योगफल की सिमा के रूप में `int_(0)^(5)(x+1)dx` का मान ज्ञात कीजिएः

Explanation

यहाँ `a=0, b=5,f(x)=x+1,h=(b-a)/(n)=(5)/(n)" या " nh=5" योगफल की सिमा की परिभाषा से, " `
`underset(0)overset(b)intdx=underset(h to 0)lim h[f(a)+f(a+h)+h(a+2h)+...+f{a+(n-1)h}]`
`underset(0)overset(5)int (x+1) dx=underset(h to 0)lim h[f(0)+f(0+h)+f(0+2h)+...+f{0+(n-1)h}]`
`underset(0)overset(5)int(x+1) dx=underset(h to 0)lim h[f(0)+f(h)+f(2h)+...+f{(n-1)h}].....(1)`
अब `f(0)=0+1=1, f(h)=h+1`
f(2h)=2h+1
`f{(n-1)h}=(n-1)h+1`
उपरोक्त मनो को समी, (1) में रखने पर `underset(0)overset(5)int(x+1) dx`
`underset(h to 0)lim h[1+(1+h)+(1+2h)+....+{1+(n-1)h}]`
`underset(h to 0)lim [1+1+1+....+1]("n times") +h{1+2+.....+(n-1)}]`
`underset(h to 0)lim [n+hxx(n(n-1))/(2)][therefore 1+2+3+....+(n-1)=(n(n-1))/(2)]`
`underset(h to 0)lim [nh+h^(2)xx(n(n-1))/(2)]`
`underset(h to 0)lim [5+(nh)((nh-h))/(2)]=underset(h to 0)lim[5+(5(5-h))/(2)]`
`underset(h to 0)lim [5+((25-5h))/(2)]=5+((25-0))/(2)`
`=5+(25)/(2)=(35)/(2)`