Question

Anti-derivative problem

Answer 1

integrate D(t) dt, with limits 0 and 60

Answer 2

this will give the total dough supplied in 1 hour

Answer 3

\[\int\limits_{0}^{60} 1.1^{-t} = [\frac{ -1.1^{-60} }{\ln 1.1 }]-[\frac{ -1.1^{0} }{ \ln 1.1 }] = 10.46\]

Answer 4

then for 30 minutes is limit 30?

Answer 5

i can not see how you integrated that

Answer 6

so its wrong? haha

Answer 7

i dont know,
how did you integrate it?

Answer 8

\[\int\limits_{?}^{?} a ^{x}dx = \frac{ a^{x} }{ lna } +c\]

Answer 9

i used this formula

Answer 10

ok, lets see if we get the same thing if we try a different method

Answer 11

\[1.1^{−t}=e^{\ln(1.1^{−t})}=e^{-t\ln (1.1)}\]

Answer 12

how did u get that?

Answer 13

\[e^{\ln x}=x\]

Answer 14

so\[\int\limits_0^{60}1.1^{-t}\mathrm dt=\int\limits_0^{60}e^{-t\ln (1.1)}\mathrm dt\]

Answer 15

ah, okay so \[\frac{ -e ^{-tln1.1} }{ \ln 1.1} + c\]

Answer 16

yeah

Answer 17

then p(0)=0, thus c = \[\frac{ 1 }{ \ln1.1 }\]

Answer 18

since u said e^lnx = x , so must be \[p(t) =\frac{ 1-1.1t }{ \ln1.1 } -> \frac{ 1-1.1^{-60} }{ \ln1.1 } = 10.46\]

Answer 19

That is right,
and i can see why the formula you used also worked and gets that same answer,
but because this is an definite integral (has limits) you should evaluate the limits this way.


\[=\left.\frac{ -e ^{-t\ln1.1} }{ \ln 1.1}\right|_0^{60}\\
=\left.\frac{ -{1.1} ^{-t} }{ \ln 1.1}\right|_0^{60}\\
=\frac{-{1.1} ^{-60} }{ \ln 1.1}-\frac{-1.1^0}{ \ln 1.1}\\
=\frac{1 -{1.1} ^{-60} }{ \ln 1.1}=10.46\]

but yeah you got the right answer

Answer 20

haha yeah the problem now is the last question.

Answer 21

its the same integral with different limits

Answer 22

but i think if i used the first method, i cannot calculate for infinity right? coz your method finds c so in the end i'll end up having \[\lim_{x \rightarrow \infty} p(t) = \frac{ 1-1.1^{\infty} }{ \ln1.1 }= \frac{ 1 }{ \ln1.1} \approx 10.49\]

Answer 23

That is correct

Answer 24

will earth be overwhelmed with spaghetti?

Answer 25

hmm, actually its the same haha.. coz i just realised \[\frac{ -1.1^{-\infty} }{ 1n1.1 } = 0\]
since -1.1^-(infinity) = 0 so ill have the last expression left which is the same as 1/ln1.1

Answer 26

yeah both methods work the same

Answer 27

the answer is no coz its only <11 kg lol.

Answer 28

thank you, now its very clear!(:

Answer 29

did you work out the amount made between 30 and 60 mins also?

Answer 30

hmm, yeah if 60 mins can produce 10.46 and the first 30 mins 9.89, then the 2nd 30 mins would be 10.46-9.89=0.57 right? coz the limit is 1 hr. what do u think? haha

Answer 31

yes, that is the easiest way to work it out,
because
\[\int\limits_{30}^{60} =\int\limits_0^{60}-\int\limits_0^{30}\]

Answer 32

oh ya, thats the mathematical way of expressing it xD thanks!