<p>These are probably easy for most of you but I could use a little assistance.</p>
<li><p>If F(x)= integral (1 to x) (dt/(1+t^2)) dt, then F’(x) =</p></li>
<li><p>Use the definition of a derivative and the Fundamental Theorem of Calculus to evaluate:</p></li>
</ol>
<p>lim h–>0</p>
<p>(int(2 to 2+h) sqrt of ( x^2+x) dx )/ h</p>
<p>Sorry if you don’t understand the notation there… just let me know if you don’t understand what I mean. Thans for the help, i appreciate it.</p>
<p>How come you have a dt in the numerator of the first one?</p>
<p>That's the way they have it written in the packet...it's there twice....</p>
<ol>
<li>1/(1+x^2) This uses the second fundemental theorm of calculus
not to sure of the second one</li>
</ol>
<p>Ok thanks. We just started integrals today and my teacher sucks and he already gave us problems like this and harder ones...im so confused</p>
<p>I could do the last one but I would need some paper probably and that would involve getting up</p>
<p>You guys just started integrals today??!! Are you taking the AP test?</p>
<p>I'm not nor are most kids in the class..we get college certified dual credit anyway so most dont take the exam...Some kids (2-3) are taking it though. We are on block schedule so we started class February. We had midterms about a week ago (or 2 due to spring break) and he likes to do first 1/2 limits and derivatives and second 1/2 integrals for AP AB Calc, but yeah, I feel bad for the kids that are taking the exam...we know nothing!!</p>
<p>for the second one, you need to use L'hopitals rule, which (basically) says that if you have the limit of a quoient where both the top and bottom are indeterminate forms the limit of the derivitive of the numerator over the derivitive of the denominater, if it exists, is equal to the limit you were originally trying to evaluate. (sorry if that doesnt make sense!!!)</p>
<p>so the derivitive of the top is (2+h)^2 + (2+h), because you are doing what you did in the first question again. the derivitve of the denominater is 1, and since these are no longer indeterminate you can evaluate the top and bottom at h=0.</p>
<p>Okay, thanks...i dont know why he would give us that considering l'hopitals comes in 4 chapters...i looked ahead. we are on chapter 5 of the book, it is covered in chapter 9...</p>
<p>Wow that sounds almost as bad as sittingbulls class.</p>
<p>yeah, that doesnt make sense, i might be wrong! thats how i'd do it, but theres probably an easier way..?</p>
<p>Either that or a more complicated way! I hate calculus though.</p>
<p>haha yes, L'hopital's rule...we've just started that this week :(</p>
<p>i like it! hehe :) but yeah its really hard, especially if you dont like math anyways.</p>
<p>But senior year, the closest thing to math I have is physics!! And I hear the physics teacher rocks unlike the calc one...thank the lord for no math as a senior!</p>