If f and g are continuous functions on the set of real nos, and the integral from 0 to x^3 of f(t) dt = the...

Question

...integral from 0 to x^2 of g(t) dt, then f(x^3)=g(x^2) for all real x.
True or False?

Thanks!

Answer 1

Since f and g are continuous, the Fundamental Theorem of Calculus implies the existence of functions F and G such that F' = f and G' = G and such that

Int (0 to x^3) f(t) dt = F(x^3) - F(0) and
Int (0 to x^2) g(t) dt = G(x^2) - G(0)

Therefore, for every real x,

F(x^3) - F(0) = G(x^2) - G(0)
F(x^3) - G(x^2) = G(0) - F(0)

If we differentiate both sides, the chain rule implies that

3x^2 f(x^3) - 2x g(x^2) = 0 , so that
3x f(x^3) = 2g(x^2) for every x <>0.

This is not the given equation, so the statement is false.

Answer 2

fake - yet dont comprehend a severe-high quality data. in case you think of of the mandatory because of the fact the section below each and every function, then purely because of the fact the part of g(x) is larger than f(x) would not recommend that the function itself unavoidably is often larger.

Answer 3

False. It should be
f(x^3)*3x^2=g(x^2)*2x.
Don't forget the chain rule!!!