Question

Evaluate \[ \int_2^5 \sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\,dx \]

• A. \(\dfrac{16}{3}\)
• B. \(\dfrac{32}{3}\)
• C. \(\dfrac{28}{3}\)
• D. \(\dfrac{4}{3}\)

Hint:

For expressions of the form \[ x\pm 2\sqrt{x-1}, \] write \(x=(x-1)+1\) and convert them into perfect squares: \[ x\pm 2\sqrt{x-1}=\left(\sqrt{x-1}\pm 1\right)^2. \]

Answer 1

The Correct Option is C

Step 1: Simplify the first square root.
We have \[ x+2\sqrt{x-1} \] Since, \[ x=(x-1)+1 \] we can write \[ x+2\sqrt{x-1}=(x-1)+1+2\sqrt{x-1} \] \[ x+2\sqrt{x-1}=\left(\sqrt{x-1}+1\right)^2 \] Therefore, \[ \sqrt{x+2\sqrt{x-1}}=\sqrt{x-1}+1 \]

Step 2: Simplify the second square root.
Similarly, \[ x-2\sqrt{x-1}=(x-1)+1-2\sqrt{x-1} \] \[ x-2\sqrt{x-1}=\left(\sqrt{x-1}-1\right)^2 \] Since \(x\in[2,5]\), \[ \sqrt{x-1}\geq 1 \] Therefore, \[ \sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1 \]

Step 3: Simplify the integrand.
So, \[ \sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}} \] \[ =\left(\sqrt{x-1}+1\right)+\left(\sqrt{x-1}-1\right) \] \[ =2\sqrt{x-1} \]

Step 4: Evaluate the integral.
Therefore, \[ \int_2^5 \sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\,dx = \int_2^5 2\sqrt{x-1}\,dx \] Let \[ u=x-1 \] Then, \[ du=dx \] When \(x=2\), \[ u=1 \] When \(x=5\), \[ u=4 \] Thus, \[ \int_2^5 2\sqrt{x-1}\,dx = 2\int_1^4 u^{\frac{1}{2}}\,du \] \[ =2\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_1^4 \] \[ =\frac{4}{3}\left[u^{\frac{3}{2}}\right]_1^4 \] \[ =\frac{4}{3}\left(4^{\frac{3}{2}}-1^{\frac{3}{2}}\right) \] \[ =\frac{4}{3}(8-1) \] \[ =\frac{28}{3} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{28}{3}} \]