Solving Summation Question with Floor & Ceiling Functions

[mohabitar]

Not exactly sure how they went from the first step to the 2nd step? Is there an easier way to solve this?

(keep in mind we're dealing with floor and ceiling functions)

 

[MHrtz]

There is an easier way to do the problem. You can use what's called integration but that is a calculus topic.

One way you can solve it is by graphing.

You can draw the graph of the function (which is linear) and find the area under the curve. When you calculate the area under the curve (which is a triangle) you only take into account the area from 0 to 30. Also, if any piece of the graph from 0 to 30 is below the x-axis then you have to subtract that portion from the area that is above the x axis.

If you have a graphing calculator I can show you an even simpler way.

 

[mohabitar]

I know calculus, I just posted it in here because I thought its a precalc question. How would I use integrals to solve this?

 

[MHrtz]

How much calculus do you know?

 

[mohabitar]

Lol I'm in Calc II now, will that be enough?

 

[MHrtz]

Oh wait...I have to apologize for being retarded. You can't solve this with integration because it's a summation (it's been awhile). Unfortunately, you have to add all those #'s together because a summation is not the exact answer but an estimation. An integral, on the other hand, is the exact answer.

 

[mohabitar]

So the question now is, I see that they did it manually in step 1, where its just the sum of all those values, but step 2 I see some multiplication going on-where did that come from?

 

[JonF]

I get: S[ (i/20 – 1/2) + (i/10 – 1/2 ) ] = S[ i/20 –1/2 + i/10 – 1/2 ] = S[ (3/20)i – 1 ] = 3/20*S -S[1] = 3*(30^2 + 30)/40 – 30 = 39.75

or

S[ (i/20 – 1/2) + (i/10 – 1/2 ) ] = S[ i/20 ] – S[1/2] + S[i/10] – S[1/2 ] = 1/20S – 1/2S[1] + 1/10S – 1/2S = 1/20*465 – 1/2*30 – 1/10*465 – 1/2*30 = 39.75