i need help with #17

- anonymous

i need help with #17

- chestercat

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- anonymous

##### 1 Attachment

- anonymous

i know the chain rule is used

- TuringTest

not really, it should just be a u-sub\[u=t+4\]\[du=dt\]

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## More answers

- anonymous

i need more help than that

- TuringTest

well if you use that sub what does this integral become\[\int_{0}^{x}\sqrt{t+4}dt\]?write it in terms of u

- anonymous

yeah im stuck at 1/2(t+4)^(-1/2)

- TuringTest

you took the derivative or something...

- TuringTest

we are integrating here, that's the opposite

- anonymous

that response was to whoever said use the power rule...
they deleted their comment

- TuringTest

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

- anonymous

im not sure how to do that

- TuringTest

\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]so our integral (indefinite for the time being) is now\[\int\sqrt udu\]which you should be able to integrate

- TuringTest

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

- anonymous

the only part im not sure how to get is the - 16 / 3

- TuringTest

what do you get after integrating?

- TuringTest

after integrating you get
________ evaluated from __ to __
fill in the blanks

- anonymous

t+4
0
x

- TuringTest

not quite...
would it help you if I reminded you that we can rewrite this as\[\int u^{1/2}du\]??

- TuringTest

then you can use the power rule for \(integration\)
(basically the opposite of the power rule for differentiation)

- TuringTest

which is\[\int x^ndx=\frac{x^{n+1}}{n+1}\]

- anonymous

can you solve the problem so i can see how its done? ill understand it better that way.

- TuringTest

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!

- anonymous

i try not to do that unless i forget a step or something

- TuringTest

\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]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?

- anonymous

not really lol

- TuringTest

ok full solution:

- TuringTest

\[\int_{0}^{x}\sqrt{t+4}dt\]\[u=t+4\implies du=dt\]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!

- TuringTest

very much*

- anonymous

i wont. thanks for your time/help!

- TuringTest

*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\[\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}=\frac23(x+4)^{3/2}-\frac23(0+4)^{3/2}\]\[=\frac23(x+4)^{3/2}-\frac23(4)^{3/2}=\frac23(x+4)^{3/2}-\frac23(8)=\frac23(x+4)^{3/2}-\frac{16}3\]

- anonymous

ok i see what was done. thanks. :D

- TuringTest

very welcome :)

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