# fundamental theorem

• Dec 9th 2006, 02:01 AM
0123
fundamental theorem
let F(x)= int from 1 to x of (e^t)/(2t+3)
which is the equation of the tangent line to the graph of F(x) at x=1?
I can't find the F(x0) since I don't know how to solve that integer. Might someone help me?thanks
• Dec 9th 2006, 02:55 AM
CaptainBlack
Quote:

Originally Posted by 0123
let F(x)= int from 1 to x of (e^t)/(2t+3)
which is the equation of the tangent line to the graph of F(x) at x=1?
I can't find the F(x0) since I don't know how to solve that integer. Might someone help me?thanks

Fundamental theorem of calculus may be written as:

$
\frac{d}{dx}\ \int_a^x f(t)\ dt=f(x)
$

so:

$
\frac{dF}{dx}(x)=(e^x)/(2x+3)
$

So the slope of $y=F(x)$ at $x=1$ is:

$
\frac{dF}{dx}(1)=f(1)=(e^1)/(2+3)=e/5
$

RonL
• Dec 9th 2006, 03:13 AM
0123
I got the idea of the slope. my problem comes when I need to solve the integer; I mean, since this is the equation y=m(x-x0)+y0, m is the slope, which is clear, and x0=1, I need to find y0 which is F(x) at x=1, and to do that I think I need to solve that integer, which I am not able to. I hope my explaination is clear enough...:(
• Dec 9th 2006, 07:39 AM
CaptainBlack
Quote:

Originally Posted by 0123
I got the idea of the slope. my problem comes when I need to solve the integer; I mean, since this is the equation y=m(x-x0)+y0, m is the slope, which is clear, and x0=1, I need to find y0 which is F(x) at x=1, and to do that I think I need to solve that integer, which I am not able to. I hope my explaination is clear enough...:(

$
F(x)= \int_1^x e^t/(2t+3)\ dt
$

so:

$
F(1)= \int_1^1 e^t/(2t+3)\ dt=0
$

RonL
• Dec 9th 2006, 07:44 AM
0123
now I see.. I mistook the very approach to the problem..:p thanks!