Sketch the region enclosed by the given curves. Decide...

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle.\ y = 2/x, y = 2/x$^2$, x = 7\ Find the area of the region.\ 1.55

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00:01 In this question, we're going to sketch the region that's enclosed by the given curve.

00:03 We're going to integrate with respect to either x or y, and then figure out what the area of the region is.

00:10 So first, i have y equals x squared minus 3x.

00:16 What shape is that? that is a parabola that opens up, since there is a positive leading coefficient on the x squared, which means these two answers can't be correct.

00:27 It's got to be either answer choice 3 or 4 down here.

00:31 If you look at y equals 3x, that's a line that's passing through the origin with a slope of 3.

00:40 It has a positive slope, and so therefore, it's got to be this option here as my graph.

00:48 Now i'm trying to find the area enclosed by these curves.

00:51 I'm going to start by figuring out where do they intersect.

00:56 So we're going to set x squared minus 3x equal to 3x.

01:01 I subtract 3x from each side.

01:03 X squared minus 6x equals 0.

01:05 I factor out an x.

01:07 X times the quantity of x minus 6 equals 0.

01:10 So i'm getting x equals 0 and x equals 6.

01:13 Now, those are my intersections.

01:17 I am going to integrate in the order of top function minus bottom function.

01:22 So my top function this time was the line 3x minus the bottom function, which is the quantity of x squared minus 3x dx...

Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle.\ y = 2/x, y = 2/x$^2$, x = 7\ Find the area of the region.\ 1.55

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