anonymous
  • anonymous
cal 1 question.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
=cos(1+x+x^2)
anonymous
  • 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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anonymous
  • anonymous
oops i meant\[cos(1+x+x^2)\] sorry.
anonymous
  • anonymous
but thats not the answer
anonymous
  • anonymous
=/
anonymous
  • anonymous
really?
amistre64
  • amistre64
we need to find the integrand from -1 to x and then derive it back
amistre64
  • amistre64
u = (1+x+x^2); du = 1+2x dx du/1+2x = dx maybe?
anonymous
  • anonymous
i am sticking with my answer. unless they changed the fundamental theorem of calculus or unless i read the problem incorrectly.
anonymous
  • anonymous
How come?! I am pretty sure it's the answer.
amistre64
  • amistre64
the integrand changes in form and then you derive the change in form down to the answer, i think
amistre64
  • amistre64
try it on a simpler problem; {S} 1+x^2 dx ; [-1,x] x + x^3/3 gets what?
amistre64
  • amistre64
x^2 +x^4/3 -(-1 + -1/3)
anonymous
  • 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
  • amistre64
Dx( (1/3)x^4 +x^2 + 4/3) (4/3) x^3 +2x right?
anonymous
  • 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
  • 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
  • anonymous
blame the software. what exactly did you type in?
anonymous
  • anonymous
and one more thing: why aren't we differentiating 1+ t+ t^2
anonymous
  • anonymous
use the newton-leibniz theorem
anonymous
  • anonymous
you are not asked for the derivative of the INTEGRAND. you are asked for the derivative of the INTEGRAL.
anonymous
  • 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
  • anonymous
\[F(x)=\int_a^x f(t)dt\] is a function of x, not of t.
anonymous
  • anonymous
derivative of f(t)dt integral from f1(x) to f2(x) is f(f1)f1' - f(f2)f2'
anonymous
  • anonymous
okay probably the problem with teh software.
anonymous
  • anonymous
and (at the risk of repeating myself) the derivative is \[f(x)\]
amistre64
  • amistre64
so was i right in my thought? :) integrate than derive the result
anonymous
  • anonymous
hey btw why arent we differentiating 1+t+t^.5
anonymous
  • anonymous
i told u guys
amistre64
  • amistre64
{S} cos(1+t+t^2) dt ; [-1,x] = what?
anonymous
  • anonymous
the answer is cos(1 +x + x^2)
anonymous
  • 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
  • amistre64
oh....
amistre64
  • amistre64
so i shoulda used another variable to test with then i spose :) thnx
anonymous
  • anonymous
Here, this is the final answer after evaluating the integral and then taking its derivative: http://bit.ly/lIvQ5I
anonymous
  • 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
  • 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
  • anonymous
this is called newton-leibniz rule...for derivatives of integral forms
anonymous
  • anonymous
GUYS IM EXPLAINING THE ANSWER 2 U..IS ANYONE LISTENING
anonymous
  • anonymous
rsaad did u get the answer????
anonymous
  • anonymous
yea i got it =) thnx
anonymous
  • anonymous
did u understand the working i explained?
anonymous
  • 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
  • anonymous
bt weve done that havent we...ive explained above..see..ive proved the rule for any general function
anonymous
  • 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
  • anonymous
oops typo. i meant to say that the derivative of the integral is the integrand, not the other way around.
anonymous
  • 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
  • anonymous
people just explain as to why the chain rule does not apply to whats in th funcyion that is whats in cos (.............)?
anonymous
  • anonymous
were using the chain rule in the second step here saad
anonymous
  • anonymous
yes. so what abt wats in the cos function
anonymous
  • anonymous
first we are integrating a function...with limits which are variable wrt another variable..and then we differentiate using chain rule
anonymous
  • anonymous
ok.
anonymous
  • 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
  • anonymous
se wht i wrote above
anonymous
  • anonymous
exactly!
anonymous
  • anonymous
um =S
anonymous
  • anonymous
plz explain it a bit more
anonymous
  • 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
  • anonymous
1 Attachment
anonymous
  • anonymous
see this file..ull understand
anonymous
  • anonymous
why not cos(1+x+x^2) x (2x+1) ?
anonymous
  • anonymous
throughout this operation we never differentiate the cos function..do u get it?
anonymous
  • anonymous
yea i get the cos part but why not wats inside it
anonymous
  • anonymous
damn i made a mistake. sorry. i will correct it.
anonymous
  • 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
  • anonymous
the chain rule says that the derivative of \[F(G(x))\] is \[F'(G(x))\times G'(x)\]
anonymous
  • 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
  • anonymous
ok
anonymous
  • anonymous
thanks everyone!
anonymous
  • anonymous
no prob mate
anonymous
  • 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
  • anonymous
right
anonymous
  • 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
  • anonymous
yeah..now righter lol;)

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