<p>Is this question insanely difficult or am I missing something? o.O</p>
<p>It’s E)</p>
<p>Apply FTC to the integral and you should get 2f(t) = g’(t)</p>
<p>g’(3) = 2f(3) = 2*-3 = -6</p>
<p>Nope, the answer is not E. And the answer that I am getting isn’t even a choice! Here is what I am doing:</p>
<p>g’(x) = 2f(2x) - f(0)</p>
<p>g’(3) = 2f(6) - 7
g’(3) = -2 - 7 = -9</p>
<p>Can anyone explain? o.O</p>
<p>Woops, you are right, it is 2f(2x) </p>
<p>f(2<em>3) = f(6) = -1 (where are you getting -7 from? -2</em> 1 = -2. It is C</p>
<p>Lol this isn’t too bad
Is the answer D?
If g(x) is the integral of the graph of f
Then g’(x) is the simply the graph of f pictured right in front of you
G’(3)=f(3)=-3</p>
<p>If I’m wrong just ignore this haha</p>
<p>I think it’s -1. </p>
<p>Let F(x) be the integral of f(x).</p>
<p>g(x) = integral f(x) from 0 to 2x</p>
<p>g(x) = F(2x) - F(0)</p>
<p>Now differentiate both sides.</p>
<p>g’(x) = f(2x) - 0</p>
<p>Since f(0) is a constant value.</p>
<p>So g’(3) = f(6) </p>
<p>Which according to the graph is -1</p>
<p>Ha, the only person that got it is Holden, very nice.</p>
<p>What I am missing is why do you not evaluate the bottom limit of integration? If you know what I mean. Like you plug in the 2x, but you just disregard the 0 on the bottom? Is that just the rule?</p>
<p>FTC says you don’t even care about what the lower liimit of integration is. It can be constant and your answer wont’ change. it doesn’t even have to do be 0. It could be 9, 2,4 , e, pi anything</p>
<p>Ok, thanks a bunch! That was annoying me. </p>
<p>And what if the bottom limit of integration also had an x term in it? Or should we not worry about venturing into that neck of the woods?</p>
<p>If it had an “x” (or a function like sin(x)), you have to split the integral. Just remember that the bottom number must always be a constant and the top function</p>
<p>Alrighty, sounds good. Thank you very much Mr. Holden Caulfield.</p>
<p>You are welcome my friendly phoney</p>
<p>It’s pretty simple to do the two limits case… S represents integration here, take the lower limit to be g(x), and the upper limit to be h(x), and say F’(t) = f(t): </p>
<p>d/dx S f(t)dt = d/dx [F(upper limit) - F(lower limit)]
= d/dx [F(h(x)) - F(g(x))]
= d/dt * dt/dx [F(h(x)) - F(g(x))] … (F is a function of time, right? The dt/dx might be an abuse of notation, but you get the idea.)
= d/dt [F(h(x))] * d(h(x))/dx - d/dt [F(g(x))] * d(g(x))/dx
= f(h(x)) * d(h(x))/dx - f(g(x)) * d(g(x))/dx … (Since F’(t) = f(t))</p>
<p>In this case, h(x) = 2x, g(x) = a, so we get: f(2x) * 2 - f(a) * 0 = 2 * f(2x). Look up differentiation under the integral sign for more information.</p>
<p>The answer is C, right?</p>
<p>don’t you have to multiply by the derivative of the upper bound? So the answer should be 2f(2x)</p>
<p>I got C.</p>
<p>Pretty easy…</p>
<p>^^Not pretty easy, extremely easy. ;)</p>
<p>well you have to remember the chain rule with the limits so it could trick some people…</p>