Prove: Integral of |f(x)|dx = 0 $\implies$ f(x) = 0 for all x in [a,b]

[theneedtoknow]

Homework Statement

PRove that, if f is continuous on [a,b] and the integral from a to b of |f(x)|dx = 0 (thats absolute value of f(x) ), then f(x) = 0 for all x in [a,b]

Homework Equations

The Attempt at a Solution

HEre is my attempt...its not very rigorous which is why I'm asking for any tips that could make this a more formal proof:

Proof by contradiction:
Assume f(x) is not equal to zero for all x in [a,b]
then there exists an x1 element of [a,b] such that f(x1) > 0
Then by properties of integrals we can integrate f(x) from a to x1 and from x1 to b and add them together
since f(x1) is greater than zero and not just a removable discontinuity (since f is stated as continusous), the area bound by x = a, x = x1, y = 0 and y = f(x) must also be greater than zero and therefore the integral from a to x1 of |f(x)|dx is also greater than zero. Since the function |f(x)| is always nonnegative, the integral from x1 to b of |f(x)|dx is also nonnegative. Since the integral from a to x1 is positive and from x1 to b is non-negative, then their sum must also be positive. But this congradicts the assumption that the integral from a to b of |f(x)|dx = 0 for all x in [a,b] , so f(x) must be 0 for all x in [a,b]

 

[HallsofIvy]

Your first statement is wrong. You are asked to prove that f(x)= 0 for all x in [a,b]. The "contradiction" of that is NOT that f(x) is non 0 for all x in [a,b], it is that f(x) is not 0 for at least one x₁ in [a,b]. Since f(x) is continuous, there exist some $\delta$ such that f(x) is non-zero (and |f(x)|> 0) for all x in [itex](x_1-\delta,x_1+\delta). Then you know the integral on [itex](x_1-\delta,x_1+\delta) is non-zero.

 

[theneedtoknow]

Ohhh I see what you mean :)
thankss, i understand it better now