# Given: f en g are continious functions on [0,2]. h is defined: h(x)=f(x)+g(x) for x is element of [0,1> ; h(x)=f(x)-g(x) for x is element of [1,2] Show the integrabiltiy of h

### 1 Answer | Add Yours

You need to remember that all continuous functions are integrable over an interval.

The problem specifies that functions f(x) and g(x) keeps the continuity over the interval [0,2], hence they are Riemann integrable over [0,2].

You need to remember that the sum of Riemann integrable functions over an interval is an integrable function over the same interval such that:

`int_0^2 h(x)dx = int_0^1 h(x)dx + int_1^2 h(x)dx`

Considering x in [0,1] yields:

`int_0^1 h(x)dx = int_0^1 (f(x) + g(x)) dx = (F(x) + G(x))|_0^1` `int_0^1 h(x)dx = F(1) + G(1) - F(0) - G(0)`

`int_1^2 h(x)dx = int_1^2 (f(x)- g(x)) dx = (F(x)- G(x))|_1^2`

`int_1^2 h(x)dx =F(2)- G(2) - F(1)+ G(1)`

`int_0^2 h(x)dx = F(1) + G(1) - F(0) - G(0) + F(2)- G(2) - F(1)+ G(1)`

Reducing like terms yields:

`int_0^2 h(x)dx = 2G(1) - F(0) - G(0) + F(2)- G(2)`

**Hence, the function h(x) is integrable over [0,2] such that `int_0^2 h(x)dx = 2G(1) - F(0) - G(0) + F(2)- G(2).` **