Question

\(5 - 32\)Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

11. \(\int_0^\infty {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx\)

Short answer

Integral is divergent.

Step-by-step solution

Definition for converging integral

Consider the integral\(\int_{\bf{a}}^{\bf{t}} {\bf{f}} {\bf{(x)dx}}\)is for the limit with the finite number\(t \ge a\). Then, the integral is written as:

\(\int_a^\infty f (x)dx = \mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\)

Change limit

In order to determine if integral converges or diverges we evaluate both integrals i.e; \(\int_a^\infty f (x)dx\,\& \,\mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\)& see if both integrals are equal or not.

Let \(u = 1 + {x^3}\)

Differentiate \(u\) with respect to\(x\).

\(\begin{aligned}{l}du &= 3{x^2}dx\\\frac{{du}}{3} &= {x^2}dx\end{aligned}\)

When\(x = t\), then\(u = 1 + {t^3}\)

When\(x = 0\), then \(u = 1\)

Evaluate integral

Using theorem written above we have:

\(\int_0^\infty {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx = \mathop {\lim }\limits_{t \to \infty } \int_0^t {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx\)

Substituting\(u = 1 + {x^3}\)& new limits we have:

\(\begin{aligned}{c}\int_0^t {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx &= \int_1^{{t^3} + 1} {\frac{1}{{\sqrt u }}} \frac{{du}}{3}\\ &= \frac{1}{3}\int_1^{{t^3} + 1} {{u^{\frac{{ - 1}}{2}}}} du\\ &= \frac{1}{3}\left( {\frac{{{u^{\frac{1}{2}}}}}{{\frac{1}{2}}}} \right)_1^{{c^3} + 1}\\ &= \frac{2}{3}\left( {{{\left( {{t^3} + 1} \right)}^{\frac{1}{2}}} - {1^{\frac{1}{2}}}} \right)\end{aligned}\)

Solve further as,

\(\int_0^t {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx = \frac{2}{3}\left( {{{\left( {{t^3} + 1} \right)}^{\frac{1}{2}}} - 1} \right)\)

\(So,\int_0^1 {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx = \frac{2}{3}\left( {{{\left( {{t^3} + 1} \right)}^{\frac{1}{2}}} - 1} \right)\)

Evaluate limit:

Solve for the integral with the limit as:

\(\begin{aligned}{c}\int_0^\infty {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx &= \mathop {\lim }\limits_{t \to \infty } \int_0^t {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx\\\int_0^\infty {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{2}{3}\left( {{{\left( {{t^3} + 1} \right)}^{\frac{1}{2}}} - 1} \right)} \right)\;\;\;\left( {\int_0^t {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx &= \frac{2}{3}\left( {{{\left( {{t^3} + 1} \right)}^{\frac{1}{2}}} - 1} \right)} \right)\\ &= \frac{2}{3}\mathop {\lim }\limits_{t \to \infty } \left( {{{\left( {{t^3} + 1} \right)}^{\frac{1}{2}}} - 1} \right)\\ &= \frac{2}{3}\mathop {\lim }\limits_{t \to \infty } \left( {\sqrt {{t^3} + 1} - 1} \right)\end{aligned}\)

Solve further as,

\(\begin{aligned}{c}\int_0^\infty {\frac{{{x^2}}}{{\sqrt {1 + {x^3}} }}} dx &= \frac{2}{3} \cdot \infty \\ &= \infty \end{aligned}\)

As limit is not finite so the integral is divergent.