Question

Compute lower and upper integrals o0f the function
f(x)
=4, 1<=x<=2
=3, 2<x<=4
=2, 4<x<=5

Is this function Riemann integrable on [1,5]? Justify.

Answer 1

Conditions

Compute lower and upper integrals off the function $f(x)$

=4, 1<=x<=2

=3, 2<=4

=2, 4<=5

Is this function Riemann integrable on $[1,5]$? Justify.

Solution

$$f = \begin{cases} 4, & 1 \leq x \leq 2 \\ 3, & 2 < x \leq 4 \\ 2, & 4 < x \leq 5 \end{cases}$$

Here we must use the Darboux integral theory.

The lower Darboux sum:

$$s(f, \tau) = \sum_{k=1}^{n} m_k (x_k - x_{k-1})$$

The upper Darboux sum:

$$S(f, \tau) = \sum_{k=1}^{n} M_k (x_k - x_{k-1})$$

Where:

$$\begin{aligned} m_k &= \inf \left\{ f(x) : x \in [x_{k-1}, x_k] \right\}, k = \overline{1, n} \\ M_k &= \sup \left\{ f(x) : x \in [x_{k-1}, x_k] \right\}, k = \overline{1, n} \\ \tau &= \left\{ x_k \right\}_{k=0}^{n} : a = x_0 < x_1 < \dots < x_{n-1} < x_n = b \end{aligned}$$

For our case:

$$\tau : a = 1 < 2 < 4 < 5 = b$$ $$s(f, \tau) = 4 \cdot (2 - 1) + 3(4 - 2) + 2(5 - 4) = 4 + 6 + 2 = 12$$ $$S(f, \tau) = 4 \cdot (2 - 1) + 3(4 - 2) + 2(5 - 4) = 12$$ $$\forall \tau \ s(f, \tau) \leq I_*(f) \leq I^*(f) \leq S(f, \tau)$$

As we have equal $s(f, \tau)$ and $S(f, \tau)$, then upper and lower integrals are equal to 12.

$$\int_{1}^{5} f(x) \, dx = \int_{1}^{2} 4 \, dx + \int_{2}^{4} 3 \, dx + \int_{4}^{5} 2 \, dx = 8 - 4 + 12 - 6 + 10 - 8 = 12$$

As

$$I^{*}(f) = I_{*}(f) = \int_{a}^{b} f(x) \, dx$$

Then function is integrable on $[1,5]$ by the Darboux criterion.