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mariomintchev
 3 years ago
i need help with #17
mariomintchev
 3 years ago
i need help with #17

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mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0i know the chain rule is used

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3not really, it should just be a usub\[u=t+4\]\[du=dt\]

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0i need more help than that

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3well if you use that sub what does this integral become\[\int_{0}^{x}\sqrt{t+4}dt\]?write it in terms of u

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0yeah im stuck at 1/2(t+4)^(1/2)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3you took the derivative or something...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3we are integrating here, that's the opposite

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0that response was to whoever said use the power rule... they deleted their comment

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3they 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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0im not sure how to do that

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu\]which you should be able to integrate

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3notice we put the \(u\) where the \(t+4\) was and the \(du\)0 where the \(dt\) was. that's the essence of how a usub in integration works

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0the only part im not sure how to get is the  16 / 3

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3what do you get after integrating?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3after integrating you get ________ evaluated from __ to __ fill in the blanks

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3not quite... would it help you if I reminded you that we can rewrite this as\[\int u^{1/2}du\]??

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3then you can use the power rule for \(integration\) (basically the opposite of the power rule for differentiation)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3which is\[\int x^ndx=\frac{x^{n+1}}{n+1}\]

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0can you solve the problem so i can see how its done? ill understand it better that way.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3I 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!

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0i try not to do that unless i forget a step or something

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]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?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]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!

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0i wont. thanks for your time/help!

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3*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\[\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}_{0}^{x}=\frac23(x+4)^{3/2}\frac23(0+4)^{3/2}\]\[=\frac23(x+4)^{3/2}\frac23(4)^{3/2}=\frac23(x+4)^{3/2}\frac23(8)=\frac23(x+4)^{3/2}\frac{16}3\]

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0ok i see what was done. thanks. :D
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