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=cos(1+x+x^2)

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?

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)

Dx( (1/3)x^4 +x^2 + 4/3)
(4/3) x^3 +2x right?

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

\[F(x)=\int_a^x f(t)dt\] 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)

oh....

so i shoulda used another variable to test with then i spose :) thnx

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?

bt weve done that havent we...ive explained above..see..ive proved the rule for any general function

were using the chain rule in the second step here saad

yes. so what abt wats in the cos function

ok.

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 \[cos(1+x+x^2)\times 3x^2\]

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.

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

right

yeah..now righter lol;)