Area of a region using the limit process

**wisezeta** · January 12th 2010, 04:27 PM

The equation is y=-2x+3 and the points are [0,1]

i simplified and got

lim n as it approaches infiniy of
-2/n^2 * n(n+1)/2 + 3/n *n

the answer is supposed to be 2 but i dont understand how to get the limits to equal 2. I cant get all the n to cancel out

**nikhil** · January 12th 2010, 06:05 PM

y=-2x+3
on integration we get RHS as
=-2((x^2)/2)+3x +c
=3x-x^2+c
now put limits from 0 to 1
so area obtained will be
=3(1)-1+c-[3(0)-0+c]
=3-1
=2 sq.units

**galactus** · January 12th 2010, 06:09 PM

${\Delta}x=\frac{1-0}{n}=\frac{1}{n}$

$x_{k}=\frac{k}{n}$

$f(x_{k})==2(\frac{k}{n})+3$

$f(x_{k}){\Delta}x=\left[-2(\frac{k}{n})+3\right]\cdot \frac{1}{n}$

$\frac{1}{n}\sum_{k=1}^{n}3-\frac{2}{n^{2}}\sum_{k=1}^{n}k$

But, $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

$\frac{1}{n}\cdot 3n-\frac{2}{n^{2}}\cdot \frac{n(n+1)}{2}$

$3-(\frac{1}{n}+1)$

Limit:

$\lim_{n\to {\infty}}\left(2-\frac{1}{n}\right)=2$

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