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Need someone to check my answer \[\int\limits\limits_{16}^{49}\frac{ \ \sqrt{lnx} dx }{ x } \]
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I got approximately 2.04
would it help if I did the steps?
yes please do
well i used substitution. \[u=\ln x\]\[du=\frac{ dx }{ x }\]
\[\int\limits_{\ln16}^{\ln49}\sqrt{u}\times du\]
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\[\frac{ 2 }{ 3 }u ^{3/2}\]
from ln 16 to ln 49
and i got 2.04
You did well then.
does this seem correct?
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THe limits are 49 and 16 tho, not ln 49 and ln 16
he did an u substitution @Meepi, hence his limits are correct: \[\Large u=\ln x \] from x=16 to x=49
Your answer is correct @MATTW20
yeah wasn't looking my bad
Thank You :)
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you're welcome
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