approximate the area under each curve over the specified interval by using the indicated number of subintervals (or rectangles) and evaluating the function at the right-hand endpoints of the subintervals. f(x)=x^3 from x=0 to x=3; 3 subintervals

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approximate the area under each curve over the specified interval by using the indicated number of subintervals (or rectangles) and evaluating the function at the right-hand endpoints of the subintervals. f(x)=x^3 from x=0 to x=3; 3 subintervals

Calculus1
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|dw:1353031457071:dw|
You basically want to find the area of the three rectangles. Make sense?
sort of? how would i find the area's ?

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Well, for one, what is the area of a rectangle?
is it from 0 to 3?
I'm talking about in the general sense
that's all that the book has. I looked at another problem and it is f(x) = 4x-x^2 from x=0 to x=2; 2 subintervals and the answer was 7 square units. how did they get that?
Okay, I just wanted to say, the area of a rectangle is length times width.
So our first task is to find the width of the rectangles.
We will split up the interval equally into sub intervals... So we know with n subintervals on the interval [a,b], the width w is:\[ nw = b-a \implies w=\frac{b-a}{n} \]In this case, b = 3, a =0, and n = 3
So what is the width of each rectangle?
ok
3
That is the width of the entire interval, not the width of each subinterval/rectangle
\[ \frac{3-0}{3} = 1 \]
oh ok. I see now b-a/n=1
Now, to find the length of the interval is a bit tricky.
We need to pick some x in each subinterval. The length is then f(x).
But which x do we pick? It depends on the problem. This problem says that you want to pick the right-hand endpoints, so we want to pick the x such that x is the highest value of x for that subinterval.
The first interval is from a to a+w, in this case 0 to 1. Since 1 is highest, right hand side, we use f(1) for the length of the interval.
So the first interval is \(1*f(1) = 1 * 1^3 = 1\)
The second interval is between a+w and (a+w)+w
So the highest point is a+w+w = 0+1+1 = 2
So the second subinterval area is \(1*f(2) = 1*2^3=1*8=8\)
Do you think you can do the third sub interval?
can you tell me the steps sorry im not good at this?
|dw:1353032610122:dw|
Okay, first step is to find with width of the rectangles. Next step is to find the length of each rectangle.
Then you add the areas of all the rectangles. Do you at least understand the high level idea of what we are doing?
are we adding the last f(3) to 8
the answer f(3) * 8
Nope. All I did was show that rectangle 1 has area 1, and rectangle 2 has area 8
So you need to find width times f(3) to find the area of rectangle 3. Then you add up the areas of all rectangles.
1*3^3=27
27+8+1= 36
can you explain how the left hand works? to the same problem? it makes sense now? thank you!!!
|dw:1353033340267:dw|
For the right hand, we use f(a+w), f(a+2w), f(a+3w), ... , f(a+(n-1)w), f(a+nw) For the left hand, we use f(a), f(a+w), f(a+1w), ... , f(a+(n-2)w), f(a+(n-1)w)
ok
How do you understand how I'm getting them?
from the picture you drawn up you use the rectangles starting from the left to the right.
You will learn later, a method to do this with infinite subintervals. It's called a definite integral.

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