Question

AP Calculus AB Help
http://prntscr.com/bdketn

Answer 1

@phi @Mehek14 @mathstudent55 @mathmate @nincompoop @Preetha @Vocaloid

Answer 2

hint: the integral of 1/t is ln(x)

Answer 3

*** i mean ln(t)

Answer 4

so you want to show that

ln(5x) - ln(2x) is a constant

any ideas how?

Answer 5

Can you integrate both?

Answer 6

not quite

Answer 7

I already integrated so the integration step is done

Answer 8

remember your log rules

Answer 9

ln(A) - ln(B) = ln(A/B)

so ln(5x) - ln(2x) = ?

Answer 10

ln(5x/2X)

Answer 11

then the x cancels

Answer 12

fantastic!

the x's cancel out giving us ln(5/2) which is a constant

Answer 13

I think you may have been able to do this using the 2nd fundamental theorem.

Answer 14

So basically by creating a constant, they're moving at the same ratE?

Answer 15

remind me how 2nd fundamental theorem works again ;_;

Answer 16

it's been a while

Answer 17

wait was i right?

Answer 18

You substitute the upper limit for the variable in the inside the integrand then you multiply that by the derivative of the upper limit.
So something like this
\[\int\limits_{2x}^{5x}\frac{ 1 }{ t }dt\] then
\[5\times \frac{ 1 }{ 5x } = \frac{ 1 }{ x }\]
although thinking about it now not sure if it works

Answer 19

ah

Answer 20

to answer tmagloire1's question i'm not entirely sure

Answer 21

what do you mean by "moving at the same rate"?

Answer 22

Yea the way you did it was the way you were supposed to. Im just adding confusion.

Answer 23

oh ok haha thanks guys!

Answer 24

\[\frac{d}{dx}\int\limits_{2x}^{5x}\frac{ 1 }{ t }dt=\frac{1}{5x}5-\frac{1}{2x}2=\frac{1}{x}-\frac{1}{x}=0\]

the derivative is zero therefore function is constant