Question

The area bounded by \( y = x^2 + 3 \) and \( y = 2x + 3 \) is (in sq. units)

• A. \( \frac{12}{7} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{8}{3} \)
• E. \( \frac{3}{8} \)

Hint:

Always check which curve lies above before subtracting—wrong order gives negative area.

Answer 1

The Correct Option is B

Concept: Area between two curves: \[ \text{Area} = \int_{a}^{b} [\text{upper} - \text{lower}]\,dx \]

Step 1: Find points of intersection
\[ x^2 + 3 = 2x + 3 \] \[ x^2 - 2x = 0 \Rightarrow x(x-2)=0 \] \[ x = 0, 2 \]

Step 2: Determine upper and lower curve
At \(x=1\): \[ y_{\text{line}} = 5,\quad y_{\text{parabola}} = 4 \] Thus: \[ \text{Upper} = 2x+3,\quad \text{Lower} = x^2+3 \]

Step 3: Form the integral
\[ \text{Area} = \int_0^2 [(2x+3)-(x^2+3)]dx \] \[ = \int_0^2 (2x - x^2)dx \]

Step 4: Integrate
\[ \int (2x - x^2)dx = x^2 - \frac{x^3}{3} \]

Step 5: Apply limits
\[ \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3} \] Final Answer: \[ \boxed{\frac{4}{3}} \]