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](Q99161.png)
$$\left.\mathrm{a}\right)\:\mathrm{divergent}\:\mathrm{integral} \\ $$$$\left.\mathrm{b}\right)\int_{−\mathrm{1}} ^{\mathrm{1}} \:\mathrm{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} \:\mathrm{dx}\:=\left[\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}}\mathrm{x}^{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}} \right]_{−\mathrm{1}} ^{\mathrm{1}} \:=\left[\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{x}}\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$=\mathrm{3}\left\{\mathrm{1}−\left(−\mathrm{1}\right)\right)\:=\mathrm{6} \\ $$
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](Q99162.png)
$$\left.\mathrm{c}\right)\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnx}}{\sqrt{\mathrm{x}}}\mathrm{dx}\:\:\mathrm{changemrnt}\:\sqrt{\mathrm{x}}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}}\left(\mathrm{2t}\right)\mathrm{dt}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{lnt}\:\mathrm{dt} \\ $$$$=\mathrm{4}\left[\mathrm{tlnt}−\mathrm{t}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{4}\left\{−\mathrm{1}\right\}\:=−\mathrm{4} \\ $$
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 =∞](Q99174.png)
$$\left(\mathrm{a}\right)\int_{\mathrm{0}} ^{\infty} \mathrm{cos}\:{xdx}\:=\:\underset{{t}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{{t}} \:\mathrm{cos}\:{xdx}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{sin}\:{x}\right]_{\mathrm{0}} ^{{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:{t}\:=\infty \\ $$