# Indefinate integrals

1. Nov 20, 2004

### mugzieee

im given a problem that gives me y=7-8x+x^(2) and it says calculate the exact area between the x-axis and the graph of y.
What i have done is tried using the fundamental theorom, by saying that F(b)=2600, and F(a)=2600. I got these values by graphing and looking at where the y graph starts and stops. but when i use the fundamental theorom i get a bogus answer. The answer given in the book is 36, can any1 point me in the correct direction please?

2. Nov 20, 2004

### vsage

do you have to use the fundamental theorem? I'm not sure where you got F(b) and F(a) from really also. Where did the 2600 come from again? I don't see how it's relevant to where the curve crosses the x axis. Integrating from where f(b) = f(a) = 0 for distinct a and b should work but I'm not sure if you're allowed to do it that way.

Last edited by a moderator: Nov 20, 2004
3. Nov 20, 2004

### mugzieee

yes i have to use the theorom, F(b) and F(a) i got from looking at the graph of y, and just taking the starting point, and the ending point of the graph, as in the interval its in. so what i did was just take the total area
from [-2600,2600]. i have no idea what the distinct a and distinct be method is sorry.

4. Nov 20, 2004

### vsage

The integral of f(x)dx from a to b is F(b) - F(a). That's the first fundamental theorem right? What's F(t)? It's the integral from 0 to t of f(x)dx. So F(b) - F(a) = integral from 0 to b of f(x)dx - integral from 0 to a of f(x)dx. You can't just take the area from an arbitrarily large value of x on the function because any area that is below the x axis gets subtracted from the area above the x axis. This is why you should choose your b and a more carefully - the two places where f(x) = 0.

5. Nov 20, 2004

### cAm

You're making this way too complicated.
Take your equation Y = x^2 - 8x + 7 and factor it to get y=(x-7)(x-1), and solve this for the points where y = 0.

so you have x=1, 7.

Take this, and do a definate integral of that equation:
Integral(1 to 7) (x^2 - 8x + 7)dx
so you get: ((x^3)/3 -4x^2 + 7x)|(1 to 7)
Plug in your numbers: (7^3)/3 - 4(7^2) + 7(7) - (1^3)/3 + 4(1^2) - 7
simplify to get: -36
since area is always positive, your area is 36.

6. Nov 20, 2004

### mugzieee

thanx cam, appreciate it

7. Nov 20, 2004

### vsage

-_- to the above. If I wanted to just do the question for you I would have like above,haha. Just a note the term "indefinite integral" applies to integrals without boundaries, which this question has - in case you ever have to mention it on a test.

8. Nov 20, 2004

### cAm

I decided to solve this one for him because i know that seeing it solved sometimes helps me a lot more than other things could, and what you were saying vsage was even confusing Me. :tongue2:

9. Nov 20, 2004

### vsage

Yeah I'm not the most coherent person in the world, sorry! -_-