Question

If \[ \alpha=\int_{0}^{2\sqrt{3}} \log_2(x^2+4)\,dx + \int_{2}^{4} \sqrt{2^x-4}\,dx, \] then \(\alpha^2\) is equal to _____.

Answer 1

Correct Answer: 52

Step 1: Understanding the Concept:
This problem relies on the property of the sum of integrals of a function and its inverse:
\[ \int_a^b f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(y) dy = b f(b) - a f(a) \]
Step 2: Key Formula or Approach:
1. Identify \(f(x) = \log_2(x^2+4)\).
2. Find its inverse \(f^{-1}(x)\).
Step 3: Detailed Explanation:
Let \(y = \log_2(x^2+4)\).
Then \(2^y = x^2 + 4 \implies x^2 = 2^y - 4 \implies x = \sqrt{2^y - 4}\).
Thus, \(f^{-1}(x) = \sqrt{2^x - 4}\).
Now check the limits:
For \(f(x)\): Lower limit \(a = 0 \implies f(0) = \log_2(4) = 2\).
Upper limit \(b = 2\sqrt{3} \implies f(2\sqrt{3}) = \log_2((2\sqrt{3})^2 + 4) = \log_2(12 + 4) = 4\).
Notice the second integral is \(\int_2^4 f^{-1}(x) dx\), which matches the limits \([f(0), f(2\sqrt{3})]\).
Applying the formula:
\[ \alpha = [x \cdot f(x)]_0^{2\sqrt{3}} = (2\sqrt{3} \times f(2\sqrt{3})) - (0 \times f(0)) \]
\[ \alpha = 2\sqrt{3} \times 4 = 8\sqrt{3} \].
The required value is \(\alpha^2\):
\[ \alpha^2 = (8\sqrt{3})^2 = 64 \times 3 = 192 \].
Step 4: Final Answer:
The value of \(\alpha^2\) is 192.