Question

find the exact atea under y=x^2 on the interval [0,2] by using an infinite number of circumscribed rectangles. (Hint: 1^2 + 2^2 + 3^2 +.... +n^2= n(n+1) (2n+1) /6.)

Answer 1

Answer:  $\bold{\dfrac{8}{3}}$

Step-by-step explanation:

$\int\limits^2_0 {x^2} \, dx =\dfrac{x^3}{3}\bigg|^2_0=\bigg(\dfrac{2^3}{3}\bigg)-\bigg(\dfrac{0^3}{3}\bigg)=\dfrac{8}{3}-0=\dfrac{8}{3}$

NOTE: Neither the Trapezoidal Rule or Reimann Sums can be used without the number of rectangles (n) you are separating the interval into.  Besides, they only provide an estimate. If you want the EXACT area, you must use integration.

Answer 2

We first partition the interval [0, 2] into $n$ sub-intervals:

$\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]$

Notice how each sub-interval has length $\dfrac2n$.

Since $y=x^2$ is an increasing function on the interval [0, 2] (which is evident from the fact that $\dfrac{\mathrm dy}{\mathrm dx}=2x>0$ for $0$), the circumscribed rectangles will have heights determined by the right endpoints of each sub-interval. This is to say, for the $i$-th sub-interval, the area of the rectangle is

$\dfrac2n\left(\dfrac{2i}n\right)^2=\dfrac{8i^2}{n^3}$

where $1\le i\le n$. So the area is approximated by the Riemann sum

$\displaystyle\frac8{n^3}\sum_{i=1}^ni^2=\frac{8n(n+1)(2n+1)}{6n^3}=\frac{4(n+1)(2n+1)}{3n^2}$

where the second equality makes use of the given hint, and we simplify from there.

The exact area is obtained by taking the limit as $n\to\infty$ (and by definition is equivalent to the definite integral over [0, 2]):

$\displaystyle\lim_{n\to\infty}\frac{4(n+1)(2n+1)}{3n^2}=\frac83$