<p>I am having a hard time understanding how to solve this problem.</p>
<p>Find the global maximum value for the function F on the interval [0,1] if F(x)=int(1/(t+1))dt. The upper bound is x and the lower bound is 0</p>
<p>I understand that by the theorem F(x)=F(V(x))*V'(x) I should get F(x)=1/(x+1). However, the answer is 0.693 at x=1, but I don't see where this could be derived from.</p>
<p>Could you make it ln(x+1)?</p>
<p>Wow, never mind-got the answer.</p>
<p>Here's how I think this one goes, and it's a good one to say the least! :D</p>
<ol>
<li>By the FTC, F'(x) = 1/(x + 1) since 1/(t + 1) was the integrand.</li>
<li>Find where F'(x) = 0 or DNE. In this case, we'd set the denominator = 0 to find where F'(x) DNE. So, x + 1 = 0 and x = -1. However, -1 is not part of the domain [0, 1], so we can discard this extreme value.</li>
<li>Now we have to look at the endpoints of the [0, 1] domain to see if those are extreme values. BEFORE WE CAN DO THAT, we must antidifferentiate/integrate indefinitely that integral so we know what F(x) is!</li>
</ol>
<p>F(x) = Integral [1/(t + 1)] dt</p>
<p>from the rule, Integral [1/u] du = ln u + C, where u is a linear function,</p>
<p>F(x) = ln (x + 1). (I am not sure why, but the + C portion of this function can be left out. I'm guessing it might have something to do w/ a given domain. GUYS, any idea as to why you can leave + C out?)</p>
<p>Just from a basic understanding of how the ln x [ln(x + 1) is just a translated function of ln x] function looks, we should see the function increasing on this interval.
ln(0 + 1) = ln 1 = 0
ln(1 + 1) = ln 2 = ~ 0.693</p>
<p>and indeed, we do!</p>
<p>Just from comparing those two endpoint values, one can readily see that the global maximum on the interval [0, 1] is ln 2, or 0.693 in a rounded, decimal form.</p>
<p>Hope this helps! Have a great weekend.</p>