question_answer

The area of the region in 1st quadrant bounded by the y-axis, y = \[\frac{x}{4}\], y = 1 + \[\sqrt{x}\], and \[y=\frac{2}{\sqrt{x}}\] is

A) 2/3 sq. units      

B) 8/3 sq. units

C) 11/3 sq. units    

D) 13/6 sq. units

Correct Answer: C

Solution :

[c] \[{{A}_{1}}=\int\limits_{0}^{1}{\left( 1+\sqrt{x}-\frac{x}{4} \right)dx}\] \[={{\left[ x+\frac{2{{x}^{3/2}}}{3}-\frac{{{x}^{2}}}{8} \right]}_{0}}^{1}=1+\frac{2}{3}-\frac{1}{8}=\frac{37}{24}.\] \[{{A}_{2}}=\int\limits_{1}^{4}{\left( \frac{2}{\sqrt{x}}-\frac{x}{4} \right)dx}\] \[={{\left[ 4\sqrt{x}=\frac{{{x}^{2}}}{8} \right]}_{1}}^{4}\] \[=\left[ 8-2-4+\frac{1}{8} \right]\] \[=\frac{17}{8}\] Or \[A={{A}_{1}}+{{A}_{2}}=\frac{88}{24}=\frac{11}{3}\] sq. units