Question

i need help with #17

Answer 1

i know the chain rule is used

Answer 2

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

Answer 3

i need more help than that

Answer 4

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

Answer 5

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

Answer 6

you took the derivative or something...

Answer 7

we are integrating here, that's the opposite

Answer 8

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

Answer 9

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

Answer 10

im not sure how to do that

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

what do you get after integrating?

Answer 15

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

Answer 16

t+4
0
x

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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!

Answer 22

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

Answer 23

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

Answer 24

not really lol

Answer 25

ok full solution:

Answer 26

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

Answer 27

very much*

Answer 28

i wont. thanks for your time/help!

Answer 29

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

Answer 30

ok i see what was done. thanks. :D

Answer 31

very welcome :)