InterviewSolution
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InterviewSolution
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योगफल की सिमा के रूप में `int_(0)^(2) e^(x)dx` का मान ज्ञात कीजिएः |
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Answer» यहाँ `=0,b=2, f(x)=e^(x), h=(b-a)/(n)=2/n" या "nh=2` योगफल की सिमा की परिभाषा से, `underset(a)overset(b)int f(x)dx=underset(h to 0)lim h[f(a+h)+f(a+2h)+....+f{a+(n-1)h}]` `Rightarrow underset(0)overset(2)int e^(x)edx=underset(h to0)lim [f(0)+f(0+h)+f(0+2h)+....+f{0+(n-1)h}]` `Rightarrow underset(0)overset(2)int e^(x)edx=underset(h to0)lim [f(0)+f(h)+f(2h)+....+f{(n-1)h}]......(1)` ltBRgt अब `f(0)=e^(0)=1, f(h)=e^(h),f(2h)=e^(2h),f{n-1)h}=e^((n-1)h)` उपरोक्त मनो को समी, (1) में रखने पर `underset(0)overset(2)inte^(x)dx=underset(h to 0)lim h[e^(0)+e^(h)+e^(2h)+....+e^((n-1)h)]` `underset(0)overset(2)inte^(x)dx=underset(h to 0)lim h[e^(0)+e^(h)+(e^(h))^(2)+....+(e^(h))^(n-1)]` `=underset(h to 0)lim h[e^(0){(e^(h))^(n)-1)/(e^(h)-1)]` `[{:(,therefore a+ar+...+ar^(n-1)=(a(r^(n-1)))/(r-1)),(,"यहाँ "r=e^(h)):}]` `=underset(h to 0)lim h {(e^(nh)-1)/(e^(h)-1)}=underset(h to 0)limh{(e^(2)-1)/((e^(h)-1))}` `=underset(h to 0)lim h ((e^(2)-1))/(((e^(2)-1)/(h)))=(e^(2)-1)/(1) " "[therefore underset(h to 0)lim (e^(h)-1)/(h)=1]` `=e^(2)-1)` |
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