To evaluate the integral
[tex]\[
\int_{-1 / 4}^0 \frac{x}{\sqrt{1-x^2}} \, dx
\][/tex]
consider the following steps:
1. Integral Setup:
- We are given the definite integral with limits of integration from [tex]\(-\frac{1}{4}\)[/tex] to [tex]\(0\)[/tex].
2. Determine the Integrand:
- The integrand is [tex]\(\frac{x}{\sqrt{1-x^2}}\)[/tex].
3. Use a Substitution:
- Let's make the substitution [tex]\(u = 1 - x^2\)[/tex].
- Then, [tex]\(du = -2x \, dx\)[/tex] or equivalently, [tex]\(dx = \frac{du}{-2x}\)[/tex].
4. Change Limits of Integration:
- When [tex]\(x = -\frac{1}{4}\)[/tex], [tex]\(u = 1 - \left(-\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16}\)[/tex].
- When [tex]\(x = 0\)[/tex], [tex]\(u = 1 - 0^2 = 1\)[/tex].
5. Rewrite the Integral:
- [tex]\[
\int_{-1 / 4}^0 \frac{x}{\sqrt{1-x^2}} \, dx = \int_{\frac{15}{16}}^1 \frac{x}{\sqrt{u}} \cdot \frac{du}{-2x}
\][/tex]
- Simplifying inside, the [tex]\(x\)[/tex] terms cancel out:
- [tex]\[
= \int_{\frac{15}{16}}^1 \frac{1}{-2\sqrt{u}} \, du
\][/tex]
- [tex]\[
= -\frac{1}{2} \int_{\frac{15}{16}}^1 \frac{1}{\sqrt{u}} \, du
\][/tex]
6. Integrate:
- The integral [tex]\(\int \frac{1}{\sqrt{u}} \, du\)[/tex] is a standard integral which equals [tex]\(2\sqrt{u}\)[/tex].
7. Evaluate the Antiderivative:
- [tex]\[
-\frac{1}{2} \left[ 2\sqrt{u} \right]_{\frac{15}{16}}^1 = -\frac{1}{2} \left( 2\sqrt{1} - 2\sqrt{\frac{15}{16}} \right)
\][/tex]
- Simplify:
- [tex]\[
= -\frac{1}{2} \left( 2 - 2\cdot\frac{\sqrt{15}}{4} \right)
\][/tex]
- [tex]\[
= -\frac{1}{2} \left( 2 - \frac{2\sqrt{15}}{4} \right)
\][/tex]
- [tex]\[
= -\frac{1}{2} \left( 2 - \frac{\sqrt{15}}{2} \right)
\][/tex]
- [tex]\[
= -\frac{1}{2} \left( 4 - \sqrt{15} \right)
\][/tex]
- [tex]\[
= -2 + \frac{\sqrt{15}}{2}
\][/tex]
8. Numerical Evaluation:
- Evaluating the numerical value, we get:
- [tex]\[
-0.0317541634481457
\][/tex]
Therefore, the value of the integral
[tex]\[
\int_{-1 / 4}^0 \frac{x}{\sqrt{1-x^2}} \, dx
\][/tex]
is
[tex]\[
-0.0317541634481457.
\][/tex]
