It says...
Find the length fo the graph and compare it to the straight-line distance between the endpoints of the graph.
f(x)=ln[sex(x)] x E [0,(pi/4)]
how would you do this?
please help
ln(sex)?. That's everyone's favorite integral. What is E?.
Do you mean arc length of $\displaystyle ln(sec(x)), \;\ [0,\frac{\pi}{4}]$
If so, $\displaystyle \frac{d}{dx}[ln(sec(x))=tan(x)$
$\displaystyle \int_{0}^{\frac{\pi}{4}}\sqrt{1+tan^{2}(x)}dx=\int _{0}^{\frac{\pi}{4}}sec(x)dx$
oops. haha sorry about that. about the E, im not sure what it means. thats what it has in my book. it doesnt exactly look like an E though, its more curved than straight. it just says f(x)= ln[sec(x)], x E [0, (pi/4)]
but i think you're right. though when you take the derivative of sec(x) it equals sec(x)tan(x) right? i dont quite understand how you got tanx from ln[sec(x)]
The derivative of ln(sec(x))=tan(x).
Chain rule:
$\displaystyle \frac{1}{sec(x)}\cdot\sec(x)tan(x)=tan(x)$
The E may be 'element of'?. Is this it, $\displaystyle \in$?.
If so, they just mean that's the limits of integration.
$\displaystyle \int_{0}^{\frac{\pi}{4}}\sqrt{1+[f'(x)]^{2}}dx$
Which we found whittles down to something easy.
Once you find your solution to the integral, compare it to:
$\displaystyle \sqrt{(\sqrt{2}-1)^{2}+(\frac{\pi}{4}-0)^{2}}$