Sketch the region enclosed by the given curves.y = 4/x, y = 16x, y =x/16, x>0
I assume we want to find the area of the region
$\displaystyle \int_0^{1/2} \left(16x-\frac{x}{16}\right) \, dx+\int_{1/2}^8 \left(\frac{4}{x}-\frac{x}{16}\right) \, dx=16 \ln 2\approx 11.0904$
The graph of y= 4/x is a hyperbola passing through (1/2, 8), (2, 2), and (8, 1/2). y= 16x is the straight line through (0, 0) and (1/2, 8). y= x/16 is the straight line through (0, 0) and (8, 1/2). The region enclosed is the "triangle" (one curved side) with vertices (0, 0), (1/2, 8), and (8, 1/2). You should have two integrals from 0 to 1/2 and from 1/2 to 8 because, for x< 1/2, the region goes from the straight line y= x/16 up to the straight line y= 16x while for 1/2< x< 8, it goes from the straight line y= x/16 up to the hyperbola y= 4/x.