InterviewSolution
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InterviewSolution
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`int_(0)^(4)(x+e^(2x))dx` |
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Answer» हम जानते हैं `int_(1)^(b)f(x)dx=underset(hrarr0)(lim)h[f(a)+f(a+h)+f(a+2h)+…+f{a+(n-1)h}]` जहाँ,`" "nh=b-a` `int_(0)^(4)(x+e^(2x))dx` के लिये, यहाँ, `a=0, b= 4` और `nh=4` और `f(x)=(x+e^(2x))` `therefore" "int_(0)^(4)f(x)dx=underset(hrarr0)(lim)h[f(0)+f(0+h)+f(0+2h)+...+f{0+(n-1)h}]` `rArr int_(0)^(4)(x+e^(2x))dx=underset(hrarr0)(lim)h[1+(h+e^(2h))+(2h+e^(4h))+...+(n-1)h+e^(2(n-1)h)]` `" "=underset(hrarr0)(lim)h{h+2h+3h+...+(n-1)h}+{1+e^(2h)+e^(4h)+...+e^(2(n-1)h)}` `=underset(hrarr0)(lim)h[{1+2+3+...+(n-1)}+{1+e^(2h)+e^(4h)+...+e^(2nh-2h)}]` `=underset(hrarr0)(lim)h[h((n-1)n)/(2)+1{((e^(2h))^(n)-1)/(e^(2h-1))}]` `=underset(hrarr0)(lim){h^(2)((n-1)n)/(2)+h((e^(2nh)-1))/(2^(2h)-1)}` `=underset(hrarr0)(lim)((nh-h)(nh))/(2)+underset(hrarr0)(lim)((e^(2))^(4)-1)/((e^(2h)-1)/(2h)xx2)` `=underset(hrarr0)(lim){((4-h)4)/(2)+(e^(8)-1)/(2)}(because underset(xrarr0)(lim)(e^(x)-1)/(x)=1)` `=(4xx4)/(2)+(e^(8)-1)/(2)` `=(8-(1)/(2))+(e^(8))/(2)=(15+e^(8))/(2)` |
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