anonymous
  • anonymous
i need help with #17
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
i know the chain rule is used
TuringTest
  • TuringTest
not really, it should just be a u-sub\[u=t+4\]\[du=dt\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
i need more help than that
TuringTest
  • 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
  • anonymous
yeah im stuck at 1/2(t+4)^(-1/2)
TuringTest
  • TuringTest
you took the derivative or something...
TuringTest
  • TuringTest
we are integrating here, that's the opposite
anonymous
  • anonymous
that response was to whoever said use the power rule... they deleted their comment
TuringTest
  • 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
  • anonymous
im not sure how to do that
TuringTest
  • 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
  • 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
  • anonymous
the only part im not sure how to get is the - 16 / 3
TuringTest
  • TuringTest
what do you get after integrating?
TuringTest
  • TuringTest
after integrating you get ________ evaluated from __ to __ fill in the blanks
anonymous
  • anonymous
t+4 0 x
TuringTest
  • TuringTest
not quite... would it help you if I reminded you that we can rewrite this as\[\int u^{1/2}du\]??
TuringTest
  • TuringTest
then you can use the power rule for \(integration\) (basically the opposite of the power rule for differentiation)
TuringTest
  • TuringTest
which is\[\int x^ndx=\frac{x^{n+1}}{n+1}\]
anonymous
  • anonymous
can you solve the problem so i can see how its done? ill understand it better that way.
TuringTest
  • 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
  • anonymous
i try not to do that unless i forget a step or something
TuringTest
  • 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
  • anonymous
not really lol
TuringTest
  • TuringTest
ok full solution:
TuringTest
  • 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
  • TuringTest
very much*
anonymous
  • anonymous
i wont. thanks for your time/help!
TuringTest
  • 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
  • anonymous
ok i see what was done. thanks. :D
TuringTest
  • TuringTest
very welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.