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योगफल की परिभाषा द्वारा `int_(0)^(1)(3x^(2)+2x+1)dx` का मान ज्ञात कीजिए । |
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Answer» `int_(0)^(1)(3x^(2)+2x+1)dx` यदि`a=0, b=1, nh=1` व `f(x)=(3x^(2)+2x+1),` तब `int_(0)^(1)(3x^(2)+2x+1)dx=underset(hrarr0)(lim)h[f(0)+f(0+h)+f(0+2h)+…+f{0+(n-1)h}]` `=underset(hrarr0)(lim)h[1+(3h^(2)+2h+1)+(3.2^(2)h^(2)+2.2h+1)+…+{3.(n-1)^(2)h^(2)+2(n-1)h+1}` `underset(hrarr0)(lim)h[n+3h^(2){1^(2)+2^(2)+3^(2)+...+(n-1)^(2)}+2h{1+2+3+...+(n-1)}]` `=underset(hrarr0)(lim)h[n+3h^(2).((n-1)n(2n-1))/(6)+2h,(n(n-1))/(2)]` `=underset(hrarr0)(lim)[nh+(1)/(2)(1-h).1.(2-h)+1.(1-h)]" जहाँ " nh=1` `=3` |
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