## tavale21 Group Title 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 one year ago one year ago

1. wio Group Title

|dw:1353031457071:dw|

2. wio Group Title

You basically want to find the area of the three rectangles. Make sense?

3. tavale21 Group Title

sort of? how would i find the area's ?

4. wio Group Title

Well, for one, what is the area of a rectangle?

5. tavale21 Group Title

is it from 0 to 3?

6. wio Group Title

I'm talking about in the general sense

7. tavale21 Group Title

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?

8. wio Group Title

Okay, I just wanted to say, the area of a rectangle is length times width.

9. wio Group Title

So our first task is to find the width of the rectangles.

10. wio Group Title

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

11. wio Group Title

So what is the width of each rectangle?

12. tavale21 Group Title

ok

13. tavale21 Group Title

3

14. wio Group Title

That is the width of the entire interval, not the width of each subinterval/rectangle

15. wio Group Title

$\frac{3-0}{3} = 1$

16. tavale21 Group Title

oh ok. I see now b-a/n=1

17. wio Group Title

Now, to find the length of the interval is a bit tricky.

18. wio Group Title

We need to pick some x in each subinterval. The length is then f(x).

19. wio Group Title

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.

20. wio Group Title

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.

21. wio Group Title

So the first interval is $$1*f(1) = 1 * 1^3 = 1$$

22. wio Group Title

The second interval is between a+w and (a+w)+w

23. wio Group Title

So the highest point is a+w+w = 0+1+1 = 2

24. wio Group Title

So the second subinterval area is $$1*f(2) = 1*2^3=1*8=8$$

25. wio Group Title

Do you think you can do the third sub interval?

26. tavale21 Group Title

can you tell me the steps sorry im not good at this?

27. wio Group Title

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28. wio Group Title

Okay, first step is to find with width of the rectangles. Next step is to find the length of each rectangle.

29. wio Group Title

Then you add the areas of all the rectangles. Do you at least understand the high level idea of what we are doing?

30. tavale21 Group Title

are we adding the last f(3) to 8

31. tavale21 Group Title

32. wio Group Title

Nope. All I did was show that rectangle 1 has area 1, and rectangle 2 has area 8

33. wio Group Title

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.

34. tavale21 Group Title

1*3^3=27

35. tavale21 Group Title

27+8+1= 36

36. tavale21 Group Title

can you explain how the left hand works? to the same problem? it makes sense now? thank you!!!

37. wio Group Title

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38. wio Group Title

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)

39. tavale21 Group Title

ok

40. wio Group Title

How do you understand how I'm getting them?

41. tavale21 Group Title

from the picture you drawn up you use the rectangles starting from the left to the right.

42. wio Group Title

You will learn later, a method to do this with infinite subintervals. It's called a definite integral.