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int_(0)^(x)(2^(t))/(2^([t]))dt, where [.] denotes the greatest integer function, and x epsilonR^(+) is equal to |
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Answer» `1/(1n2)([x]+2^({x})-1)` `I=int_(0)^(x)(2^(t))/(2^([t]))DT=int_(0)^(n)2^({t})dt+int_(n)^(x)2^({t})dt` `=n int_(0)^(1)2^({t})dt+int_(n)^(x)2^({t})dt` `=n int_(0)^(1)2^(t)dt+int_(n)^(x)2^(t-n)dt` `=n(2^(t))/(In2)|_(0)^(1)+1/(2^(n)) (2^(t))/(In2)|_(n)^(x)` `=n/(In2)(2-1)+1/(2^(n)In2)(2^(x)-2^(n))` `=n/(In2)+1/(In2)(2^(x-n)-1)` `=([x]+2^({x})-1)/(In2)` |
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