Question Number 26771 by julli deswal last updated on 29/Dec/17
![∫_( 0) ^( [x]) (2^x /2^([x]) ) dx =](Q26771.png)
$$\underset{\:\mathrm{0}} {\overset{\:\:\:\left[{x}\right]} {\int}}\frac{\mathrm{2}^{{x}} }{\mathrm{2}^{\left[{x}\right]} }\:{dx}\:= \\ $$
Commented by prakash jain last updated on 29/Dec/17
![∫_0 ^([x]) (2^x /2^([x]) )dx=Σ_(i=0) ^([x]−1) ∫_i ^(i+1) (2^x /2^i )dx Σ_(i=0) ^([x]−1) (1/2^i )∫_i ^(i+1) 2^x dx =Σ_(i=0) ^([x]−1) (1/2^i )[(2^x /(ln 2))]_i ^(i+1) =(1/(ln 2))Σ_(i=0) ^([x]−1) (1/2^i )[2^(i+1) −2^i ] =(1/(ln 2))Σ_(i=0) ^([x]−1) 1 =(([x])/(ln 2)) pls recheck this answer.](Q26806.png)
$$\int_{\mathrm{0}} ^{\left[{x}\right]} \frac{\mathrm{2}^{{x}} }{\mathrm{2}^{\left[{x}\right]} }{dx}=\underset{{i}=\mathrm{0}} {\overset{\left[{x}\right]−\mathrm{1}} {\sum}}\int_{{i}} ^{{i}+\mathrm{1}} \frac{\mathrm{2}^{{x}} }{\mathrm{2}^{{i}} }{dx} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\left[{x}\right]−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} }\int_{{i}} ^{{i}+\mathrm{1}} \mathrm{2}^{{x}} {dx} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\left[{x}\right]−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} }\left[\frac{\mathrm{2}^{{x}} }{\mathrm{ln}\:\mathrm{2}}\right]_{{i}} ^{{i}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}\underset{{i}=\mathrm{0}} {\overset{\left[{x}\right]−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} }\left[\mathrm{2}^{{i}+\mathrm{1}} −\mathrm{2}^{{i}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}\underset{{i}=\mathrm{0}} {\overset{\left[{x}\right]−\mathrm{1}} {\sum}}\mathrm{1} \\ $$$$=\frac{\left[{x}\right]}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\mathrm{pls}\:\mathrm{recheck}\:\mathrm{this}\:\mathrm{answer}. \\ $$