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Question Number 99154 by bramlex last updated on 19/Jun/20

Answered by bramlex last updated on 19/Jun/20

Answered by abdomathmax last updated on 19/Jun/20

$$\mathrm{a})\mathrm{divergent}\mathrm{integral}$$$$\mathrm{b}){\int }_{-1}^1{\mathrm{x}}^{-\frac{2}{3}}\mathrm{dx}={\left[\frac{1}{1-\frac{2}{3}}{\mathrm{x}}^{1-\frac{2}{3}}\right]}_{-1}^1={\left[3{}^3\sqrt{\mathrm{x}}\right]}_{-1}^1$$$$=3\{1-(-1))=6$$

Answered by abdomathmax last updated on 19/Jun/20

$$\mathrm{c})\mathrm{I}={\int }_0^1\frac{\mathrm{lnx}}{\sqrt{\mathrm{x}}}\mathrm{dx}\mathrm{changemrnt}\sqrt{\mathrm{x}}=\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_0^1\frac{\ln \left({\mathrm{t}}^2\right)}{\mathrm{t}}\left(2\mathrm{t}\right)\mathrm{dt}=4{\int }_0^1\mathrm{lnt}\mathrm{dt}$$$$=4{[\mathrm{tlnt}-\mathrm{t}]}_0^1=4\{-1\}=-4$$

Answered by Rio Michael last updated on 19/Jun/20

$$\left(\mathrm{a}\right){\int }_0^{\mathrm{\infty }}\cos xdx=\underset{t\to \mathrm{\infty }}{\lim }{\int }_0^t\cos xdx$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[\sin x\right]}_0^t$$$$=\underset{t\to \mathrm{\infty }}{\lim }\sin t=\mathrm{\infty }$$

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