Use left endpoints and right endpoints with 6 rectangles to find two approximations of the area of the region between the graph of f(x) = 2x squared minus x minus 1 and the x-axis over the interval from 2 to 5.

Step-by-step solution

1. 1

   Confirm the region really is the area under the curve. On $[2,5]$ the roots of $f$ are $x=1$ and $x=-\tfrac12$, both to the left of the interval, and $f(2)=5>0$. So $f>0$ throughout and the rectangles measure area directly, with no sign flip to worry about.

2. 2

   Compute the width of each rectangle. Splitting an interval into $n$ equal pieces gives

   $\Delta x=\frac{b-a}{n}=\frac{5-2}{6}=\frac12$

   so the partition points are $2,\ \tfrac52,\ 3,\ \tfrac72,\ 4,\ \tfrac92,\ 5$.

3. 3

   Evaluate $f$ once at every partition point. The left and right sums reuse the same seven values, so computing them once saves half the work:

   $f(2)=5,\quad f\!\left(\tfrac52\right)=9,\quad f(3)=14,\quad f\!\left(\tfrac72\right)=20,\quad f(4)=27,\quad f\!\left(\tfrac92\right)=35,\quad f(5)=44$

4. 4

   Build the left sum from the first six values. The left rule uses the value at the left edge of each strip, so it drops $f(5)$:

   $L_6=\tfrac12\,(5+9+14+20+27+35)=\tfrac12(110)=55$

5. 5

   Build the right sum from the last six values. The right rule drops $f(2)$ and picks up $f(5)$:

   $R_6=\tfrac12\,(9+14+20+27+35+44)=\tfrac12(149)=74.5$

   Notice $R_6-L_6=\tfrac12\big(f(5)-f(2)\big)=\tfrac12(39)=19.5$ — the two sums differ only in their first and last terms.

6. 6

   Interpret the two numbers as a bracket. $f$ is increasing on $[2,5]$, so every left rectangle sits under the curve and every right rectangle overshoots it. The true area therefore satisfies $55<A<74.5$. Integrating exactly gives

   $A=\int_2^5\left(2x^2-x-1\right)dx=\frac{129}{2}=64.5$

   which indeed lies between them.