<p>Hello, I have a couple questions in AP Calculus, and I was wondering if anyone could help out.</p>
<li> Using a graphing utility to graph f(x)=x^4 - 6x^3 + 11x^2 - 6x. Then use ummer sums to approximate the area of the region in the first quadrant bounded by f and the x-axis using four sub intervals. ( I tried putting it into the rieman program on the calculator but i kept on coming up with a negative number. I do’nt know what to use for left and ride interval.</li>
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<p>2.Determine if F(x)= 5/2x-3 is integrable on [0,2]. Give reasons.</p>
<p>No idea what you mean for 1, you're not giving any upper limit. Since lim x->+inf f(x) = +inf, wouldn't the area under the curve be infinite.</p>
<p>For 2, it't cause there's a vertical asymptote at x=1.5. Now you're obviously way ahead of us in AP calculus, but ive done some integration in AP Physics C and it seems intuitively impossible to find the definite inteagral of a function across an interval which isn't continuous. I'd bet my firstborn that it cannot be done and that there is no doubt a rule that says as much.</p>
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<li><p>Using the graph, you are looking for the upper sum from 1 to 2:
find the maximum values for f(x) for these intervals: [1,5/4],[5/4,3/2],[3/2,7/4],[7/4,2]. For sum of the values, the max will be an end-point. For others, you need to use the first derivative to find the critical point. Then, find the upper sum from the definition (i.e., sum of (max of f on interval)(interval length) for each interval).</p></li>
<li><p>5/(2x-3) is discontinuous at 1.5. The function may be integrable though...</p></li>
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<p>well idk....dy/dx of 5/2[ln(2x-3)] = 5/(2x-3) so that function should be integrable</p>
<p>I didn't see any question resembling the 2nd one in the book. Anyone else know the answer with a reason to support it ?</p>
<p>Question 2 is worded badly.</p>
<p>Is it integrable? Your book is probably looking for "no, it's discontinuous at x=1.5." And that has some logical backing, since a definite integral is only defined over a continuous interval.</p>
<p>However, you also have not learned improper integrals yet, and this integral CAN be integrated improperly. However, the integral does not converge over that interval anyway. </p>
<p>Anyway, the answer you're looking for is probably "no."</p>