This is false. Consider the function `f(t)=t,` which is integrable from `-1` to `1.` There are (at least) two ways to see this.

Geometrically, the integral from `-1` to `x` is the area under the curve from `t=-1` to `t=x`  (a portion of `t=x` being the red line). If `x<0,` as is the case in the figure below, then this area is negative. The integral from `x` to `1` is the remaining area under the curve, which must be positive because the negative part doesn't "make up" for the larger positive part. Actually, by the same reasoning, this is also the case when `x=0` and when `x>0,` showing that this is indeed a counterexample.

The algebraic approach is to note that by the Fundamental Theorem of Calculus,

`int_-1^x t dt=(x^2-1)/2`   and `int_x^1 t dt=(1-x^2)/2.`

For these to be equal, `x^2-1=1-x^2=> x=+-1.` But these points are not in the open interval `(-1,1),` again showing that this is a counterexample.