#### To determine

The given statement is true or false.

#### Answer

False

#### Explanation

Let,

fx=x+2 , if-1≤x<01, if 0≤x≤1

This function fx has discontinuity at x=0 But

∫-11f(x)dx=∫-10f(x)dx+∫01f(x)dx

∫-11f(x)dx=∫-10(x+2)dx+∫01(1)dx

By integrating,

∫-11f(x)dx=x22+2x-10+x01

Using fundamental theorem of calculus,

∫-11f(x)dx=02+20--122+2-1+1-0

=0--32+1=32+1=52

∫-11f(x)dx=52

Though f has a discontinuity at 0 ∫-11fxdx exist. Notice that we could do this because discontinuity was jump discontinuity.

**Conclusion:**

Therefore, the given statement is false.