Question Number 174420 by ali009 last updated on 31/Jul/22
$${determine}\:{whether}\:{the}\:{following}\:{integral} \\ $$$${are}\:{convergence}\:{or}\:{divergence} \\ $$$$\left.\mathrm{1}\right)\int_{−\infty} ^{\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}−\mathrm{2}{x}}{dx} \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{8}−{x}^{\mathrm{3}} }}{dx} \\ $$
Answered by floor(10²Eta[1]) last updated on 31/Jul/22
$$\left.\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{2}}\int_{−\infty} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{x}−\mathrm{2}}=−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{x}−\mathrm{2}\mid_{−\infty} ^{\mathrm{1}} \\ $$$$−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{ln}\mid−\mathrm{1}\mid−\mathrm{ln}\mid−\infty\mid\right) \\ $$$$−\frac{\mathrm{3}}{\mathrm{2}}\left(−\mathrm{ln}\left(\infty\right)\right)=\infty \\ $$$$ \\ $$
Answered by floor(10²Eta[1]) last updated on 31/Jul/22
![2)u=8−x^3 ⇒du=−3x^2 dx −(1/3)∫_1 ^8 (dx/( (√u)))=−(1/3)∫_1 ^8 u^(−1/2) =−(2/3)[(√u)]_1 ^8 =−(2/3)(2(√2)−1)=((2−4(√2))/3)](Q174423.png)
$$\left.\mathrm{2}\right)\mathrm{u}=\mathrm{8}−\mathrm{x}^{\mathrm{3}} \Rightarrow\mathrm{du}=−\mathrm{3x}^{\mathrm{2}} \mathrm{dx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{8}} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{u}}}=−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{8}} \mathrm{u}^{−\mathrm{1}/\mathrm{2}} =−\frac{\mathrm{2}}{\mathrm{3}}\left[\sqrt{\mathrm{u}}\right]_{\mathrm{1}} ^{\mathrm{8}} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right)=\frac{\mathrm{2}−\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}\: \\ $$$$ \\ $$