- #1

- 55

- 0

- Thread starter barksdalemc
- Start date

- #1

- 55

- 0

- #2

matt grime

Science Advisor

Homework Helper

- 9,395

- 3

Why did you think this? It is not correct. The limit of

[tex] \int_c^x f(t)dt[/tex]

as x tends to c is zero.

What you shuld be thinkig about is

[tex]\frac{1}{h} (\int_c^{x+h} f(t)dt - \int_c^x f(t)dt)[/tex]

as h tends to zero.

[tex] \int_c^x f(t)dt[/tex]

as x tends to c is zero.

What you shuld be thinkig about is

[tex]\frac{1}{h} (\int_c^{x+h} f(t)dt - \int_c^x f(t)dt)[/tex]

as h tends to zero.

Last edited:

- #3

- 55

- 0

- #4

- 55

- 0

I just saw the second equation you posted. That makes it 100% clear to me. Thanks.

- Replies
- 3

- Views
- 13K

- Replies
- 7

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 3K

- Replies
- 2

- Views
- 531

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 7

- Views
- 4K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 774

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 9

- Views
- 1K