# Find the lower sum for the region bounded by f(x) = 9 - x^2 and the x-axis between x=0 and x=3.

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embizze | Certified Educator

The lower sum is found by finding the sum of the areas of the rectangles of minimum height on each interval. Since f(x) is decreasing on the given interval, we use the right endpoint of each interval.

`s_n=lim_(n->oo)sum_(i=1)^nf(c_i)Delta x_i`

We can use a partition with rectangles of equal width. So the width of each rectangle will be `Deltax=3/n` . Then `c_i=a_1+iDeltax=0+(3i)/n=(3i)/n` .So:

`s_n=lim_(n->oo)sum_(i=1)^n(9-((3i)/n)^2)(3/n)`

`=lim_(n->oo)sum_(i=1)^n (9-(9i^2)/n^2)(3/n)`

`=lim_(n->oo)sum_(i=1)^n [27/n-(27i^2)/n^3]`

`=lim_(n->oo)[27/n sum_(i=1)^n1-27/n^3sum_(i=1)^ni^2]`

`=lim_(n->oo)[27/n*n-27/n^3(n^3/3+n^2/2+n/6)]`

`=lim_(n->oo)[27-9-27/(2n)-27/(6n^2)]`

`=27-9`

`=18`