Question
The area enclosed by the curve $${x^2}y = 36,$$ the $$x$$-axis and the lines $$x = 6$$ and $$x = 9$$ is :
A.
6
B.
1
C.
4
D.
2
Answer :
2
Solution :
$$y = \frac{{36}}{{{x^2}}},\,x = 6,\,x = 9$$
Desired area $$ = \int\limits_6^9 {\frac{{36}}{{{x^2}}}dx} = - 36\left[ {\frac{1}{9} - \frac{1}{6}} \right] = 2$$
$$y = \frac{{36}}{{{x^2}}},\,x = 6,\,x = 9$$
Desired area $$ = \int\limits_6^9 {\frac{{36}}{{{x^2}}}dx} = - 36\left[ {\frac{1}{9} - \frac{1}{6}} \right] = 2$$