Question:medium

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

Updated On: Apr 3, 2026
  • $2\sqrt{3}$
  • $4\sqrt{3}$
  • $6\sqrt{3}$
  • $8\sqrt{3}$
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
The expression is a sum of two integrals. Let's check if the functions are inverses of each other.
If \( y = f(x) \), then \( \int f(x) dx + \int f^{-1}(y) dy \) has a standard property.
Step 2: Key Formula or Approach:
Identity: \( \int_{a}^{b} f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(y) dy = b f(b) - a f(a) \).
Step 3: Detailed Explanation:
Let \( f(x) = \log_2(x^2 + 4) \).
Find the inverse: \( x^2 + 4 = 2^y \implies x = \sqrt{2^y - 4} \).
This matches the second integrand exactly!
Now check the limits:
For first integral, \( a = 0, b = 2\sqrt{3} \).
\( f(a) = f(0) = \log_2(0^2 + 4) = 2 \).
\( f(b) = f(2\sqrt{3}) = \log_2( (2\sqrt{3})^2 + 4 ) = \log_2(12 + 4) = \log_2(16) = 4 \).
The limits of the second integral are indeed \( f(a)=2 \) and \( f(b)=4 \).
Using the property:
\( I = (2\sqrt{3}) \cdot f(2\sqrt{3}) - (0) \cdot f(0) \)
\( I = 2\sqrt{3} \cdot 4 - 0 = 8\sqrt{3} \).
Step 4: Final Answer:
The value of the integral is \( 8\sqrt{3} \).
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