#7 and #8 UVOD f(x) 7x X2 + 8 1 SXS 3 lim n- 1 Need...

Q7. f(x)= 7x/(x²+8), the area is required to be determined in the interval [1,3]

The length of the interval is 3-1=2. Let this be divided into 'n' subintervals of equal length. The length of each subinterval would be 2/n.

The area can be determined using the Riemann sums. The area under f(x) would be divided into n rectangles each of width 2/n.

The length of the rectangles can be considered either at the left end points or the right end points of these rectangles.

Here we are considering the length of these intervals at the right end points. The are of these rectangles can be summed up as

(2/n)[ f(1+2/n) +f(1 +4/n) +f(1+6/n) +...........f(1+2n/n)]

(2/n)[7{(1+2/n)/((1+2/n)² +8) +(1+4/n)/((1+4/n)²+8) +...............}]

The area under f(x) would then be Lim _n-> (2/n)[7{(1+2/n)/((1+2/n)² +8) +(1+4/n)/((1+4/n)² +8) +,,,,,,,,,,, } ].....................Ans

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