Question

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

5. \(\int_3^\infty {\frac{1}{{{{(x - 2)}^{3/2}}}}} dx\)

Short answer

Integral is convergent & have value 2.

Step-by-step solution

Definition.

If\(\int_a^t f (x)dx\)exists for every number\(t \ge a\), then\(\int_a^\infty f (x)dx = \mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\)provided this limit exists (as a finite number).

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 \(x - 2 = u\)

\( \Rightarrow dx = du\)

The new limits of integrations are,

Lower limit:

\(\begin{aligned}{l}u &= x - 2\\u &= 3 - 2\\u &= 1\end{aligned}\)

Upper limit:

\(\begin{aligned}{l}u &= x - 2\\u &= \infty - 2\\u &= \infty \end{aligned}\)

Hence, the new limits are\(({\bf{1}},\infty )\)

Evaluate\(\int_a^\infty  f (x)dx\).

Substituting\(x - 2 = u\)& change limit to evaluate the integral.

\(\int_3^\infty {\frac{1}{{{{(x - 2)}^{\frac{3}{2}}}}}} dx = \int_1^\infty {\frac{1}{{{u^{\frac{3}{2}}}}}} du\)

The integral is convergent, since\(\frac{3}{2} > 1\) using p test.

So, now integral can be evaluated.

\(\begin{aligned}{c}\int_1^\infty {\frac{1}{{{u^{\frac{3}{2}}}}}} du &= \mathop {\lim }\limits_{t \to \infty } \int_1^t {\frac{1}{{{u^{\frac{3}{2}}}}}} du\\ &= \mathop {\lim }\limits_{t \to \infty } \int_1^t {{u^{\frac{{ - 3}}{2}}}} du\\ &= \left( {\mathop {\lim }\limits_{t \to \infty } - 2{u^{ - \frac{1}{2}}}} \right)_1^t\end{aligned}\)

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

It is know that\(\mathop {\lim }\limits_{t \to \infty } \frac{1}{{\sqrt t }} = {\bf{0}}\)

Therefore,

\(\begin{aligned}{c}\,\mathop {\lim }\limits_{t \to \infty } \left( {\frac{{ - 2}}{{\sqrt u }}} \right)_1^t &= - 2\mathop {\lim }\limits_{t \to \infty } \left( {\frac{1}{{\sqrt u }}} \right)_1^t\\ &= - 2\left( {\mathop {\lim }\limits_{t \to \infty } \frac{1}{{\sqrt t }} - \frac{1}{{\sqrt 1 }}} \right)\\ &= - 2(0 - 1)\\ 7= 2\end{aligned}\)

So, limit exists & is a finite number and so the improper integral is convergent.