योगफल कि सीमा के रूप में ` int _(0)^(4) (x+e^(2x))dx` का मान ज्ञात कीजिए ।

Explanation

परिभाषा से , ` int _(a)^(b) f(x) dx = lim _(h to 0) sum _(r=1)^(n) h (a+rh) " "` …(1)
जहाँ, `nh=b-a` तथा `n to oo` यहाँ ` f(x) = x+e^(2x), a = 0 , b = 4`
` :. nh = b-a = 4-0 = 4`
` :. f(a+rh)=f(0+rh)=f(rh)=rh+e^(2rh)`
अब ( 1) से , ` int _(0)^(4) (x+e^(2x)) dx = lim_(h to 0) sum _(r=1)^(n) hf (a+rh)`
` lim _( h to 0) sum_( r=1)^(n) h ( rh + e^(2rh)) = lim_( h to 0) (h^(2) sum_(r=1)^(n) r+h sum_(r=1)^(n) e^(2rh))`
` = lim _( h to 0) [ h^(2) (1+2 +....+ n) + h (e^(2h)+e^(4h)+...+ e^(2nh))]`
`=underset(hrarr0)(lim)[h^(2)(n(n+1))/(2)+(e^(2h){(e^(2h))^(n)-1})/(e^(2h)-1)]`
` [ :' " G.P के n terms का योगफल " = (a(r^(n)-1))/(r-)] `
` = lim_(h to 0) [ (nh(nh+h))/2 + h (e^(2h)(e^(2nh)-1))/(e^(2h)-1)]`
` = lim _( hto 0) [ (4(4+h))/2 + h. (e^(2h)(e^(2xx4)-1))/(((e^(2h)-1)/(2h))2h)]`
` = (4 xx4) /2 + (e^(0).(e^(8)-1))/(1xx2) " " [ :' lim_(h to 0) (e^(2h)-1)/(2h) = 1 ] ` ,
` = 8+ (e^(8)-1)/2 = 1/2 (16+e^(8)-1)= 1/2 (15+e^(8))`