Definite IntegrationeasyFree

∫ (√(x+√(12x−36))+√(x−√(12x−36))) dx | JEE

JEE Maths question with a full step-by-step solution.

Question

The value of

\displaystyle\int_{3}^{6}\left(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}\right)dx

6\sqrt3

correct

4\sqrt3

12\sqrt3

2\sqrt3

Solution

Step 1: Substitute

x-3=t

, so

dx=dt

and

12x-36=12t

, with

t:0\to 3

. The nested radicals become

\sqrt{12t}=2\sqrt{3t}

, and

x=t+3

x\pm\sqrt{12x-36}=(t+3)\pm 2\sqrt{3t}.

Step 2: Recognise perfect squares. Since

(\sqrt t+\sqrt3)^{2}=t+3+2\sqrt{3t}

and

(\sqrt t-\sqrt3)^{2}=t+3-2\sqrt{3t}

\sqrt{x+\sqrt{12x-36}}=\sqrt t+\sqrt3,\qquad \sqrt{x-\sqrt{12x-36}}=\big|\sqrt t-\sqrt3\big|.

Step 3: Resolve the modulus on

t\in[0,3]

. Here

\sqrt t\le\sqrt3

, so

\big|\sqrt t-\sqrt3\big|=\sqrt3-\sqrt t

, and the sum telescopes:

\big(\sqrt t+\sqrt3\big)+\big(\sqrt3-\sqrt t\big)=2\sqrt3.

Step 4: The integrand is the constant

2\sqrt3

\int_{0}^{3}2\sqrt3\,dt=2\sqrt3\cdot 3=6\sqrt3.

Correct answer: (1)

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