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Question Number 28903 by amit96 last updated on 01/Feb/18

Commented by abdo imad last updated on 01/Feb/18

$$letputI={\int }_{\frac{1}{2}}^1\left[\frac{1}{x^2}\right]dxandusethech.\frac{1}{x^2}=t\iff x^2=\frac{1}{t}$$$$\iff x=\frac{1}{\sqrt{t}}\Rightarrow dx=\frac{-1}{2t\sqrt{t}}dtandI=-{\int }_1^4\left[t\right]\frac{-1}{2t\sqrt{t}}dt$$$$=\frac{1}{2}{\int }_1^4\frac{\left[t\right]}{t\sqrt{t}}dt=\frac{1}{2}\sum _{k=1}^{3}k{\int }_k^{m+1}\frac{dt}{t\sqrt{t}}$$$$=\frac{1}{2}\sum _{k=1}^{3}k\frac{1}{-\frac{3}{2}+1}{\left[t^{-\frac{3}{2}+1}\right]}_k^{k+1}(t^{\frac{-3}{2}}$$$$=-\sum _{k=1}^{3}k({(k+1)}^{-\frac{1}{2}}-k^{-\frac{1}{2}})=\sum _{k=1}^{3}k(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}})$$$$=1-\frac{1}{\sqrt{2}}+2(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}})+3(\frac{1}{\sqrt{3}}-\frac{1}{2})$$$$=1-\frac{1}{\sqrt{2}}+\sqrt{2}-\frac{2}{\sqrt{3}}+\frac{3}{\sqrt{3}}-\frac{3}{2}=-\frac{1}{2}+\sqrt{2}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}$$$$=-\frac{1}{2}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}andtheanswerisA.$$$$$$

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