योगफल कि सीमा के रूप ` int _(0)^(1) x^(2) dx ` का मान ज्ञात कीजिए ।

Explanation

` int _(a)^(b) f(x) dx = lim_(h to 0) h sum_(r=1)^(n) f(a+rh)` ,
जहाँ `nh = b-a " तथा " n to oo " यहाँ ", f (x)^(2) , ` a = 0 तथा b= 1
` :. f(a+rh) = f(0+rh) = f(rh) = (rh)^(2) = r^(2) h^(2)`
तथा ` nh = b-a = 1- 0 =1 `
अब , ` int _(0)^(1) f(x) dx = lim_(h to 0) h sum_(r=1)^(n) f(a+rh) = lim_(h to 0) h sum _(r=1)^(n) r^(2)h^(2) `
` = lim _(h to 0 ) h^(3) sum _(r =1)^(n) r^(2) = lim _(h to 0 ) (h^(3) n(n+1) (2n+1))/6`
` = lim _(h to 0 ) (nh (nh+h) (2nh+h))/6 = lim _(h to 0 ) (1.(1+h) (2 xx 1 + h))/6 = 1/3 " " [ :' nh = 1 ] `
अतः ` int _(0)^(1) x^(2) dx = 1/3`