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:
$\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
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...