InterviewSolution
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InterviewSolution
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योगफल की सिमा के रूप में `underset(1)overset(3)(3x^(2)+1)dx` का मान ज्ञात कीजिएः |
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Answer» यहाँ a=1, b=3, `f(x)=3x^(2)+1,h=(b-a)/(n)=2/n"या " nh=2` योगफल की सिमा की परिभाषा से, `underset(0)overset(b)int f(x)dx=underset(h to 0)lim h[(f(a)+f(a+h)+f(a+2h)+...+f{a+(n-1)h}]` `underset(1)overset(3)int (3x^(2)+1)dx=underset(h to 0)lim h[(f(1)+f(1+h)+f(1+2h)+...+f{1+(n-1)h}]........(1)` अब `f(1)=3(1)^(2)+1=4` `f(1+h)=3(1+h)^(2)+1=3(1+2h+h^(2))+1` `=3+6h+3h^(2)+1=4+6h+3h^(2)` `f(1+2h)=3(1+2h)^(2)+1=3(1+4h+4h^(2))+1` `=3+12h+12h^(2)+1=4+12h+12h^(2)` `f(1+(n-1)h]=3[{1+(n-1)h}]^(2)+1` `=3(1+2(n-1)h+(n-1)^(2)h+(n-1)^(2)h^(2)]+1` `=4+6h(n-1)+3h^(2)(n-1)^(2)` उपरोक्त मनो को समी, में रखने पर `underset(1)overset(3)int (3x^(2)+1)dx` `=underset(h to 0)limh[4+(4+6h+3h^2)+(4+12h+12h^(2))+{4+6h(n-1)+3h^(2)(n-1)^(2)}]` `=underset(h to 0)limh[4+4+....+4("n बार ")+{6h+12h+....+6h(n-1)}+{3h^(2)+12h^(2)+....+3h^(2)(n-1)^(2)}]` `=underset(h to 0)limh[4+4+....+4("n बार ")+6h(1+2+...+(n-1)}+3h^(2){1^(2)+2^(2)+.....+(n-1)^(2)}]` `=underset(h to 0)lim h[4n+6h xx (n(n-1))/(2)+3h^(2)xx(n(n-1)(2n-1))/(6)]` `=underset(h to 0)lim h[4n+6(nh(nh-h))/(2)+3(nh(nh-h)(2nh-h))/(6)]` `=underset(h to 0)lim h[4xx2+(6xx2xx(2-h))/(2)+(3xx2xx(2-h)(2xx2-h))/(6)]` `=8+(12(2-0))/(2)+(6(2-0)(4-0))/(6)` =8+12+8=28 |
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