# Integral problem

Wicketer
Is there a non-zero continuous function, f, R -> R such that ∫f(x) dx over (c, c^2+1) = 0, for all c in R? Hope this makes sense. I've been trying to find a more formal way of approaching this rather than just visualizing it with graphs. Any thoughts?

Many thanks.

## Answers and Replies

Gold Member
Try to see if you can built a primitive F(c) such that F(c²+1)=F(c) .
If you can find one, then f is its derivative.

I guess it might be useful to observe that c²+1 > c .
Defining F in the interval [0,1] might be all that is necessary.
Will F be continuous, will its derivative be continuous, ???

Did you give the full statement of the question?

Try to graph such a function.
Give it a try.

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Generally, the "zero-function" is f(x)= 0 for all x so that "non- zero" means "not equal to 0 for at least one x". But we can have f(x)= 0 for x= 1, non-zero for x not equal to 0, and its Riemann integral, $\int_a^b f(x)dx$, from any a to any b, is non-zero. If you use the Lebesque function, then we can have f(x) equal to zero on any set of measure 0 and have $\int_a^b f(x)d\mu$ non-zero.