Question

If $$ \int_{0}^{a} \frac{1}{4 + x^2} \, dx = \frac{\pi}{6}, $$ then the value of \( a \) is: 

• A. \( \frac{\sqrt{3}}{2} \)
• B. \( 2\sqrt{3} \)
• C. \( \sqrt{3} \)
• D. \( \frac{1}{\sqrt{3}} \)
• E.

Hint:

For integrals involving \( \frac{1}{a^2 + x^2} \), use the standard result \( \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \).

Answer 1

The Correct Option is B

The given integral is: \[ \int_{0}^{a} \frac{1}{4 + x^2} \, dx = \frac{\pi}{6}. \] 1. Standard Integral Form: The integral of \( \frac{1}{4 + x^2} \) is of the form: \[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C. \] Here, \( a^2 = 4 \), so \( a = 2 \). The integral becomes: \[ \int \frac{1}{4 + x^2} \, dx = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right). \] 2. Evaluate the Definite Integral: Substitute the limits \( 0 \) to \( a \): \[ \int_{0}^{a} \frac{1}{4 + x^2} \, dx = \left[\frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)\right]_{0}^{a}. \] This simplifies to: \[ \frac{1}{2} \tan^{-1}\left(\frac{a}{2}\right) - \frac{1}{2} \tan^{-1}(0). \] Since \( \tan^{-1}(0) = 0 \), we have: \[ \frac{1}{2} \tan^{-1}\left(\frac{a}{2}\right) = \frac{\pi}{6}. \] 3. Solve for \( a \): Multiply through by \( 2 \): \[ \tan^{-1}\left(\frac{a}{2}\right) = \frac{\pi}{3}. \] Take the tangent on both sides: \[ \frac{a}{2} = \tan\left(\frac{\pi}{3}\right). \] The value of \( \tan\left(\frac{\pi}{3}\right) \) is \( \sqrt{3} \), so: \[ \frac{a}{2} = \sqrt{3}. \] Multiply through by \( 2 \): \[ a = 2\sqrt{3}. \] Hence, the value of \( a \) is (B) \( 2\sqrt{3} \).