Find the centroid of the region bounded by y=lnx, y=0, x=e Solution Sketch your

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Sketch your region: The natural log asymptotes with the y-axis, so start @ below the x-axis, then come up above it, then sort of flatten out (but always rising). Then draw a vertical line...so this is the region, we need to find the bounds (x AND y bounds): See where y = ln(x) and y = 0 intersect: ln(x) = 0 --> raise to the e: e^(ln(x) = e^0 ---> x = 1 --> (1,0) of course y = 0 and x = e intersect at (e, 0) Then where to y = ln(x) and x = e intersect (the third cornder)? y = ln(e) = ln(e¹) = 1 So our bounds are: to the left: (1,0) to the right: (e, 0) up-right: (e, 1) Now to find the area, just tile the region (same thing we will be doing for when we find the centroid). Draw vertical strips that cover the entire region. These strips are the dA's that we will add up to find the area: Find the area of ONE strip (any strip...so let's say the strip @ x): The width of the strip = dx (the differential length). What is the height of the strip: Easy, it goes from y = 0 to y = ln(x) so the height = ln(x) - 0 So the area of one strip: dA = ln(x)dx A = ?dA * ?ln(x)dx, from 1 to e u = ln(x) du = dx/x dv = 1 v = x --> ?ln(x)dx = xln(x) - ?dx = x * (ln(x) - 1) + C So we have: e * (ln(e) - 1) - 1 * (ln(1) - 1) --> e * (1 - 1) - (0 - 1) = 0 + 1 = 1 So the area is 1. To find the centroid you HAVE to remember the center of mass formula: cm = (Sxi*mi)/(Smi) So all you do is sum all of the masses and multiply by their x position, then divide by the TOTAL mass. In our case, we have an area...so we are going to pretend that area = mass, of course it doesn't, in fact, in general people will you the mass density function...but here we assume that the mass is evenly distributed throughout the area, so the mass is DIRECTLY proportional to the area. Well go back to your strips that you drew earlier, each strip is at the same x value, so we could do this: dxl = x * dA Then we would just sum all of these up (an integral like above)...but what about summing up the masses?...WE ALREADY DID THAT, A = the sum of all the masses (remember that when we do the y below). So now just integrate: ?xln(x)dx, from 1 to e u = ln(x) --> du = (1/x)dx dv = x -----> v = (1/2)x² ?xln(x)dx = (1/2)x² * ln(x) - (1/2)?xdx ...so finally: (1/2 * (x²ln(x) - 1/2x²) = (x²/4) * (2ln(x) - 1) So now plug in your bounds: (e²/4) * (2ln(e) - 1) - (1²/4) * (ln(1) - 1) --> (e²/4) * (2*1 - 1) - 1/4 * (0 - 1) = e²/4 + 1/4 = (e² + 1)/4 This is the x coordinate of the centroid. For the y we have to do something a little different. Redraw the region and now tile it with horizontal strips. Now each of these horizontal strips has the SAME y value, so we can do something very similar as we did with x: y * dA But now we have to go back and find dA: Now the width is dy, and the height of the rectangles (or length) is: Well you go from x = e down to y = ln(x)...but that doesn't tell us the x values which is what we want: but y = ln(x) --> x = exp(y) (as in e^y) So it goes from exp(y) on the left to x = e on the right. dA = dy * (e - exp(y)) So now we have: ?y * (e - exp(y))dy = e?ydy - ?yexp(y)dy (the first one is EASY --> e/2y²) For the second: u = y --------------->du = dy dy = exp(y)dy --> y = exp(y) ?yexp(y)dy = yexp(y) - ?exp(y)dy --> y * exp(y) - exp(y) + C = exp(y) * (y - 1) So we have: (e/2)y² - exp(y) * (y - 1) + C But WHAT are the limits??? Remember it goes from y = 0 up to y = 1. So we have: e/2 - {exp(1) * (1 - 1) - exp(0) * (0 - 1)} --> e/2 - 0 + 1*-1 = e/2 - 1 So the centroid is at: ((e² + 1)/4, e/2 - 1) ~ (2.097, .359)

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