Daily Question 109

Daily Question 109

$\textbf{Q109.}$ $\displaystyle\int\limits_{1}^{5}\left(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\right)=$

(A) $\dfrac{16}{3}$

(B) $\dfrac{32}{3}$

(C) $\dfrac{34}{3}$

(D) $\dfrac{8}{3}$

$\textbf{Ans.} (C)$

$\textbf{Sol.}$ Let $I=\displaystyle\int\limits_{1}^{5}\left[\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\right]\mathbb{d}x$

Substitute $\sqrt{x-1}=u\to\mathbb{d}x=2\sqrt{x-1}\mathbb{d}u$

$I=\displaystyle\int\limits_{0}^{2}2u\left[\sqrt{u^2+2u+1}+\sqrt{u^2-2u+1}\right]\mathbb{d}u$

$I=\displaystyle\int\limits_{0}^{2}2u\left[|u+1|+|u-1|\right]\mathbb{d}u$

$I=\displaystyle\int\limits_{0}^{1}2u\left[u+1-u+1|\right]\mathbb{d}u+\int\limits_{1}^{2}2u\left[u+1+u-1|\right]\mathbb{d}u$

$I=\displaystyle\int\limits_{0}^{1}4u\mathbb{d}u+\int\limits_{1}^{2}4u^2\mathbb{d}u$

$I=\left[2u^2\right]_{0}^{1}+\left[\dfrac{4u^3}{3}\right]_{1}^{2}$

$I=\dfrac{34}{3}$

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