question_answer

Area under the curve \[y=\sqrt{3x+4}\] between \[x=0\] and \[x=4,\] is                                               [AI CBSE 1979, 80]

A)            \[\frac{56}{9}\] sq. unit   

B)            \[\frac{64}{9}\] sq. unit

C)            8 sq. unit                                

D)            None of these

Correct Answer: D

Solution :

                   Area \[=\int_{0}^{4}{\sqrt{3x+4}}dx=\left| \frac{{{(3x+4)}^{3/2}}}{3.(3/2)} \right|_{0}^{4}\]                             \[=\frac{2}{9}\times 56=\frac{112}{9}sq.\]unit.