Question Number 173836 by mathlove last updated on 19/Jul/22
Commented by aleks041103 last updated on 21/Jul/22
![I=∫_(−2) ^(−1) (10(x^6 )^(1/4) −2x)dx= =10∫_(−2) ^(−1) (x^6 )^(1/4) dx − ∫_(−2) ^(−1) 2xdx= =10J−[x^2 ]_(−2) ^(−1) =16(√2)−(1−4)=3+16(√2) x^6 is even⇒J=∫_(−2) ^(−1) (x^6 )^(1/4) dx=∫_1 ^( 2) (x^6 )^(1/4) dx=∫_1 ^2 x^(3/2) dx= =[(2/5)x^(5/2) ]_1 ^2 =((8(√2))/5)](https://www.tinkutara.com/question/Q173938.png)
$${I}=\int_{−\mathrm{2}} ^{−\mathrm{1}} \left(\mathrm{10}\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} }−\mathrm{2}{x}\right){dx}= \\ $$$$=\mathrm{10}\int_{−\mathrm{2}} ^{−\mathrm{1}} \sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} }{dx}\:−\:\int_{−\mathrm{2}} ^{−\mathrm{1}} \mathrm{2}{xdx}= \\ $$$$=\mathrm{10}{J}−\left[{x}^{\mathrm{2}} \right]_{−\mathrm{2}} ^{−\mathrm{1}} =\mathrm{16}\sqrt{\mathrm{2}}−\left(\mathrm{1}−\mathrm{4}\right)=\mathrm{3}+\mathrm{16}\sqrt{\mathrm{2}} \\ $$$${x}^{\mathrm{6}} \:{is}\:{even}\Rightarrow{J}=\int_{−\mathrm{2}} ^{−\mathrm{1}} \sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} }{dx}=\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} }{dx}=\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{3}/\mathrm{2}} {dx}= \\ $$$$=\left[\frac{\mathrm{2}}{\mathrm{5}}{x}^{\mathrm{5}/\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{5}} \\ $$$$ \\ $$
Answered by CElcedricjunior last updated on 19/Jul/22
![I=∫_(−2) ^(−1) (10(√x^3 )−2x)dx=∫_(−2) ^(−1) (10x^(3/2) −2x)dx or (√x^3 ) n′est definie en [−2;−1] alors I n′existe pas dans R mais dansC](https://www.tinkutara.com/question/Q173851.png)
$$\boldsymbol{\mathrm{I}}=\int_{−\mathrm{2}} ^{−\mathrm{1}} \left(\mathrm{10}\sqrt{\boldsymbol{{x}}^{\mathrm{3}} }−\mathrm{2}\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\int_{−\mathrm{2}} ^{−\mathrm{1}} \left(\mathrm{10}\boldsymbol{\mathrm{x}}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{2}\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}} \\ $$$${or}\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }\:\:\boldsymbol{\mathrm{n}}'\boldsymbol{\mathrm{est}}\:\boldsymbol{\mathrm{definie}}\:\boldsymbol{\mathrm{en}}\:\left[−\mathrm{2};−\mathrm{1}\right] \\ $$$$\boldsymbol{\mathrm{alors}}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{n}}'\boldsymbol{\mathrm{existe}}\:\boldsymbol{\mathrm{pas}}\:\boldsymbol{\mathrm{dans}}\:\mathbb{R} \\ $$$$\boldsymbol{\mathrm{m}{ais}}\:\boldsymbol{{dans}}\mathbb{C} \\ $$
Answered by a.lgnaoui last updated on 19/Jul/22
![I=10∫_(−2) ^(−1) x^(3/2) dx−2∫xdx I=10[(x^((3/2)+1) /((3/2)+1))]_(−2) ^(−1) −2[(x^2 /2)]_(−2) ^(−1) =4[x^2 (√x) ]_(−2) ^(−1) −[x^2 ]_(−2) ^(+1) =[x^2 (4(√x)−1)]_(−2) ^(−1) x<0⇒(√x)=(√(∣x∣i^2 ))=i(√(∣x∣)) I=[x^2 (4i(√(∣x∣))−1)]_(−2) ^(−1) =4i−1−16(√2)i+4 I=3+(4−16(√2))i=3−18,56i](https://www.tinkutara.com/question/Q173855.png)
$$ \\ $$$$\mathrm{I}=\mathrm{10}\int_{−\mathrm{2}} ^{−\mathrm{1}} {x}^{\frac{\mathrm{3}}{\mathrm{2}}} {dx}−\mathrm{2}\int{xdx} \\ $$$$\mathrm{I}=\mathrm{10}\left[\frac{{x}^{\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}} }{\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}}\right]_{−\mathrm{2}} ^{−\mathrm{1}} −\mathrm{2}\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{2}} ^{−\mathrm{1}} =\mathrm{4}\left[{x}^{\mathrm{2}} \sqrt{{x}}\:\right]_{−\mathrm{2}} ^{−\mathrm{1}} −\left[{x}^{\mathrm{2}} \right]_{−\mathrm{2}} ^{+\mathrm{1}} \:\: \\ $$$$=\left[{x}^{\mathrm{2}} \left(\mathrm{4}\sqrt{{x}}−\mathrm{1}\right)\right]_{−\mathrm{2}} ^{−\mathrm{1}} \:\:\:\:{x}<\mathrm{0}\Rightarrow\sqrt{{x}}=\sqrt{\mid{x}\mid{i}^{\mathrm{2}} }={i}\sqrt{\mid{x}\mid}\: \\ $$$$\mathrm{I}=\left[{x}^{\mathrm{2}} \left(\mathrm{4}{i}\sqrt{\mid{x}\mid}−\mathrm{1}\right)\right]_{−\mathrm{2}} ^{−\mathrm{1}} =\mathrm{4}{i}−\mathrm{1}−\mathrm{16}\sqrt{\mathrm{2}}{i}+\mathrm{4} \\ $$$$\mathrm{I}=\mathrm{3}+\left(\mathrm{4}−\mathrm{16}\sqrt{\mathrm{2}}\right){i}=\mathrm{3}−\mathrm{18},\mathrm{56}{i} \\ $$