Question

Value of definite integral $I = \int_{0}^{2\sqrt{3}} \log_2(x^2 + 4) dx + \int_{2}^{4} \sqrt{2^x - 4} dx$

• A. $2\sqrt{3}$
• B. $4\sqrt{3}$
• C. $6\sqrt{3}$
• D. $8\sqrt{3}$

Answer 1

The Correct Option is D

Step 1: Identify the relationship between the two integrals.
Let $f(x) = \log_2(x^2 + 4)$.
We want to find its inverse $f^{-1}(x)$.
$y = \log_2(x^2 + 4) \implies 2^y = x^2 + 4 \implies x = \sqrt{2^y - 4}$.
So, $f^{-1}(x) = \sqrt{2^x - 4}$.


Step 2: Check boundaries.
In the first integral, the limits are $a = 0$ to $b = 2\sqrt{3}$.
$f(0) = \log_2(0+4) = 2$.
$f(2\sqrt{3}) = \log_2((2\sqrt{3})^2 + 4) = \log_2(12 + 4) = \log_2(16) = 4$.
Notice these are the limits of the second integral.


Step 3: Apply the property of inverse function integrals.
Property: $\int_{a}^{b} f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dx = b f(b) - a f(a)$.
Substituting our values:
$I = (2\sqrt{3} \cdot f(2\sqrt{3})) - (0 \cdot f(0))$
$I = (2\sqrt{3} \cdot 4) - 0 = 8\sqrt{3}$.
The answer is $8\sqrt{3}$ (Option 4).