Question

The area bounded by the curve \( y = x^2 + 3, y = x, x = 3 \) and \( y \)-axis is

• A. \( \frac{9}{2} \) sq. units
• B. 18 sq. units
• C. \( \frac{27}{2} \) sq. units
• D. \( \frac{27}{3} \) sq. units

Hint:

Area $= \int (f(x) - g(x)) dx$. Always verify which curve is "upper" in the given interval.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The area is bounded by a parabola, a line, and vertical boundaries \( x = 0 \) (\( y \)-axis) and \( x = 3 \).
Step 2: Key Formula or Approach:
Area \( = \int_{a}^{b} [f(x) - g(x)] dx \), where \( f(x) \ge g(x) \).
Step 3: Detailed Explanation:
On the interval \( [0, 3] \), \( x^2 + 3 \) is always greater than \( x \).
Area \( = \int_{0}^{3} [(x^2 + 3) - x] dx \)
\[ = \left[ \frac{x^3}{3} + 3x - \frac{x^2}{2} \right]_0^3 \] \[ = \left( \frac{27}{3} + 3(3) - \frac{9}{2} \right) - 0 = 9 + 9 - 4.5 = 18 - 4.5 = 13.5 = \frac{27}{2} \] Step 4: Final Answer:
The area is \( \frac{27}{2} \) sq. units.