anonymous
  • anonymous
"Evaluate the integral of t^4ln(2t)dt from 3 to 7." I got a rather complicated answer that I'll amend to the question but I don't know where I went wrong.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Psymon
  • Psymon
What was your answer by chance? The answer is a bit crazy, lol.
anonymous
  • anonymous
This crashed twice while i was trying to type what i have done, but the final answer i got was 8442.85 which my online homework said was wrong. im still trying to type out where that came from
anonymous
  • anonymous
parts... yeah?

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Psymon
  • Psymon
Thats actually not far off from me just looking up the answer real fast. Maybe only a slight error.
anonymous
  • anonymous
maybe. when i plugged in and simplified the numbers before entering it in (because my homework simplifies for me) i had \[\frac{ 7^5 }{ 5 }\ln(14)-\frac{ 3^5 }{ 5 }\ln(6)-\frac{ 7^5+3^5 }{ 50 }\]
anonymous
  • anonymous
and my answer before evaluating at all was \[\frac{ t^5 }{ 5 }\ln(2t)-\frac{ t^5 }{ 50 }\]
anonymous
  • anonymous
is the exponent on the second term correct?
anonymous
  • anonymous
I can upload a picture of my work, just a sec. it looked right to me because i was integrating t^4
anonymous
  • anonymous
u = ln (2t) du = 1/t... dv = t^4 dt, v = t^5 / 5 uv-int vdu = t^5 /5 * ln (2t) - int t^4/5 dt so denominator should be 25 on second term
anonymous
  • anonymous
\[\frac{t^{5} \ln{2t}}{5}-\frac{t^{5}}{25}\]
anonymous
  • anonymous
if you integrate vdu though shouldn't that be t^5/5*1/2t? that's where i ended up with t^4/10 that eventually led to me having 50 where i probably shouldn't... but i dont understand what you're using for vdu
anonymous
  • anonymous
no du = 1/t dt remember d/dt (ln 2t) = 2/2t = 1/t
anonymous
  • anonymous
vdu = t^5/5 * 1/t dt
anonymous
  • anonymous
oh i guess it's been a while since i've derived anything other than plain old ln(x). thanks for pointing that out
anonymous
  • anonymous
no worries... i was very rusty at all of this until i started coming to this site and helping (hopefully) people. this is a good place to come and get sharp again!

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