# (Warning: Tough question ahead) I'm stuck on this question and would love some help through it. - Thanks

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The limit process divides the given area into a number of rectangles and the area under the curve is calculated as the sum of areas of all such rectangles.

Here, f(x) = `3-x^2 , -1 <=x >= 1`

by definition, area = `lim_(n-> oo) sum_(i=1)^n f(c_i)Delta x = lim_(n->oo) sum_(i=1)^n [3-(-1+(2i)/n)^2]((1-(-1))/n)`

`= lim_(n->oo) sum_(i=1)^n [3-1-4(i^2 /n^2) + 4i/n] (2/n) `

`= lim_(n->oo) (2/n) sum_(i=1)^n [2 - 4(i^2/n^2) +4i/n]`

`= lim_(n->oo) (2/n) [sum_(i=1)^n 2 - (4/n^2) sum_(i=1)^n (i^2) + (4/n) sum_(i=1)^n i ]`

`= lim_(n->oo) (2/n) [ 2n - (4/n^2) (n(n+1)(2n+1))/6 +(4/n) (n(n+1))/2]`

`= lim_(n->oo) (16n^2 -4)/(3n^2)`

`= lim_(n->oo) [((n^2)(16-4/n^2))/(3n^2)]`

`= lim_(n->oo) (16-4/n^2) /3`

`= 16/3`

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Hence the area under the curve and x-axis in the given interval is **16/3**

The same can also be calculated by integrating the equation of curve between the given interval and we will get the same answer.

Hope this helps.