Question

41-46 use the comparison theorem to determinewhether the integral function is convergent or divergent.

\(\int_0^1 {\frac{{{{\sec }^2}x}}{{x\sqrt x }}} dx\)

Short answer

The integral function is divergent.

Step-by-step solution

Definition of comparison theorem

Comparison Theorem:

Assume that\(f\)and\(g\)are continuous functions with\(f(x) \ge g(x) \ge 0\)for\(x \ge a\).

(a) If\(\int_a^\infty f (x)dx\)is convergent, then\(\int_a^\infty f (x)dx\)is convergent.

(b) If\(\int_a^\infty f (x)dx\)is divergent, then\(\int_a^\infty f (x)dx\)is divergent

Substitute 1 for x in equation (2) and (3)

Given\(\int_0^1 {\frac{{{{\sec }^2}x}}{{x\sqrt x }}} dx \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Consider the value of the function\(f(x) = \frac{{{{\sec }^2}x}}{{x\sqrt x }} \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)\)

Consider the value of the function\(g(x) = \frac{1}{{{x^{\frac{3}{2}}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 3 \right)\)

Calculate the value of\(f(x)\)within limits (0,1) using Equation (2).

Substitute 1 for x in Equation (2),

\(\begin{aligned}{l}f(x) &= \frac{{{{\sec }^2}(1)}}{{1\sqrt 1 }}\\f(x) &= 3.4255\end{aligned}\)

Calculate the value of\(g(x)\)within limits (0,1) using Equation (3).

Substitute 1 for x in Equation (3)

\(\begin{aligned}{c}g(x) &= \frac{1}{{{x^{\frac{3}{2}}}}}\\g(x) &= \frac{1}{{{1^{\frac{3}{2}}}}}\\g(x) &= 1\end{aligned}\)

Compare the value of \(f(x)\)and\(g(x)\)within limits (0,1).

Get \(f(x) \ge g(x)\)

Check the integral function is either   convergent or divergent

Compare the value of\(f(x)\)and\(g(x)\)within limits (0,1).

Get\(f(x) \ge g(x)\)

Check whether the integral function\(\int_0^1 g (x)dx\)is convergent or divergent.\(\begin{aligned}{l}\int_0^1 g (x)dx &= \int_0^1 {\frac{1}{{{x^{\frac{3}{2}}}}}} dx\\\int_0^1 g (x)dx &= \int_t^1 {{x^{ - \frac{3}{2}}}} dx\\\int_0^1 g (x)dx &= - \frac{3}{2}\left( {{x^{ - \frac{3}{2} - 1}}} \right)_0^\infty \\\int_0^1 g (x)dx &= - \frac{3}{2}\left( {{x^{ - \frac{5}{2}}}} \right)_0^\infty \\\int_0^1 g (x)dx &= - \frac{3}{2}\left( {{\infty ^{ - \frac{5}{2}}} - 0} \right)\\\int_0^1 g (x)dx &= \infty \end{aligned}\)

The integral function\(\int_0^1 g (x)dx\)is divergent.

Refer to part (a) of the Comparison Theorem.

The integral function\(f(x)\)and\(g(x)\)are continuous function and\(f(x) \ge g(x)\).

The integral function\(\int_0^1 g (x)dx\) is divergent, therefore\(\int_0 f (x)dx\) is divergent.

Hence, the integral function is divergent.