rectangle was told to you by @satellite73 , take the domain of x we are looking at and divide it by the number of rectangles\[2/8=0.25=w\]the height of each rectangle is the y value at a particular x.

3 years ago
OpenStudy (anonymous):

oh ok. I hadn't understood. Thank you

3 years ago
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