Question

Evaluate: \( \int_{0}^{4} \frac{x + 2}{\sqrt{4x - x^2}}\, dx \)

• A. \( 2\pi \)
• B. \( 0 \)
• C. \( \pi \)
• D. \( \frac{\pi}{2} \)

Hint:

Use trigonometric substitution for integrals involving \( \sqrt{ax - x^2} \) or similar quadratic square roots.

Answer 1

The Correct Option is C

Let: \[ \begin{align} I = \int_0^4 \frac{x+2}{\sqrt{4x - x^2}}\, dx = \int_0^4 \frac{x+2}{\sqrt{-(x^2 - 4x)}}\, dx = \int_0^4 \frac{x+2}{\sqrt{-(x - 2)^2 + 4}}\, dx \] Make substitution: \[ x = 2 + 2\sin\theta \Rightarrow dx = 2\cos\theta\, d\theta \] Then: \[ \sqrt{4 - (x - 2)^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos\theta \] Bounds: When \( x = 0 \Rightarrow \theta = -\frac{\pi}{2} \), and \( x = 4 \Rightarrow \theta = \frac{\pi}{2} \) So the integral becomes: \[ \begin{align} I = \int_{-\pi/2}^{\pi/2} \frac{2\sin\theta + 4}{2\cos\theta} \cdot 2\cos\theta\, d\theta = \int_{-\pi/2}^{\pi/2} (2\sin\theta + 4)\, d\theta \] \[ \begin{align} = \int_{-\pi/2}^{\pi/2} 2\sin\theta\, d\theta + \int_{-\pi/2}^{\pi/2} 4\, d\theta = 0 + 4\pi/2 = \pi \]