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give me edals
so the answer is?
oh that was annoying i cldnt type

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Other answers:

give me medals free medals for everyone
can you write the equation again please :D
\[ \frac{d \ }{dx} \int_{17}^x \sin(t^2)/t \ dt \]
d/dx(sinx^2/x) - d/dx(sin(17)/17)
No
where did i go wrong?
look at the earlier example with ln
k i will
wasn't it d/dx(F(x)-F(17))
oh i see where i went wrong!!!!
Yes, but F is not the integrand.
F(x)=-cosx/x???
Nooo
K just tell me the answer
Look at the ln example again.
In the example we subsituted t for x
In the example we subsituted t for x
\[ \frac{d \ }{dx} \int_2^x \ln(t^2+1) \ dt \]
What is the value of that expression?
i dont know i am feeling very stupid!!!!!! dF/dx=ln(x^2+1)
If \[ F(t) = \int \ln(t^2+1) \ dt \] then \[ \int_2^x \ln(t^2+1) \ dt = F(x) - F(2) \]. Hence \[\frac{d \ }{dx} \int_2^x \ln(t^2+1) \ dt = \frac{d \ }{dx} (F(x) - F(2)) \]
righ
and by the Fund Theorem of Calculus, dF/dx = ln(x^2 + 1). Also (d/dx)F(2) = 0 because F(2) is a number, a constant. Hence the entire expression is just ln(x^2+1)
James do you think you help me please?
So, return now to the next example where the integrand has a sin function. What is the value of that derivative?
@mft, I'll try and get to it. unfortunately you're not the first to ask.
dont take james away form me:(
i can wait
quick rld613. I want to move on with you to yet another example afterwards
rld said she didn't need OS anymore
so (d/dx) int_17^x sin(t^2)/t dt = ...
i didnt get that why is 17 to th epower of x
\[ (d/dx) \int_{17}^x \sin(t^2)/t \ dt \]
yes
evaluate. What is the that equal to.
y is it wrong what i showed u b4?
plz be nice to me and show me all the step? :D
No. You write out the steps, following line by the line the example we just used for ln. I'm going to help someone else for a while and expect the right answer when I get back ;-)
LOL see u later :D
I'm serious. And I'll be back in 5 minutes, so hurry.
??
u did u come back already!!!
What's the answer?
K can u walk me thru the steps cuz like i bever learnt this stuff? YOu r so scary LOL
I was hoping i wld have the answer b4 u wld return
teh answer wld sinx/x?
Exactly as before, exactly, let \[ F(t) = \int \sin(t^2)/t \ dt \] Then \[ \int_{17}^x \sin(t^2)/t \ dt = F(x) - F(17) \] Thus the derivative of this definite integral \[ \frac{d\ }{dx} (F(x) - F(17)) = ... \]
iSo what is F(x)?????
That is where i am going wrong
oh i know si(x)?
d/dx(si(x)?
like we've seen, we actually don't need an explicit formula for F(x). We just need to know it is this integral and then use the Fundamental Theorem of Calculus.
What is dF/dx here? What is it equal to?
By the FTC, dF/dx = .... what?
You are forsure ready to kill me!!!!!!!!!!!!!!!
You are forsure ready to kill me!!!!!!!!!!!!!!!
shakes head
dF/dx= sinx^2/x
correct. Hence the derivative wrt (with respect to) x of the entire integral equals what?
hey james can you help me with math?
This thing ... what's it equal to?
sinx^2/x????? this is a guess
it shouldn't be guess. it is exactly right.
oh ya !!!!!!!!
That is what i thought in the first place but i left out by accident the squared
I can be so annoying sometimes
Now one more to make sure you really understand. Evaluate this: \[ \frac{d \ }{dx} \int_x^{x^2} \sqrt{t^3+1} \ dt \]
oh u r so funny!!!
Now write out the steps. This problem isn't exactly the same as the others.
ok give me a sec
\[d/dx=(\sqrt{x ^{6}+1}-\sqrt{x ^{3}+1})\]
This one i guessed
I think it was suppossed to be dF/dx
The expression is equal to \[ \frac{d \ }{dx} \left( F(x^2) - F(x) \right) \ \ \ \ \ \ --(*)\] where by the Fund Theorem of Calculus, \[ dF/dx = \sqrt{x^3+1} \] Now given that, evaluate the first expression, (*).
hint: you need to use the chain rule.
ok so let me try that
james u gotta help me out on this one
you think about it for a bit and write the answer sometime in the next day when you know what's going on.
alrighty thanks james
I will call you back when i get the
I got the answer :D
\[2x \sqrt{x ^{6}+1} - \sqrt{x ^{3}+1}\] Here is my answer. Thanks james for your help. I really appreciate it
that's it. Good
One more variation they might throw at you in the exam. Evaluate: \[ \frac{d^2 \ }{dx^2} \int_2^x \ln(\cos t) \ dt \]
Oh this is very simple there is just one small catch that you have to differentiate it twice
d/dx(ln(cosx)) (1/cosx)*-sin(x)= -sin(x)/cos(x) -tan(x)
exactly, good.
I think that is the answer. I took my final exam already and it was basically simple except ofr a related rates problem. Going on vacation for a month So see u then. Thanks for your help james
oh u are here LOL
Thanks for helping me out
sure. have a good vacation.
U better take a vacation from OS tooo!!! :D
I might just do that.
Ya u need a break to refresh again
OS wldnt survive without James myinaya satellite and Hero

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