Question

The area (in square units) enclosed by the curve $x ^ { 2 } y = 36$ the x-axis and the lines x = 6 and x = 9 is

• A. 2
• B. 1
• C. 4
• D. 3

Answer 1

Answer: (A)

2

Answer 2

Required area $= \int _ { 6 } ^ { 9 } y d x = \int _ { 6 } ^ { 9 } \frac { 36 } { x ^ { 2 } } d x$

[Given $x ^ { 2 } y = 36 \Rightarrow y = \frac { 36 } { x ^ { 2 } }$]

$= \left[ - \frac { 36 } { x } \right] _ { 6 } ^ { 9 } = - \left[ \frac { 36 } { 9 } - \frac { 36 } { 6 } \right] = - [ 4 - 6 ] = 2$.