If α = ∫_0^ 2 3 _2(x^2 + 4)\,dx + ∫_2^4 2^x - 4 \,dx , then α^2 is...

Let $f(x) = \log_2(x^2 + 4)$ for $x \ge 0$.

To find the inverse function $f^{-1}(x)$, we set $y = \log_2(x^2 + 4)$ and solve for $x$:

$2^y = x^2 + 4$

$x^2 = 2^y - 4$

$x = \sqrt{2^y - 4}$

Thus, $f^{-1}(x) = \sqrt{2^x - 4}$.

The given expression is $\alpha = \displaystyle\int_0^{2\sqrt{3}} f(x)\,dx + \displaystyle\int_2^4 f^{-1}(x)\,dx$.

We observe the limits of the first integral and evaluate the function at these points:

$f(0) = \log_2(0 + 4) = 2$

$f(2\sqrt{3}) = \log_2((2\sqrt{3})^2 + 4) = \log_2(12 + 4) = \log_2(16) = 4$

These exactly match the limits of the second integral. Using the standard property of definite integrals for inverse functions:

$\displaystyle\int_a^b f(x)\,dx + \displaystyle\int_{f(a)}^{f(b)} f^{-1}(x)\,dx = b f(b) - a f(a)$

Substituting $a = 0$ and $b = 2\sqrt{3}$:

$\alpha = 2\sqrt{3} \cdot f(2\sqrt{3}) - 0 \cdot f(0)$

$\alpha = 2\sqrt{3} \cdot 4 - 0 = 8\sqrt{3}$

Therefore, $\alpha^2 = (8\sqrt{3})^2 = 64 \times 3 = 192$.

Answer: $192$