Question

What? I don't even....
https://i.gyazo.com/c8e7baefa826eaa192989cd431857b66.png

TBH pretty good with calc but this one has me stumped any help?

Answer 1

do you know how to integrate 1/t ?

if not all is not over... do you know how to differentiate the given integral ? if we can show the derivative is 0 then that means we have shown the original is constant

Answer 2

What a second, if i remember correcty changing this into a function of x simple means we can ignore the lower value of the integral and do a U substitution of 4x in, giving me the integral of 1/u all times 4 since that is put out front?

Answer 3

but that gives me 4*(ln|4x|+c)

Answer 4

I do not understand your substitution


but anyways I can work with part of what you said ... you do know the antiderivative of 1/t is ln(t)
so I will work that...
\[F(x)=\int\limits_x^{4x} \frac{1}{t} dt= \ln|t||_x^{4x}\]

Answer 5

now enter in your limits
plug in upper limit then minus plug in lower limit

Answer 6

so that is ln|4x|-ln|x|

Answer 7

right

Answer 8

now we are given x>0

Answer 9

so you can write this as ln(4x)-ln(x)

Answer 10

use properties of log to rewrite the ln(4x)

Answer 11

or you can use properties of log to rewrite ln(4x)-ln(x)
either way

Answer 12

two choices:
\[\text{ use } \ln(ab)=\ln(a)+\ln(b) \\ \text{ or use } \ln(\frac{a}{b})=\ln(a)-\ln(b)\]

Answer 13

either one will work

Answer 14

Well since the second one has a minus sign and it proves that our function is constant i think I will choose that one
also ln(4)

Answer 15

either one would have worked you can write ln(4x) as ln(4)+ln(x)

so ln(4x)-ln(x)
=ln(4)+ln(x)-ln(x)
=ln(4)+0
=ln(4)

Answer 16

anyways there is another way to do this problem

Answer 17

\[F(x)=\int\limits_x^{4x} \frac{1}{t} dt \\ F'(x)= (4x)'\frac{1}{4x}-(x)' \frac{1}{x} \\ F'(x)=4 \frac{1}{4x}-1 \frac{1}{x} \\ F'(x)=\frac{1}{x}-\frac{1}{x} \\ F'(x)=0 \\ \implies F(x)=C \\ \text{ where } C \text{ is a constant }\]

Answer 18

this is just using the other part of the fundamental theorem of calculus

Answer 19

TY very much for your help!

Answer 20

if this confused you can always select an arbitrary constant between x and 4x
\[F(x)=\int\limits_x^{4x} \frac{1}{t} dt \\ F(x)=\int\limits_x^{1} \frac{1}{t} dt+\int\limits_1^{4x} \frac{1}{t} dt \\ F(x)=-\int\limits_1^x \frac{1}{t} dt +\int\limits_1^{4x} \frac{1}{t} dt \\ F'(x)=[ -\int\limits_1^x \frac{1}{t} dt +\int\limits_1^{4x} \frac{1}{t} dt]' \\ F'(x)=[- \int\limits_1^{x} \frac{1}{t} dt]'+[\int\limits_1^{4x} \frac{1}{t} dt]' \\ \\ F'(x)=-[\int\limits_1^x \frac{1}{t} dt ]'+(4x)' \frac{1}{4x} \\ F'(x)=- \frac{1}{x}+4 \frac{1}{4x} \\ F'(x)=\frac{-1}{x}+\frac{1}{x} \\ F'(x)=0\]
you know if you are used to using fundamental theorem of calculus when the upper limit is is a variable and the lower limit is a constant

Answer 21

but yeah you can skip all of those steps :p