Question

@Michele_Laino

Answer 1

please not that y have to integate t function:
\[\frac{ 6t }{ 1+2t }\]
over the interval 0-3

Answer 2

namely:
\[\int\limits_{0}^{3}\frac{ 6t }{ 1+2t }dt\]

Answer 3

hint:
\[\frac{ 6t }{ 1+2t }=3-\frac{ 3 }{ 1+2t }\]

Answer 4

hint:
\[\int\limits_{0}^{3}3dt=3t|_{0}^{3}=3(3-0)=9\]

Answer 5

now, what is?\[-3\int\limits_{0}^{3}\frac{ 1 }{ 1+2t }dt=...\]

Answer 6

im not sure @Michele_Laino

Answer 7

dear @mondona please try to understand me, I can not give you the solution directly, since the code of conduct!

Answer 8

nt have my calc with me

Answer 9

have you got the table of integrals on your textbook?

Answer 10

no

Answer 11

ill ask someone else

Answer 12

ok! good idea, try please!

Answer 13

k

Answer 14

please note that, from the tables of integrals, you can read this:
\[\int\limits \frac{ 1 }{ x }dx=\ln(x)+C\]

Answer 15

i got -6.095

Answer 16

now, if I try this substitution: x=1+2t, I can write:
dx=2 dt
so the above integral can be rewritten as below:

Answer 17

please, wait a moment...
I give you all the steps

Answer 18

okay @Michele_Laino

Answer 19

\[-3\int\limits\limits_{0}^{3}\frac{ dt }{ 1+2t }=-\frac{ 3 }{ 2 }\int\limits_{1}^{7}\frac{ dx }{ x }=\]
\[=-\frac{ 3 }{ 2 }\ln (1+2t)|_{0}^{3}=-\frac{ 3 }{ 2 }\ln 7\]
so your answer is:

Answer 20

\[3-\frac{ 3 }{ 2 }\ln 7=0.0811\]

Answer 21

better is 0.081
is it right?