Inspired by Abdalrahman Gamal

Calculus Level 2

True or false :

If $\displaystyle \int_a^b f(x) \, dx = 0$ for all real $a$ and $b$ , then $f(x) = 0$ for all real $x$ .

False True

1 solution

Guillermo Templado
Apr 15, 2016

Let the function $f: \mathbb{R} \longrightarrow \mathbb{R} \space / \space f(x) = 1$ if $x = 0$ and $f(x) = 0 \space \forall x \in \mathbb{R} - \{0\}$ . This function is Riemman integrable due to its set of points of discontinuity is zero Lebesgue measure or due to Riemman infimum(upper) and supremumum(lower) sums match,and satisfies the condition required being distinct to 0

Moderator note:

Indeed. Changing one (finitely many) value of $f(x)$ doesn't affect the value of the integral.

What additional condition do we need to make this statement true?

Indeed. Changing one (finitely many) value of $f(x)$ doesn't affect the value of the integral.

What additional condition do we need to make this statement true?

Calvin Lin Staff - 5 years, 1 month ago

For example, other function fullfiling the requirements is $f:\mathbb{R} \longrightarrow \mathbb{R} \space / \space f(x) = 2016$ if $x \in \mathbb{Z}$ and $f(x) = 0$ if $x \in \mathbb{R} -\{\mathbb{Z}\}$ . If $f$ is a real continuous function and $\int_{a}^b f(x) dx = 0$ for all real numbers $a,b$ then $f(x) = 0$ for all real x .

Or if you like harmony, $f$ is a real continuous function and $\int_{a}^b f(x) dx = 0$ for all real numbers $a,b$ $\iff$ $f(x) = 0$ for all real x

Guillermo Templado - 5 years, 1 month ago

Inspired by Abdalrahman Gamal

1 solution

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