Split up the interval [0, 8] into 4 equally spaced subintervals:
[0, 2], [2, 4], [4, 6], [6, 8]
Take the right endpoints, which form the arithmetic sequence
![r_i=2+\dfrac{8-0}4(i-1)=2i](https://tex.z-dn.net/?f=r_i%3D2%2B%5Cdfrac%7B8-0%7D4%28i-1%29%3D2i)
where 1 ≤ <em>i</em> ≤ 4.
Find the values of the function at these endpoints:
![f(r_i)=-4{r_i}^2+32r_i=-16i^2+64i](https://tex.z-dn.net/?f=f%28r_i%29%3D-4%7Br_i%7D%5E2%2B32r_i%3D-16i%5E2%2B64i)
The area is given approximately by the Riemann sum,
![\displaystyle\int_0^8f(x)\,\mathrm dx\approx\sum_{i=1}^4f(r_i)\Delta x_i](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E8f%28x%29%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D1%7D%5E4f%28r_i%29%5CDelta%20x_i)
where
; so the area is approximately
![\displaystyle2\sum_{i=1}^4(-16i^2+64i)=-32\sum_{i=1}^4i^2+128\sum_{i=1}^4i=-32\cdot\frac{4\cdot5\cdot9}6+128\cdot\frac{4\cdot5}2=\boxed{320}](https://tex.z-dn.net/?f=%5Cdisplaystyle2%5Csum_%7Bi%3D1%7D%5E4%28-16i%5E2%2B64i%29%3D-32%5Csum_%7Bi%3D1%7D%5E4i%5E2%2B128%5Csum_%7Bi%3D1%7D%5E4i%3D-32%5Ccdot%5Cfrac%7B4%5Ccdot5%5Ccdot9%7D6%2B128%5Ccdot%5Cfrac%7B4%5Ccdot5%7D2%3D%5Cboxed%7B320%7D)
where we use the formulas,
![\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Eni%3D%5Cfrac%7Bn%28n%2B1%29%7D2)
![\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Eni%5E2%3D%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D6)