Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

mariomintchevBest ResponseYou've already chosen the best response.0
i know the chain rule is used
 one year ago

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

mariomintchevBest ResponseYou've already chosen the best response.0
i need more help than that
 one year ago

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

mariomintchevBest ResponseYou've already chosen the best response.0
yeah im stuck at 1/2(t+4)^(1/2)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
you took the derivative or something...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
we are integrating here, that's the opposite
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
that response was to whoever said use the power rule... they deleted their comment
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
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
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
im not sure how to do that
 one year ago

TuringTestBest 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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
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 usub in integration works
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
the only part im not sure how to get is the  16 / 3
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
what do you get after integrating?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
after integrating you get ________ evaluated from __ to __ fill in the blanks
 one year ago

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

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

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

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

TuringTestBest ResponseYou've already chosen the best response.3
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!
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
i try not to do that unless i forget a step or something
 one year ago

TuringTestBest 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?
 one year ago

TuringTestBest 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!
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
i wont. thanks for your time/help!
 one year ago

TuringTestBest 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\[\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}_{0}^{x}=\frac23(x+4)^{3/2}\frac23(0+4)^{3/2}\]\[=\frac23(x+4)^{3/2}\frac23(4)^{3/2}=\frac23(x+4)^{3/2}\frac23(8)=\frac23(x+4)^{3/2}\frac{16}3\]
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
ok i see what was done. thanks. :D
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.