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`
No comments:
Post a Comment