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Question Number 167037 by HongKing last updated on 05/Mar/22

If I = ∫_(-1) ^( -8) (1/(x^(2/3) − x)) dx then what is the value of 81e^I ?

$$\mathrm{If}\:\:\:\:\:\mathrm{I}\:=\:\int_{-\mathrm{1}} ^{\:-\mathrm{8}} \:\frac{\mathrm{1}}{\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \:−\:\mathrm{x}}\:\mathrm{dx} \\ $$$$\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\:\mathrm{81}\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{I}}} \:? \\ $$

Answered by ghakhan last updated on 05/Mar/22

I = [−3ln∣(x)^(1/3) −1∣]_(−1) ^(−8) = −3(ln(3)−ln(2))=−3ln((3/2)) 81e^I =81e^(ln((3/2))^(−3) ) =((81∙8)/(27)) =24

$$\mathrm{I}\:=\:\left[−\mathrm{3ln}\mid\sqrt[{\mathrm{3}}]{\mathrm{x}}−\mathrm{1}\mid\right]_{−\mathrm{1}} ^{−\mathrm{8}} \: \\ $$$$=\:−\mathrm{3}\left(\mathrm{ln}\left(\mathrm{3}\right)−\mathrm{ln}\left(\mathrm{2}\right)\right)=−\mathrm{3ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\mathrm{81e}^{\mathrm{I}} =\mathrm{81e}^{\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{−\mathrm{3}} } =\frac{\mathrm{81}\centerdot\mathrm{8}}{\mathrm{27}}\:=\mathrm{24} \\ $$

Answered by qaz last updated on 05/Mar/22

I=∫_(−1) ^(−8) (dx/(x^(2/3) −x))=−∫_1 ^8 (dx/(x^(2/3) +x))=−∫_1 ^8 (dx/(x^(2/3) (1+x^(1/3) ))) =−3∫_1 ^8 ((d(x^(1/3) ))/(1+x^(1/3) ))=3ln(1+x^(1/3) )∣_8 ^1 =ln(8/(27)) ⇒81e^I =81e^(ln(8/(27))) =81×(8/(27))=24

$$\mathrm{I}=\int_{−\mathrm{1}} ^{−\mathrm{8}} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}/\mathrm{3}} −\mathrm{x}}=−\int_{\mathrm{1}} ^{\mathrm{8}} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}/\mathrm{3}} +\mathrm{x}}=−\int_{\mathrm{1}} ^{\mathrm{8}} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}/\mathrm{3}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{1}/\mathrm{3}} \right)} \\ $$$$=−\mathrm{3}\int_{\mathrm{1}} ^{\mathrm{8}} \frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{1}/\mathrm{3}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{1}/\mathrm{3}} }=\mathrm{3ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{1}/\mathrm{3}} \right)\mid_{\mathrm{8}} ^{\mathrm{1}} =\mathrm{ln}\frac{\mathrm{8}}{\mathrm{27}} \\ $$$$\Rightarrow\mathrm{81e}^{\mathrm{I}} =\mathrm{81e}^{\mathrm{ln}\frac{\mathrm{8}}{\mathrm{27}}} =\mathrm{81}×\frac{\mathrm{8}}{\mathrm{27}}=\mathrm{24} \\ $$

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