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Integrate 2xSqrt{1-5x} dx
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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\[\int\limits_{}^{}2x \sqrt{1-5x} ~~dx\]
Zarkon
  • Zarkon
\[u=1-5x\]
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i tried but i couldnt get the answer

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heres my work
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u^2=1-5x , x=(u^2-1) / (-5) 2udu=-5dx \[∫2(\frac{u^2-1}{5})u(-\frac{2}{5}u)du\] \[-\frac{2}{5}∫(u^2-1)u^2du\] \[-\frac{2}{5}∫(u^4-u^2)du\] \[-\frac{2}{5}[\frac{u^5}{5}-\frac{u^3}{3}]\] \[\huge -\frac{2}{75}[3u^{5}-5u^3]\] \\[\huge -\frac{2}{75}u^3[3u^{2}-5]\]\] \[\huge -\frac{2}{75}(1-5x)^{3/2}[3(1-5x)-5]\] \[\huge -\frac{2}{75}(1-5x)^{3/2}[-2-15x]\]
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@zar
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@Zarkon
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@LagrangeSon678
anonymous
  • anonymous
?
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can you check my working...
Zarkon
  • Zarkon
\[∫2(\frac{u^2-1}{5})u(-\frac{2}{5}u)du=-\frac{4}{25}\int\left(u^2-1\right)u^2du\]
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ahh i found my mistake but , is it wrong to substitute like , e.g. u^2=1-5x
Zarkon
  • Zarkon
you can do that...but I think the substitution \(u=1-5x\) is better
Calculator
  • Calculator
1 more thing, is u^2=1-5x ---------> x=(u^2-1) / (-5) correct?
TuringTest
  • TuringTest
\[\int2x \sqrt{1-5x} dx\]\[u=1-5x\implies x={1-u\over5}\]\[du=-5dx\implies dx=-\frac15du\]\[-\frac2{25}\int(1-u)u^{1/2}du\]is probably a little more direct
Zarkon
  • Zarkon
@Calculator yes, but you forgot the - sign when you did the integration
TuringTest
  • TuringTest
but I don't see anything illegal with your rather complicated approach
Zarkon
  • Zarkon
\[∫2(\frac{u^2-1}{5})u(-\frac{2}{5}u)du\] should be \[∫2(\frac{u^2-1}{-5})u(-\frac{2}{5}u)du\]
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its a typo ,my actual working is okay, but you caught my mistakes at multiplying fractons, thanks!!!!!!!! @Zarkon @TuringTest

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