# I need serious help with this problem, it's about area and distance

1. Apr 26, 2007

### afcwestwarrior

1. The problem statement, all variables and given/known data

by reading values from the given graph of f, use five rectangles to find a lower estimate and an upper estimate for the area under the given graph of from x=0 to x=10. in each case sketch the rectangles that you use.

3. The attempt at a solution
since their are there are 5 rectangles, the width was 2 for each of them

and when i tried to figure out the lower and upper estimate i did this

L5= 2*O^2+ 2*2^2 ..... 2 * 8^2 =128

and the answer in the back of the book is 40

and for Right end= 2*2^2 ..... 2*10^2= 440

and the answer is 52

2. Apr 26, 2007

### afcwestwarrior

i already did sketched the rectangle thing, and it doesn't tell me what type of graph it is, it just tells me y= f(x)

3. Apr 26, 2007

### Dick

The area of a rectangle is length*width. There is no 'square' in the formula.

4. Apr 26, 2007

### afcwestwarrior

ok, so then i dont have to square them,

5. Apr 26, 2007

### afcwestwarrior

i was looking at an example that was squaring them, no wonder, thanks man, i didn't think about the y= f(X)

6. Apr 26, 2007

### afcwestwarrior

ok so what do i do then

7. Apr 26, 2007

### afcwestwarrior

i didnt square the left end points and i got 40
but i did the same for the right end and i got 60

8. Apr 26, 2007

### Dick

You must be able to determine the values of f(0), f(2) etc from your graph. Those are the heights. The widths are all 2.

9. Apr 26, 2007

### afcwestwarrior

how do i do that, what formula do i use

10. Apr 26, 2007

### Dick

If you are going to get a numerical answer like 52 then you must be able to find f(0) etc. It's the y value of the graph when x=0 etc.

11. Apr 26, 2007

### afcwestwarrior

i did it, i multiplied 2 * the hieght, for every one

12. Apr 26, 2007

### Dick

And got 52 for one of them?

13. Apr 26, 2007

### afcwestwarrior

yup, i got 53, but then i took a closer look at the heights and got 52

14. Apr 26, 2007

### afcwestwarrior

thanks for the help

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