Question

The graph below illustrates approximating rectangles with left endpoints for $f(x) = (16/x)$ on the interval $[2, 6]$. The estimated area based on these rectangles is and this sum is an overestimate of the area of the region enclosed by $y = f(x)$, the x-axis, and the vertical lines $x = 2$ and $x = 6$. Left endpoint Riemann sum for $y = (16/x)$ on $[2, 6]$ The graph below illustrates approximating rectangles with right endpoints for $f(x) = (16/x)$ on the interval $[2, 6]$. The estimated area based on these rectangles is and this sum is an underestimate of the area of the region enclosed by $y = f(x)$, the x-axis, and the vertical lines $x = 2$ and $x = 6$. Right endpoint Riemann sum for $y = (16/x)$ on $[2, 6]$

          The graph below illustrates approximating rectangles with left endpoints for $f(x) = (16/x)$ on the interval $[2, 6]$.
The estimated area based on these rectangles is and this sum is an overestimate of the area of the region enclosed by $y = f(x)$, the x-axis, and the vertical lines $x = 2$ and $x = 6$.
Left endpoint Riemann sum for $y = (16/x)$ on $[2, 6]$

The graph below illustrates approximating rectangles with right endpoints for $f(x) = (16/x)$ on the interval $[2, 6]$.
The estimated area based on these rectangles is and this sum is an underestimate of the area of the region enclosed by $y = f(x)$, the x-axis, and the vertical lines $x = 2$ and $x = 6$.
Right endpoint Riemann sum for $y = (16/x)$ on $[2, 6]$

the graph below illustrates approximating rectangles with left endpoints for fx 16x on the interval 2 6 the estimated area based on these rectangles is and this sum is an overestimate of the 33318

Added by Francisco Javier N.

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