## anonymous 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. anonymous

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2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

is it from 0 to 3?

6. anonymous

I'm talking about in the general sense

7. 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?

8. anonymous

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

9. anonymous

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

10. 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

11. anonymous

So what is the width of each rectangle?

12. anonymous

ok

13. anonymous

3

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. 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.

20. 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.

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

Do you think you can do the third sub interval?

26. anonymous

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

27. anonymous

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28. anonymous

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

29. anonymous

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

30. anonymous

are we adding the last f(3) to 8

31. anonymous

32. anonymous

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

33. 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.

34. anonymous

1*3^3=27

35. anonymous

27+8+1= 36

36. anonymous

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

37. anonymous

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38. 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)

39. anonymous

ok

40. anonymous

How do you understand how I'm getting them?

41. anonymous

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

42. anonymous

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