`int_(2)^(3)x^(2)dx`

Explanation

हम जानते हैं
`int_(a)^(b)f(x)=underset(hrarr0)(lim)h[f(a)+f(a+h)+f(a+2h)+…+f{a+(n-1)h}]`
जहाँ `" "nh=b-a`
यहाँ,`" "a=2, b= 3` और `nh=b-a=3-2=1`
`f(x)=x^(2),f(2)=2^(2),f(2+h)=(2+h)^(2)`
`f(2+2h)=(2+2h)^(2),…,f{2+(b-1)h=[2+(n-1)h]^(2)`
`therefore int_(2)^(3)x^(2)dx=underset(hrarr0)(lim)h[f(2)+f(2+h)+f(2+2h)+...+f{2+(n-1)h}]`
`=underset(hrarr0)(lim)h[2^(2)+(2+h)^(2)+(2+2h)^(2)+...+{2+(n-1)h}^(2)]`
`=underset(hrarr0)(lim)h[2^(2)+(2^(2)+h^(2)+4h)+(2^(2)+2^(2)h^(2)+8h)+...+{2^(2)+(n-1)^(2)h^(2)+4(n-1)h}]`
`=underset(hrarr0)(lim)h[(2^(2)+2^(2)+..+n" बार ")+h^(2){1^(2)+2^(2)+3^(3)+...+(n-1)^(2)}+4h{1+2+...+(n-1)}]`
`=underset(hrarr0)(lim){4nh+h^(3)((n-1)n(2n-1))/(6)+4h^(2)((n-1)n)/(2)}`
`=underset(hrarr0)(lim){4nh+((hn-h)hn(2nh-h))/(6)+2(nh-h)(nh)}`
`=underset(hrarr0)(lim){4xx1+((1-h).1.(2xx1-h))/(6)+2(1-h).1}`
`" "(becausenh=1)`
`=underset(hrarr0)(lim)[4+((1-h)(2-h))/(6)+2(1-h)]`
`=4+(2)/(6)+2=6+(1)/(3)=(19)/(3)`