<p>Shouldn’t you add 2 instead of subtracting 2?</p>
<p>Yeah, two should be added, I interpreted the problem wrong, I thought the marble would go through the door.</p>
<p>Tomorrow is judgment day. Let’s do this.</p>
<p>LET THE SUPERBOWL OF MATHEMATICS COMMENCE!</p>
<p>~I wish luck to all of you!~</p>
<p>Lets go guys. We got this.</p>
<p>are you guys still practicing now?</p>
<p>i need help with this!!!</p>
<p>x(t)=t^2-6t from t=2 to t=4.</p>
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<p>let’s review.</p>
<p>Theagent, </p>
<p>Is that just an integral problem? If so, you integrate that from 2 to 4, so you find the antiderivative, which would be t^3/3 - 3t^2, and go f(4) - f(2)</p>
<p>i just realized that i forgot to put in really important information to that question…</p>
<p>the question was how far does teh particle travel between 2 and 4 seconds given: x(t)=t^2-6t</p>
<p>Oh, okay… How far is it traveling… So is it asking for displacement, maybe?</p>
<p>What I’d do is integrate the absolute value of the derivative of x(t) on the interval [2,4]</p>
<p>When you add +C to an antiderivative, is it okay to make it something like +/-sqrt(C!)? Because it’s still a constant.</p>
<p>Another question: can you put at the top of a FR something like “X=☻” and then do all of your calculus with a cute little African-American smiley?</p>
<p>i’m scareeddddd</p>
<p>@elijah: I think you would get extra points for that</p>
<p>Why would you derive the function just to integrate it??? That would waste time. Don’t you just evaluate that function at the points x=2 and x=4 and subtract them.
x(4)=16-24= -8
x(2)=4-12= -8</p>
<p>x(4)-x(2)= -8 - -8= -16<br>
Take the absolute value for it and you get 16.</p>
<p>x(4)-x(2)= -8 - -8= -16</p>
<p>////
I don’t know what problem your working on but…</p>
<p>x(4)-x(2)= -8 - -8= 0</p>
<p>Whoops, the answer is zero then.</p>
<p>x(t) = t^2-6t</p>
<p>x’(t) = 2t - 6</p>
<p>∫|2t - 6|dt [2,4]</p>
<p>=2</p>
<p>really really dumb question here but my mind is blanking. so if a question asks to find a local max/min; I know you take the derivative and set it equal to zero but how you do the sign chart to tell whether it’s a max/min at that point is really confusing me. problem from my PR book:
For what value of x does the function f(x) = x^3 - 9x^2 -120x + 6 have a local min
help please!</p>