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Question-99154




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
a) divergent integral  b)∫_(−1) ^1  x^(−(2/3))  dx =[(1/(1−(2/3)))x^(1−(2/3)) ]_(−1) ^1  =[3^3 (√x)]_(−1) ^1   =3{1−(−1)) =6
$$\left.\mathrm{a}\right)\:\mathrm{divergent}\:\mathrm{integral} \\ $$$$\left.\mathrm{b}\right)\int_{−\mathrm{1}} ^{\mathrm{1}} \:\mathrm{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} \:\mathrm{dx}\:=\left[\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}}\mathrm{x}^{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}} \right]_{−\mathrm{1}} ^{\mathrm{1}} \:=\left[\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{x}}\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$=\mathrm{3}\left\{\mathrm{1}−\left(−\mathrm{1}\right)\right)\:=\mathrm{6} \\ $$
Answered by abdomathmax last updated on 19/Jun/20
c) I =∫_0 ^1  ((lnx)/( (√x)))dx  changemrnt (√x)=t give  I =∫_0 ^1  ((ln(t^2 ))/t)(2t)dt =4 ∫_0 ^1  lnt dt  =4[tlnt−t]_0 ^1  =4{−1} =−4
$$\left.\mathrm{c}\right)\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnx}}{\:\sqrt{\mathrm{x}}}\mathrm{dx}\:\:\mathrm{changemrnt}\:\sqrt{\mathrm{x}}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}}\left(\mathrm{2t}\right)\mathrm{dt}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{lnt}\:\mathrm{dt} \\ $$$$=\mathrm{4}\left[\mathrm{tlnt}−\mathrm{t}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{4}\left\{−\mathrm{1}\right\}\:=−\mathrm{4} \\ $$
Answered by Rio Michael last updated on 19/Jun/20
(a)∫_0 ^∞ cos xdx = lim_(t→∞) ∫_0 ^t  cos xdx                                = lim_(t→∞)  [sin x]_0 ^t                               = lim_(t→∞)  sin t =∞
$$\left(\mathrm{a}\right)\int_{\mathrm{0}} ^{\infty} \mathrm{cos}\:{xdx}\:=\:\underset{{t}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{{t}} \:\mathrm{cos}\:{xdx}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{sin}\:{x}\right]_{\mathrm{0}} ^{{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:{t}\:=\infty \\ $$

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