anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1353031457071:dw|
anonymous
  • anonymous
You basically want to find the area of the three rectangles. Make sense?
anonymous
  • anonymous
sort of? how would i find the area's ?

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

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