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 sq unit
• B. 1 sq unit
• C. 4 sq unit
• D. 3 sq unit

Hint:

Area under a curve between two vertical lines: $A = \displaystyle\int_a^b y\,dx$. Simplify the integrand before integrating.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Express $y$ in terms of $x$ and integrate over the given interval.
Step 2: Detailed Explanation:
$y = \dfrac{36}{x^2}$.
Area $= \displaystyle\int_6^9 \frac{36}{x^2}\,dx = 36\left[-\frac{1}{x}\right]_6^9 = 36\left(\frac{1}{6} - \frac{1}{9}\right) = 36 \times \frac{1}{18} = 2$ sq units.
Step 3: Final Answer:
Area $= 2$ sq units.