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i need help with #17

Mathematics
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i know the chain rule is used
not really, it should just be a u-sub\[u=t+4\]\[du=dt\]

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i need more help than that
well if you use that sub what does this integral become\[\int_{0}^{x}\sqrt{t+4}dt\]?write it in terms of u
yeah im stuck at 1/2(t+4)^(-1/2)
you took the derivative or something...
we are integrating here, that's the opposite
that response was to whoever said use the power rule... they deleted their comment
they just didn't quite understand your question I think \[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]now rewrite the integral in terms of u
im not sure how to do that
\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu\]which you should be able to integrate
notice we put the \(u\) where the \(t+4\) was and the \(du\)0 where the \(dt\) was. that's the essence of how a u-sub in integration works
the only part im not sure how to get is the - 16 / 3
what do you get after integrating?
after integrating you get ________ evaluated from __ to __ fill in the blanks
t+4 0 x
not quite... would it help you if I reminded you that we can rewrite this as\[\int u^{1/2}du\]??
then you can use the power rule for \(integration\) (basically the opposite of the power rule for differentiation)
which is\[\int x^ndx=\frac{x^{n+1}}{n+1}\]
can you solve the problem so i can see how its done? ill understand it better that way.
I will try to guide you a little farther than I normally would, but I hope you don't ask the exact same question again later!
i try not to do that unless i forget a step or something
\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu=\int u^{1/2}du=\frac23u^{3/2}\]now put it back in terms of t and put in the evaluation\[\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}\]evaluate this \(carefully\). Now do you see where the last term comes from?
not really lol
ok full solution:
\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu=\int u^{1/2}du=\frac23u^{3/2}\]now put it back in terms of t and put in the evaluation\[\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}=\frac23(x+4)-\frac23(0+4)^{3/2}\]\[=\frac23(x+4)-\frac23(4)^{3/2}=\frac23(x+4)-\frac23(8)=\frac23(x+4)-\frac{16}3\]please don't ask me to give the final answer again, it's very buch not my style... I hope this makes sense to you after looking it over. Good luck!
very much*
i wont. thanks for your time/help!
*typo above (I dropped the ^3/2 in the last few steps)\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu=\int u^{1/2}du=\frac23u^{3/2}\]now put it back in terms of t and put in the evaluation\[\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}=\frac23(x+4)^{3/2}-\frac23(0+4)^{3/2}\]\[=\frac23(x+4)^{3/2}-\frac23(4)^{3/2}=\frac23(x+4)^{3/2}-\frac23(8)=\frac23(x+4)^{3/2}-\frac{16}3\]
ok i see what was done. thanks. :D
very welcome :)

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