Question

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

\(\int_0^\pi {\frac{{{{\sin }^2}x}}{{\sqrt x }}} dx\)

Short answer

The integral function is convergent.

Step-by-step solution

Use the 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 \(\pi \) for x in equation (2) and (3)

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

Consider the value of the function\(g(x) = \frac{{{{\sin }^2}x}}{{\sqrt x }} \cdot \cdot \cdot \cdot \cdot \cdot (2)\)

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

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

Substitute\(\pi \)for x in Equation (2).

\(\begin{aligned}{l}g(x) &= \frac{{{{\sin }^2}(\pi )}}{{\sqrt \pi }}\\g(x) &= 0\end{aligned}\)

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

Substitute\(\pi \)for x in Equation (3).

\(\begin{aligned}{l}f(x) &= \frac{1}{{{x^{\frac{1}{2}}}}}\\f(x) &= \frac{1}{{{\pi ^{\frac{1}{2}}}}}\\f(x) &= \frac{1}{{{{(3.14)}^{\frac{1}{2}}}}}\\f(x) &= 0.564\end{aligned}\)

Check the integral function is 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.

\(\begin{aligned}{l}\int_0^\pi f (x)dx &= \int_0^\pi {\frac{1}{{{x^{\frac{1}{2}}}}}} dx\\\int_0^\pi f (x)dx &= \mathop {\lim }\limits_{t \to {0^ + }} \int_t^\pi {{x^{ - \frac{1}{2}}}} dx\\\int_0^\pi f (x)dx &= \mathop {\lim }\limits_{t \to {0^ + }} \left( {\frac{{{x^{ - \frac{1}{2} + 1}}}}{{ - \frac{1}{2} + 1}}} \right)_t^\pi \\\int_0^\pi f (x)dx &= \mathop {\lim }\limits_{t \to {0^ + }} \left( {2{x^{\frac{1}{2}}}} \right)_t^\pi \\\int_0^\pi f (x)dx &= 2\mathop {\lim }\limits_{t \to {0^ + }} \left( {{\pi ^{\frac{1}{2}}} - {t^{\frac{1}{2}}}} \right)\\\int_0^\pi f (x)dx &= 2\left( {{\pi ^{\frac{1}{2}}} - 0} \right)\\\int_0^\pi f (x)dx &= 2{\pi ^{\frac{1}{2}}}\end{aligned}\)

The integral function\(\int_0^\pi f (x)dx\) is convergent.

Refer to part (a) the Comparison Theorem.

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

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

Hence, the integral function is convergent.