Question

Bound the area of the shaded region by approximating the upper and lower sums; use rectangles of width 1. [See attachment]

Answer 1

First calculate the area of one square.

Answer 2

2. count the number of boxes.
and multiply it with the area of one square.. lol i am hoping this is not related with integrals? or is it

Answer 3

i am going to guess at about fourteen and a half

Answer 4

it looks like the parts in the first two not completed squares is about 1
also for the last two
and the middle one looks to be about one half

Answer 5

Yeah, this a bit hard for me, since you're only given a graph of the function. but, from the graph the interval is [0,4] ; the number of subintervals (rectangles) is 4; and the width is going to be (b-a)n = (4-0)/4 which is 1.

Answer 6

I know that the length of any of the rectangles is going the function f evaluated at the endpoint. What's throwing me off is not knowing what the function is, and then the whole sigma notation bit. I tried to solve the problem, but I was way off. :/

Answer 7

Correct me if I'm wrong, but aren't all of the left endpoints going to be:

\[a + (x-i)\Delta x\]

Answer 8

@nubeer -> It's in the same chapter as integrals? :x

Answer 9

But, you aren't use the whole integral notation stuff.

Answer 10

lol it's long since i did with that other method.. now forgot how to do with that rectangle method.

Answer 11

For the lower sum, just read the values of the graph at the lower endpoints of the subintervals, which would be 0, 1, 2 and 3 in this case, and multiply those by delta x

Looking at the graph I would say the lower sum is about 1*1 + 3*1 + 4*1 + 4.5*1 = 12.5, and for the upper sum you should check the right endpoints of the subintervals

And the area(A) underneath would be bounded as : lower sum < A < upper sum

Answer 12

Wow. :/ I just overcomplicated this, like always.