calculus help
find \[d/dx \int\limits_{2}^{x^3}\ln(x^2)dx\]
thats what i have so far but idk whts next
@mrm
Since you're taking the derivative of an integral, there is a shortcut here. A derivative will undo an integral, you just need to know what the limits of integration are an plug them into the expression inside the integral.
Remember to do the chain rule.
I forgot what the chain rule is.. I'll try to do a step
ln(x^3)^2
\[d/dx \int\limits_{2}^{x^3}\ln(x^2)dx = f(x)\] we want to find f(x) let \[G(x) = \int\limits_{2}^{x}\ln(x^2)dx\] then \[G(x^3) = \int\limits_{2}^{x^3}\ln(x^2)dx \] then we can say \[\frac{d}{dx} G(x^3) = f(x)\]
Then do i plug in 2 into the the equation?
as formulated, the answer is zero https://i.gyazo.com/bad010ce977ce9953dba65406af53d16.png
I put it into my calc and got 18.5 as the answer, but I haveto show my work
You will not get a number. You will get some function of x... f(x).
You see that one of your limits of integration is a function of x, \[x^3\]
thanks Prof!! we just need to switch the letters about, if i can be so bold as to make a recommendation
when you differential the integral, you get \[\frac{d}{dx}G(x^3) = ln(x^6)dx\]
right. so now we plug in 2?
then by the chain rule we have to take the derivative inside G... what is the derivative of \[x^3\]
3x^2
so our answer is \[f(x) = 3x^{2}\ln(x^6)\]
and that's the simplest form?
Yep.
Thanks for your help!
No problem.
that's just nonsense it's actually a meaningless integral \(\dfrac{d}{dx} \int ANYTHING ~ dx = ANYTHING\) [ftc] the limits don't matter, and they mean nothing. so the answer is \(\ln (x^2)\) i was wrong to be so flippant, it just looked silly, but this is wrong too.....IMHO!!!
Irish has a point. It should be something like \[\Large d/dx \int\limits\limits_{2}^{x^3}\ln(t^2)dt\]not the way it was originally posted.
yep! no-one cares though!!
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