Question

Use Property 8 to estimate the value of the integral.

\(\int_1^2 {\frac{1}{x}} dx\)

Short answer

The value of the integral \(\int_1^2 {\frac{1}{x}} dx\) is \(\frac{1}{2} \le \int_1^2 {\frac{1}{x}} dx \le 1\)

Step-by-step solution

Property Used for the integral

Comparison property of the integral:

Let\(f\)be an integrable function with the limits\(a\)and\(b\)with\(a \le b\). If\(m \le f(x) \le M\)for\(a \le x \le b\), then,\(m(b - a) \le \int_a^b f (x)dx \le M(b - a)\).

Where, \(m\) is the minimum value of \(f\) and \(M\) is the maximum value of \(f\)."

Calculate the first derivative of the function \(f\).

First, define \(f(x): = \frac{1}{x}\). Because \(f\) is continuous function on the closed interval \(\left( {1,2} \right),\)by the extreme value theorem we conclude that \(f\) must attain a maximum and a minimum on \(\left( {1,2} \right).\)

Now we calculate the first derivative:

\(f'(x) = - \frac{1}{{{x^2}}}\)

Since \({f^\prime }\) is obviously negative function on \(R\)

we conclude that \(f\) is decreasing function. Because of that, we have

\(M: = f(1) = 1 \ge f(x) \ge \frac{1}{2} = f(2) = :m,\quad \forall x \in (1,2)\)

Using the comparison property of the integral

Now, use the comparison property of the integral: If \(a \le x \le b\) and \(m \le f(x) \le M\) then

\(m(b - a) \le \int_a^b f (x)dx \le M(b - a)\)

So, in our case \(a = 1\) and \(b = 2\), and so

\( \Rightarrow \frac{1}{2}(2 - 1) \le \int_1^2 f (x)dx \le 1(2 - 1)\)

\( \Rightarrow \frac{1}{2} \le \int_1^2 {\frac{1}{x}} dx \le 1\)Hence, the value of the integral \(\int_1^2 {\frac{1}{x}} dx\) is \(\frac{1}{2} \le \int_1^2 {\frac{1}{x}} dx \le 1\).