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

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- Dec 9th 2006, 01:01 AM0123fundamental 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, 01:55 AMCaptainBlack
Fundamental theorem of calculus may be written as:

$\displaystyle

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

$

so:

$\displaystyle

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

$

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

$\displaystyle

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

$

RonL - Dec 9th 2006, 02:13 AM0123
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, 06:39 AMCaptainBlack
- Dec 9th 2006, 06:44 AM0123
now I see.. I mistook the very approach to the problem..:p thanks!