dessyj1
  • dessyj1
Integration
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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dessyj1
  • dessyj1
\[\int\limits_{0}^{4} \frac{ x^3 }{ \sqrt{x^2 +1} } dx\]
dessyj1
  • dessyj1
Sending a picture of my work...
anonymous
  • anonymous
put\[\sqrt{x^2+1}=t,x2+1=t^2,2x~dx=2t~dt,x~dx=t~dt\] when x=0,t=1 when x=4,\[t=\sqrt{17}\] \[L=\int\limits_{0}^{4}\frac{ x^2*x~dx }{ \sqrt{x^2+1} }\] i hope now you can solve.

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dessyj1
  • dessyj1
dessyj1
  • dessyj1
I cant actually sub sqrtx^2+1 in for u
anonymous
  • anonymous
you are also correct.. \[\frac{ 1 }{ 2 }\int\limits_{0}^{17}\left\{ u ^{\frac{ 1 }{ 2 }}-u ^{\frac{ -1 }{ 2 }} \right\}du\]
anonymous
  • anonymous
correction write 1 in place of 0 in the limit
dessyj1
  • dessyj1
Is that a legal move? Subbing in u for the denominator?
anonymous
  • anonymous
it is simple algebra.
dessyj1
  • dessyj1
I see. I confused my self. I got it now. Thanks.
anonymous
  • anonymous
yw

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