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anonymous
 5 years ago
cal 1 question.
anonymous
 5 years ago
cal 1 question.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fundamental 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops i meant\[cos(1+x+x^2)\] sorry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but thats not the answer

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we need to find the integrand from 1 to x and then derive it back

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0u = (1+x+x^2); du = 1+2x dx du/1+2x = dx maybe?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am sticking with my answer. unless they changed the fundamental theorem of calculus or unless i read the problem incorrectly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How come?! I am pretty sure it's the answer.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the integrand changes in form and then you derive the change in form down to the answer, i think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0try it on a simpler problem; {S} 1+x^2 dx ; [1,x] x + x^3/3 gets what?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x^2 +x^4/3 (1 + 1/3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0Dx( (1/3)x^4 +x^2 + 4/3) (4/3) x^3 +2x right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am curious as to what the answer is supposed to be. an integral of the form asked about is not an antiderivative, 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 antiderivatives to compute integrals.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0blame the software. what exactly did you type in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and one more thing: why aren't we differentiating 1+ t+ t^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use the newtonleibniz theorem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you are not asked for the derivative of the INTEGRAND. you are asked for the derivative of the INTEGRAL.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[F(x)=\int_a^x f(t)dt\] is a function of x, not of t.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0derivative of f(t)dt integral from f1(x) to f2(x) is f(f1)f1'  f(f2)f2'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay probably the problem with teh software.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and (at the risk of repeating myself) the derivative is \[f(x)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so was i right in my thought? :) integrate than derive the result

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey btw why arent we differentiating 1+t+t^.5

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0{S} cos(1+t+t^2) dt ; [1,x] = what?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is cos(1 +x + x^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre, i am afraid that is not correct. the definition of the integral \[\int_a^x f(t)dt\] is not an "antiderivative" 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
 5 years ago
Best ResponseYou've already chosen the best response.0so i shoulda used another variable to test with then i spose :) thnx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here, this is the final answer after evaluating the integral and then taking its derivative: http://bit.ly/lIvQ5I

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{g(x)}^{h(x)}f'(t)dt = f[h(x)]  f[g(x)\] now if u differentiate it \[f'[h(x)]h'(x)  f'[g(x)]g'(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0this is called newtonleibniz rule...for derivatives of integral forms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0GUYS IM EXPLAINING THE ANSWER 2 U..IS ANYONE LISTENING

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0rsaad did u get the answer????

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did u understand the working i explained?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes 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
 5 years ago
Best ResponseYou've already chosen the best response.0bt weve done that havent we...ive explained above..see..ive proved the rule for any general function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0him1618 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 antiderivative. you are stating a (correct) theorem which assumes you have an antiderivative 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 antiderivative. that weird looking thing after the "dx" on the first line includes C and S which are in fact themselves integrals.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops typo. i meant to say that the derivative of the integral is the integrand, not the other way around.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes 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
 5 years ago
Best ResponseYou've already chosen the best response.0people just explain as to why the chain rule does not apply to whats in th funcyion that is whats in cos (.............)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0were using the chain rule in the second step here saad

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes. so what abt wats in the cos function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first we are integrating a function...with limits which are variable wrt another variable..and then we differentiate using chain rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh 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
 5 years ago
Best ResponseYou've already chosen the best response.0plz explain it a bit more

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the derivative of \[\int_{1}^{x^3} cos(1+t+t^2)dt\] would be \[cos(1+x+x^2)\times 3x^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see this file..ull understand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why not cos(1+x+x^2) x (2x+1) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0throughout this operation we never differentiate the cos function..do u get it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i get the cos part but why not wats inside it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0damn i made a mistake. sorry. i will correct it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0the chain rule says that the derivative of \[F(G(x))\] is \[F'(G(x))\times G'(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if i put \[F(x)=\int_{1}^x cos(1+t+t^2)dt\] then its derivative is \[cos(1+x+x^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative of the outside function evaluated at the inside function times the derivative of the inside function. i made a mistake earlier.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah..now righter lol;)
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