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cal 1 question.

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1 Attachment
=cos(1+x+x^2)
fundamental theorem of calc says: the derivative of the integral is the integrand. so in the simplest language the answer is the function being integrated, but be sure to replace t by x (since the integral is a function of x, not t) \[cos(1+t+t^2)\] easy right?

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oops i meant\[cos(1+x+x^2)\] sorry.
but thats not the answer
=/
really?
we need to find the integrand from -1 to x and then derive it back
u = (1+x+x^2); du = 1+2x dx du/1+2x = dx maybe?
i am sticking with my answer. unless they changed the fundamental theorem of calculus or unless i read the problem incorrectly.
How come?! I am pretty sure it's the answer.
the integrand changes in form and then you derive the change in form down to the answer, i think
try it on a simpler problem; {S} 1+x^2 dx ; [-1,x] x + x^3/3 gets what?
x^2 +x^4/3 -(-1 + -1/3)
i am afraid i disagree. i read the problem again and it clearly asks for the derivative of the integral. if \[F(x)=\int_a^x f(t)dt\] then \[F'(x)=f(x)\]
Dx( (1/3)x^4 +x^2 + 4/3) (4/3) x^3 +2x right?
i am curious as to what the answer is supposed to be. an integral of the form asked about is not an anti-derivative, it is a reimann sum, aka the limit of a sum. the derivative of such an animal is in fact the integrand. not too hard to prove this. that is the reason why you get to use anti-derivatives to compute integrals.
i know i got the same answer BUT thats not the answer. the software only shows when an aswer is correct so i don not know the correct answer
blame the software. what exactly did you type in?
and one more thing: why aren't we differentiating 1+ t+ t^2
use the newton-leibniz theorem
you are not asked for the derivative of the INTEGRAND. you are asked for the derivative of the INTEGRAL.
I checked with a software, I got the same answer. Trust me the answer that satellite73 said earlier is right. This is the fundamental theorem of calculus; the first theorem you study in integration.
\[F(x)=\int_a^x f(t)dt\] is a function of x, not of t.
derivative of f(t)dt integral from f1(x) to f2(x) is f(f1)f1' - f(f2)f2'
okay probably the problem with teh software.
and (at the risk of repeating myself) the derivative is \[f(x)\]
so was i right in my thought? :) integrate than derive the result
hey btw why arent we differentiating 1+t+t^.5
i told u guys
{S} cos(1+t+t^2) dt ; [-1,x] = what?
the answer is cos(1 +x + x^2)
amistre, i am afraid that is not correct. the definition of the integral \[\int_a^x f(t)dt\] is not an "anti-derivative" and in fact \[\int_{-1}^x x + x^2 dx\] doesn't make any sense. you are using x both as the upper limit of integration, and as the 'dummy variable' in the integrand.
oh....
so i shoulda used another variable to test with then i spose :) thnx
Here, this is the final answer after evaluating the integral and then taking its derivative: http://bit.ly/lIvQ5I
\[\int\limits_{g(x)}^{h(x)}f'(t)dt = f[h(x)] - f[g(x)\] now if u differentiate it \[f'[h(x)]h'(x) - f'[g(x)]g'(x)\]
you seem to be asking for a closed form of the integral, which then you wish to differentiate. i have no idea what the closed form of the integral is, nor do i care. it probably doesn't even have one. the probability that you can find an 'anti derivative' for a randomly chosen function is 0. we only work with special cases where we show off by saying "wow look, i found a function whose derivative is the one given!" but in general this is impossible to do. most integration is done using numerical methods
this is called newton-leibniz rule...for derivatives of integral forms
GUYS IM EXPLAINING THE ANSWER 2 U..IS ANYONE LISTENING
rsaad did u get the answer????
yea i got it =) thnx
did u understand the working i explained?
yes i got that BUT as far as i can remember, in ftoc 1, u also differentiate wats inside the finction.....cuz of chain rule
bt weve done that havent we...ive explained above..see..ive proved the rule for any general function
him1618 you are certainly correct IF f is an anti derivative of f' etc. the point is that the derivative of the integrand is the integral irrespective of whether or not you can find an anti-derivative. you are stating a (correct) theorem which assumes you have an anti-derivative of f. in this case we do not, nor does it matter. if you look at the attachment from Anwar A you will see that mathematica didn't even find an anti-derivative. that weird looking thing after the "dx" on the first line includes C and S which are in fact themselves integrals.
oops typo. i meant to say that the derivative of the integral is the integrand, not the other way around.
yes of course having an antiderative for a function is prerequisite for doing this..bt i guess the question setter didnt care to go deep enough to check whether this function ws even integrable or not...so i guess they just want rsaad2 to solve the question...u have a point satellite..but i think the setter hasnt gone that deep with his question...so he just needs answers..;)
people just explain as to why the chain rule does not apply to whats in th funcyion that is whats in cos (.............)?
were using the chain rule in the second step here saad
yes. so what abt wats in the cos function
first we are integrating a function...with limits which are variable wrt another variable..and then we differentiate using chain rule
ok.
oh the chain rule does apply but just remember that \[\int_a^x f(t)dt\] is a function of x, not a function of t. so a composite function would look something like \[\int_a^{h(x)} f(t)dt\]
se wht i wrote above
exactly!
um =S
plz explain it a bit more
so the derivative of \[\int_{-1}^{x^3} cos(1+t+t^2)dt\] would be \[cos(1+x+x^2)\times 3x^2\]
1 Attachment
see this file..ull understand
why not cos(1+x+x^2) x (2x+1) ?
throughout this operation we never differentiate the cos function..do u get it?
yea i get the cos part but why not wats inside it
damn i made a mistake. sorry. i will correct it.
we integrate the cos fn with two separate functions of x as limits first, and then we differentiate the integral of the cos fn, which gives us the function back, note that we are differentiating the inside part, but after replacing t by f(x) and g(x) in ur case x and -1
the chain rule says that the derivative of \[F(G(x))\] is \[F'(G(x))\times G'(x)\]
if i put \[F(x)=\int_{-1}^x cos(1+t+t^2)dt\] then its derivative is \[cos(1+x+x^2)\]
ok
thanks everyone!
no prob mate
but if i write \[\int_{-1}^{x^3} cos(1+t+t^2)dt\] then its derivative is \[cos(1+x^3+(x^2)^3)\times 3x^2\]
right
the derivative of the outside function evaluated at the inside function times the derivative of the inside function. i made a mistake earlier.
yeah..now righter lol;)

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