Question Number 28903 by amit96 last updated on 01/Feb/18
Commented by abdo imad last updated on 01/Feb/18
![let put I= ∫_(1/2) ^1 [(1/x^2 )]dx and use the ch. (1/x^2 )=t ⇔x^2 =(1/t) ⇔x=(1/(√t)) ⇒ dx=((−1)/(2t(√t)))dt and I= −∫_1 ^4 [t]((−1)/(2t(√t)))dt = (1/2) ∫_1 ^4 (([t])/(t(√t)))dt =(1/2) Σ_(k=1) ^3 k ∫_k ^(m+1) (dt/(t(√t))) =(1/2) Σ_(k=1) ^3 k (1/(−(3/2)+1)) [ t^(−(3/2)+1) ]_k ^(k+1) ( t^((−3)/2) =− Σ_(k=1) ^3 k( (k+1)^(−(1/2)) −k^(−(1/2)) )=Σ_(k=1) ^3 k((1/(√k)) − (1/(√(k+1)))) =1−(1/(√2)) +2( (1/(√2)) −(1/(√3))) +3( (1/(√3)) −(1/2)) =1−(1/(√2)) +(√2) −(2/(√3)) +(3/(√3)) −(3/2)=−(1/2) +(√2) −(1/(√2)) +(1/(√3)) =−(1/2) +(1/(√2)) +(1/(√3)) and the answer is A.](Q28910.png)
$${let}\:{put}\:{I}=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\left[\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right]{dx}\:\:\:{and}\:{use}\:{the}\:{ch}.\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={t}\:\Leftrightarrow{x}^{\mathrm{2}} =\frac{\mathrm{1}}{{t}} \\ $$$$\Leftrightarrow{x}=\frac{\mathrm{1}}{\sqrt{{t}}}\:\Rightarrow\:{dx}=\frac{−\mathrm{1}}{\mathrm{2}{t}\sqrt{{t}}}{dt}\:\:\:{and}\:\:{I}=\:−\int_{\mathrm{1}} ^{\mathrm{4}} \left[{t}\right]\frac{−\mathrm{1}}{\mathrm{2}{t}\sqrt{{t}}}{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{4}} \:\:\:\frac{\left[{t}\right]}{{t}\sqrt{{t}}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{\mathrm{3}} \:{k}\:\int_{{k}} ^{{m}+\mathrm{1}} \:\frac{{dt}}{{t}\sqrt{{t}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{\mathrm{3}} \:{k}\:\frac{\mathrm{1}}{−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}}\:\left[\:{t}^{−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}} \right]_{{k}} ^{{k}+\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:{t}^{\frac{−\mathrm{3}}{\mathrm{2}}} \right. \\ $$$$=−\:\sum_{{k}=\mathrm{1}} ^{\mathrm{3}} \:{k}\left(\:\:\left({k}+\mathrm{1}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:−{k}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)=\sum_{{k}=\mathrm{1}} ^{\mathrm{3}} {k}\left(\frac{\mathrm{1}}{\sqrt{{k}}}\:−\:\frac{\mathrm{1}}{\sqrt{{k}+\mathrm{1}}}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:+\mathrm{2}\left(\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:+\mathrm{3}\left(\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\:+\sqrt{\mathrm{2}}\:−\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:+\frac{\mathrm{3}}{\sqrt{\mathrm{3}}}\:−\frac{\mathrm{3}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}}\:+\sqrt{\mathrm{2}}\:−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\:\:{and}\:{the}\:{answer}\:{is}\:{A}. \\ $$$$ \\ $$