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Evaluate the definite integral...
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#72
All I see is black.
use \(u=x^3+8)\)
and change the limits of integration at the same time
From -2 to 4 of \[x^2(x^3 + 8)^2 \]
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So the limits change to 0 to 72?
yes
\[\frac{1}{3}\int_{0}^{72}u^2du\]
I got 576 as final.
idk i didn't do it you want me to check?
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Sure. I got a different answer when I plugged it into the calculator.
i got a much bigger number
41472?
should be \[\frac{1}{9}72^3\]
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Yeah I got it, just squared instead of cubed. Thank you!
yw
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