Question

cal 1 question.

Answer 1

=cos(1+x+x^2)

Answer 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?

Answer 3

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

Answer 4

but thats not the answer

Answer 5

=/

Answer 6

really?

Answer 7

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

Answer 8

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

du/1+2x = dx maybe?

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

try it on a simpler problem;

{S} 1+x^2 dx ; [-1,x]

x + x^3/3 gets what?

Answer 13

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

Answer 14

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

Answer 15

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

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

Answer 16

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.

Answer 17

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

Answer 18

blame the software. what exactly did you type in?

Answer 19

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

Answer 20

use the newton-leibniz theorem

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

okay probably the problem with teh software.

Answer 26

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

Answer 27

so was i right in my thought? :)

integrate than derive the result

Answer 28

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

Answer 29

i told u guys

Answer 30

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

Answer 31

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

Answer 32

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.

Answer 33

oh....

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

rsaad did u get the answer????

Answer 41

yea i got it =) thnx

Answer 42

did u understand the working i explained?

Answer 43

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

Answer 44

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

Answer 45

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.

Answer 46

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

Answer 47

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..;)

Answer 48

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

Answer 49

were using the chain rule in the second step here saad

Answer 50

yes. so what abt wats in the cos function

Answer 51

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

Answer 52

ok.

Answer 53

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

Answer 54

se wht i wrote above

Answer 55

exactly!

Answer 56

um =S

Answer 57

plz explain it a bit more

Answer 58

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

Answer 59

see this file..ull understand

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

ok

Answer 68

thanks everyone!

Answer 69

no prob mate

Answer 70

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

Answer 71

right

Answer 72

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

Answer 73

yeah..now righter lol;)