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Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0if \[f(x) = \int\limits_{0}^{x} f(t) dt \] then f = 0

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1This is not a true statement.

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0Michael Spivak claims it is

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0The man who wrote my textbook.

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1Whoever he might be, tell him Newton has another opinion :P

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1This is a direct use of the fundamental theorem of calculus.

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int_0^x 2t dt= t^2 \] \[x^2\]

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0haha im not going anywhere with this one

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0hes claiming f(x) = 0 for any x

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0I certinally dont see it

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.0is there derivative sign infront of integral?

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0Nope that is the whole question

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1I didn't read the question well at first. This means that f is an antiderivative of itself, if I'm seeing this right.

across
 3 years ago
Best ResponseYou've already chosen the best response.0We know that\[f(x)=\int f(x)dx\implies f(x)=e^x\]^^

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0I didnt think about e^x with this one

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1Then differentiate both sides you get, f'(x)=f(x), which is an ODE that has the solution \(f(x)=ce^{x}\).

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1Your statement is still not correct :P

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0the e^x makes senese with f'(x) = f(x) but this function is an integral

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1You're saying c=0 @Zarkon.

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.4\[f(0) = \int\limits_{0}^{0} f(t) dt=0\]

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0It wants me to prove the function is 0

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1That's right! I'm a loser!! :(

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0No you're not! So basically because f'(x) = f(x) I can say f(x) = ce^x and then show that f(0) = 0 implying that c=0 so f = 0

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0makes perfect sense! Im just be a loser now but how do we know there is no other function s.t. f'(x) = f(x)

across
 3 years ago
Best ResponseYou've already chosen the best response.0Mr. Michale Spivak did a good job, that is, to elicit eager students to congregate and think this one through. xd

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0I hate michale spivak :P

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0I'm just kidding its just a challening course for me

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1Lol @across. @Wall, this is a first order homogeneous equation and had only this solution, \(i.e f(x)=ce^{x}\).

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1homogeneous differential equation* and it has* *_*

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0ahhh yes! I really like this problem! I can't believe I didnt realize f(x) had to be some form of e^x Thanks for all the help everyone!!!!

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1You're welcome! Thanks for fanning me :D
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