Question

Evaluate \[ \int_{0}^{4}\frac{x+2}{\sqrt{4x-x^2}}\,dx \] is:

• A. \(2\pi\)
• B. \(0\)
• C. \(\pi\)
• D. \(4\pi\)

Hint:

For integrals involving \[ \sqrt{4x-x^2}, \] complete the square: \[ 4x-x^2=4-(x-2)^2 \] and then shift the variable using \(t=x-2\).

Answer 1

The Correct Option is D

Step 1: Simplify the expression under the square root.
\[ 4x-x^2=4-(x-2)^2 \] Let \[ t=x-2 \] Then, \[ dx=dt \] When \[ x=0,\quad t=-2 \] and when \[ x=4,\quad t=2 \] Also, \[ x+2=t+4 \]

Step 2: Rewrite the integral.
\[ \int_{0}^{4}\frac{x+2}{\sqrt{4x-x^2}}\,dx = \int_{-2}^{2}\frac{t+4}{\sqrt{4-t^2}}\,dt \] \[ = \int_{-2}^{2}\frac{t}{\sqrt{4-t^2}}\,dt + 4\int_{-2}^{2}\frac{1}{\sqrt{4-t^2}}\,dt \]

Step 3: Use odd-even property.
The function \[ \frac{t}{\sqrt{4-t^2}} \] is odd, so \[ \int_{-2}^{2}\frac{t}{\sqrt{4-t^2}}\,dt=0 \] Also, \[ \int_{-2}^{2}\frac{1}{\sqrt{4-t^2}}\,dt=\pi \] Therefore, \[ \int_{0}^{4}\frac{x+2}{\sqrt{4x-x^2}}\,dx = 4\pi \]

Step 4: Final conclusion.
Hence, \[ \boxed{4\pi} \]