cal 1 question.

- anonymous

cal 1 question.

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- schrodinger

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- anonymous

##### 1 Attachment

- anonymous

=cos(1+x+x^2)

- anonymous

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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## More answers

- anonymous

oops i meant\[cos(1+x+x^2)\] sorry.

- anonymous

but thats not the answer

- anonymous

=/

- anonymous

really?

- amistre64

we need to find the integrand from -1 to x and then derive it back

- amistre64

u = (1+x+x^2); du = 1+2x dx
du/1+2x = dx maybe?

- anonymous

i am sticking with my answer. unless they changed the fundamental theorem of calculus or unless i read the problem incorrectly.

- anonymous

How come?! I am pretty sure it's the answer.

- amistre64

the integrand changes in form and then you derive the change in form down to the answer, i think

- amistre64

try it on a simpler problem;
{S} 1+x^2 dx ; [-1,x]
x + x^3/3 gets what?

- amistre64

x^2 +x^4/3 -(-1 + -1/3)

- anonymous

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 \[F'(x)=f(x)\]

- amistre64

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

- anonymous

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.

- anonymous

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

- anonymous

blame the software. what exactly did you type in?

- anonymous

and one more thing: why aren't we differentiating 1+ t+ t^2

- anonymous

use the newton-leibniz theorem

- anonymous

you are not asked for the derivative of the INTEGRAND. you are asked for the derivative of the INTEGRAL.

- anonymous

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.

- anonymous

\[F(x)=\int_a^x f(t)dt\] is a function of x, not of t.

- anonymous

derivative of f(t)dt integral from f1(x) to f2(x)
is f(f1)f1' - f(f2)f2'

- anonymous

okay probably the problem with teh software.

- anonymous

and (at the risk of repeating myself) the derivative is
\[f(x)\]

- amistre64

so was i right in my thought? :)
integrate than derive the result

- anonymous

hey btw why arent we differentiating 1+t+t^.5

- anonymous

i told u guys

- amistre64

{S} cos(1+t+t^2) dt ; [-1,x] = what?

- anonymous

the answer is cos(1 +x + x^2)

- anonymous

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.

- amistre64

oh....

- amistre64

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

- anonymous

Here, this is the final answer after evaluating the integral and then taking its derivative:
http://bit.ly/lIvQ5I

- anonymous

\[\int\limits_{g(x)}^{h(x)}f'(t)dt = f[h(x)] - f[g(x)\]
now if u differentiate it
\[f'[h(x)]h'(x) - f'[g(x)]g'(x)\]

- anonymous

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

- anonymous

this is called newton-leibniz rule...for derivatives of integral forms

- anonymous

GUYS IM EXPLAINING THE ANSWER 2 U..IS ANYONE LISTENING

- anonymous

rsaad did u get the answer????

- anonymous

yea i got it =) thnx

- anonymous

did u understand the working i explained?

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

oops typo. i meant to say that the derivative of the integral is the integrand, not the other way around.

- anonymous

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..;)

- anonymous

people just explain as to why the chain rule does not apply to whats in th funcyion that is whats in cos (.............)?

- anonymous

were using the chain rule in the second step here saad

- anonymous

yes. so what abt wats in the cos function

- anonymous

first we are integrating a function...with limits which are variable wrt another variable..and then we differentiate using chain rule

- anonymous

ok.

- anonymous

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
\[\int_a^{h(x)} f(t)dt\]

- anonymous

se wht i wrote above

- anonymous

exactly!

- anonymous

um =S

- anonymous

plz explain it a bit more

- anonymous

so the derivative of \[\int_{-1}^{x^3} cos(1+t+t^2)dt\]
would be \[cos(1+x+x^2)\times 3x^2\]

- anonymous

##### 1 Attachment

- anonymous

see this file..ull understand

- anonymous

why not cos(1+x+x^2) x (2x+1) ?

- anonymous

throughout this operation we never differentiate the cos function..do u get it?

- anonymous

yea i get the cos part but why not wats inside it

- anonymous

damn i made a mistake. sorry.
i will correct it.

- anonymous

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

- anonymous

the chain rule says that the derivative of \[F(G(x))\] is \[F'(G(x))\times G'(x)\]

- anonymous

if i put \[F(x)=\int_{-1}^x cos(1+t+t^2)dt\] then its derivative is
\[cos(1+x+x^2)\]

- anonymous

ok

- anonymous

thanks everyone!

- anonymous

no prob mate

- anonymous

but if i write \[\int_{-1}^{x^3} cos(1+t+t^2)dt\]
then its derivative is
\[cos(1+x^3+(x^2)^3)\times 3x^2\]

- anonymous

right

- anonymous

the derivative of the outside function evaluated at the inside function times the derivative of the inside function. i made a mistake earlier.

- anonymous

yeah..now righter lol;)

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