Question

(a) Use the table of integrals to evaluate\(F(x) = \int f (x)dx\), where

\(f(x) = \frac{1}{{x\sqrt {1 - {x^2}} }}\)What is the domain of \(f\)and\(F\)?

(b) Use a CAS to evaluate\(F(x)\). What is the domain of the function\(F\) that the CAS produces? Is there a discrepancy between this domain and the domain of the function\(F\)that you found in part (a)?

Short answer

1. \(\int {\frac{{dx}}{{x\sqrt {1 - {x^2}} }}} = - \ln \left| {\frac{{1 + \sqrt {1 - {x^2}} }}{x}} \right| + C\), the domain is\(x \in ( - 1,0) \cup (0,1)\)
2. For this domain is\(x \in (0,1)\) & It doesn’t include negative value in its domain.

Step-by-step solution

Definition

Integrationis the process of finding the area of the region under the curve.

Evaluate integral(a)

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

Use the formula\(\int {\frac{{du}}{{u\sqrt {{a^2} - {u^2}} }}} = - \frac{1}{a}\ln \left| {\frac{{a + \sqrt {{a^2} - {u^2}} }}{u}} \right| + C\)to evaluate the integral.

Here \(u = x\)and\(a = 1\).

\(\int {\frac{{dx}}{{x\sqrt {1 - {x^2}} }}} = - \ln \left| {\frac{{1 + \sqrt {1 - {x^2}} }}{x}} \right| + C\)

Find domain

Domain for both \(f\)and\(\;F\).

\(x \ne 0{\rm{ and }}1 - {x^2} \ge 0\)

For\(1 - {x^2} \ge 0\) :

\({x^2} \le 1\)

Therefore\(x \in ( - 1,0) \cup (0,1)\)

Using calculator(b)

In CAS put, integral \(\left( {\frac{1}{{x\sqrt {1 - {x^2}} }}} \right)\) & press enter.

The Computer algebra system produces:

\(\int {\frac{{dx}}{{x\sqrt {1 - {x^2}} }}} = \ln (x) - \ln \left( {\sqrt {1 + {x^2}} + 1} \right) + C\)

The domain for\(F\) is\(x > 0\)and\(1 - {x^2} \ge 0\)

For\(1 - {x^2} \ge 0\):

\({x^2} \le 1\)

Therefore\(x \in (0,1)\)

It doesn’t include negative value in its domain.[PT1]

[PT1]Write producer for CAS