hiralpatel121
I need help with an attached problem please.



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hiralpatel121
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SkykhanFalcon
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2x^35=u
6x^2dx = du> integral(sqrt(u)*du/2
=u^(3/2)/2 + constant

SkykhanFalcon
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\[\int\limits_{}^{}1/\sqrt{u}\]

hiralpatel121
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Shouldn't it be u^(3/2)/3+C?

SkykhanFalcon
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yes yes my mistake :/

Valpey
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You can \[let\ u = 2x^35;\ du=6x^2\]
\[\int3x^2\sqrt{2x^35} = \frac{1}{2}\int\sqrt{u}\ du\]

SkykhanFalcon
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(u^3/2)*2/3*1/2 so (u^3/2)*1/3

Valpey
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\[=\frac{1}{3}u^{\frac{3}{2}}+C\]

hiralpatel121
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Ugh I got: