The rectangles in the graph below illustrate a left endpoint Riemann sum for $\displaystyle f(x) = \frac{-x^{2}}{4}+2x$ on the interval $\lbrack 2, 6 \rbrack$.
The value of this left endpoint Riemann sum is , and it is the area of the region enclosed by $\displaystyle y = f(x)$, the x-axis, and the vertical lines x = 2 and x = 6.

 Left endpoint Riemann sum for $y = \frac{-x^{2}}{4}+2x$ on $\lbrack 2, 6 \rbrack$

The rectangles in the graph below illustrate a right endpoint Riemann sum for $\displaystyle f(x) = \frac{-x^{2}}{4}+2x$ on the interval $\lbrack 2, 6 \rbrack$.
The value of this right endpoint Riemann sum is , and it is an the area of the region enclosed by $\displaystyle y = f(x)$, the x-axis, and the vertical lines x = 2 and x = 6.

 Right endpoint Riemann sum for $y = \frac{-x^{2}}{4}+2x$ on $\lbrack 2, 6 \rbrack$

Using left and right Riemann sums based on the diagrams above, we definitively conclude that

$\displaystyle \leq \int_{2}^{4} \frac{-x^{2}}{4}+2x \, dx \leq$

$\displaystyle \leq \int_{4}^{6} \frac{-x^{2}}{4}+2x \, dx \leq$

$\displaystyle \leq \int_{2}^{6} \frac{-x^{2}}{4}+2x \, dx \leq$

Hint: For the last integral, you should consistently choose either to underestimate or overestimate the area. This may require that you use the left Riemann sum for some x-intervals and the right Riemann sum for other x-intervals.