## tavale21 3 years ago 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

1. wio

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2. wio

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

3. tavale21

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

4. wio

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

5. tavale21

is it from 0 to 3?

6. wio

I'm talking about in the general sense

7. tavale21

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

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

9. wio

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

10. wio

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

So what is the width of each rectangle?

12. tavale21

ok

13. tavale21

3

14. wio

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

15. wio

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

16. tavale21

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

17. wio

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

18. wio

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

19. wio

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

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

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

22. wio

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

23. wio

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

24. wio

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

25. wio

Do you think you can do the third sub interval?

26. tavale21

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

27. wio

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28. wio

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

29. wio

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

are we adding the last f(3) to 8

31. tavale21

32. wio

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

33. wio

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

1*3^3=27

35. tavale21

27+8+1= 36

36. tavale21

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

37. wio

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38. wio

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

ok

40. wio

How do you understand how I'm getting them?

41. tavale21

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

42. wio

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