Calculate Definite Integral with Logarithmic and Exponential Inverse Functions

To evaluate the value of $\alpha^2$, where $\alpha = \int_{0}^{2\sqrt{3}} \log_2 (x^2+4) \, \mathrm{d}x + \int_{2}^{4} \sqrt{2^x-4} \, \mathrm{d}x$

we can analyze the relationship between the two functions in the integrands.

Step 1: Identify the Inverse Relationship

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

To find its inverse function, $y = f(x)$: $y = \log_2 (x^2+4)$ $2^y = x^2 + 4$ $x^2 = 2^y - 4$ $x = \sqrt{2^y - 4}$

Thus, the inverse function is $g(y) = f^{-1}(y) = \sqrt{2^y - 4}$.

Step 2: Check the Limits of Integration

Now, let's check the values of $f(x)$ at the lower and upper limits of the first integral:

• For $x = 0$: $f(0) = \log_2(0^2 + 4) = \log_2(4) = 2$
• For $x = 2\sqrt{3}$: $f(2\sqrt{3}) = \log_2((2\sqrt{3})^2 + 4) = \log_2(12 + 4) = \log_2(16) = 4$

We observe that as $x$ ranges from $0$ to $2\sqrt{3}$, $y = f(x)$ ranges from $2$ to $4$, which correspond precisely to the limits of the second integral.

Step 3: Simplify the Integral using the Inverse Function Identity

We know the standard definite integral property for inverse functions: $\int_{a}^{b} f(x) \, \mathrm{d}x + \int_{f(a)}^{f(b)} f^{-1}(y) \, \mathrm{d}y = b \cdot f(b) - a \cdot f(a)$

Substitute $y = f(x)$ into the second integral $I_2 = \int_{2}^{4} f^{-1}(y) \, \mathrm{d}y$: $I_2 = \int_{0}^{2\sqrt{3}} x f'(x) \, \mathrm{d}x$

Applying Integration by Parts on $I_2$: $I_2 = \left[ x f(x) \right]_{0}^{2\sqrt{3}} - \int_{0}^{2\sqrt{3}} f(x) \, \mathrm{d}x$

Substitute $I_2$ back into the expression for $\alpha$: $\alpha = \int_{0}^{2\sqrt{3}} f(x) \, \mathrm{d}x + \left( \left[ x f(x) \right]_{0}^{2\sqrt{3}} - \int_{0}^{2\sqrt{3}} f(x) \, \mathrm{d}x \right)$ $\alpha = \left[ x f(x) \right]_{0}^{2\sqrt{3}}$ $\alpha = 2\sqrt{3} \cdot f(2\sqrt{3}) - 0 \cdot f(0)$ $\alpha = 2\sqrt{3} \cdot 4 - 0 = 8\sqrt{3}$

Step 4: Calculate $\alpha^2$

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