# Is the following function riemann integrable?

Is the following function Riemann integrable on [-5,5] and if so, what is the value of;

$$\int_{0}^5 f(x)\;dx$$

f(x) = \left\{\begin{aligned} &1 &&: |x| \ge 1\\ &0 &&: Otherwise \end{aligned} \right.

Must I prove this is Riemann integrable by finding L and U and comparing? Im lost with what must seem like a trivial problem any help greatly appreciated.

• This is a step function, and the integral can be calculated directly by a Riemann sum. $L$ and $U$ will then be equal (and equal to that sum) Feb 8, 2015 at 14:27
• If this integral were not $\>=4$ we would have given up integration long ago. Feb 8, 2015 at 19:11

$$f(x)=\chi_{[1,5]}(x)+\chi_{[-5,-1]}(x)$$that's the sum of integrable functions.

To compute the integral:$$\int_0^1f(x)=\int_0^10=0$$$$\int_1^5f(x)=\int_1^51=4$$ and therefore$$\int_0^5f(x)=4$$

I don't know whether you know any theorems about integration yet, but if you do, you can simplify this a good deal. Let $$g(x) = \begin{cases} 0 & |x| \ge 1 \\ 1 & Otherwise\end{cases}$$ Then $f$ and $g$ sum to the constant function 1. So $$\int_0^5 f(x) + g(x) dx = 5$$ because the integral of a constant function is the constant value times the length of the interval (since that's the exact value of all upper and lower sums).

I'll now show that $\int_0^5 g(x) dx$ exists (and compute its value), and hence so does $\int_0^5 -g(x) dx$; that'll show that
$$\int_0^5 f(x) + g(x) dx + \int_0^5 -g(x) = \int_0^5 f(x) dx$$ exists, since the sum of integrable functions is integrable. Simplifying, that says that $$5 - \int_0^5 g(x) = \int_0^5 f(x) dx$$

To finish up, I need to integrate $g$.

Now $$\int_0^5 g(x) dx = \int_0^1 g(x) dx + \int_1^5 g(x) dx$$ The rightmost integral is zero, because $g$ is zero on that interval; the middle integral is the integral of the constant function 1 on an interval of length 1...so it's 1. So we get $$\int_0^5 g(x) dx = 1$$ and hence $$\int_0^5 f(x) dx = 5 - 1 = 4.$$

• The "And since..." step is invalid unless you first prove $f$ is Riemann integrable.
– Pedro
Feb 8, 2015 at 14:45
• You're right. I'll edit slightly to make the logic a bit better. Feb 8, 2015 at 15:41