Question:
The area bounded by the curve \( y = x^2 + 3, y = x, x = 3 \) and \( y \)-axis is
The area bounded by the curve \( y = x^2 + 3, y = x, x = 3 \) and \( y \)-axis is
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Area $= \int (f(x) - g(x)) dx$. Always verify which curve is "upper" in the given interval.
Updated On: Apr 30, 2026
- \( \frac{9}{2} \) sq. units
- 18 sq. units
- \( \frac{27}{2} \) sq. units
- \( \frac{27}{3} \) sq. units
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The Correct Option is C
Solution and Explanation
Step 1: Define Integral Limits
The $y$-axis is $x=0$. The upper limit is $x=3$.
Step 2: Set up Area Integral
Area $= \int_0^3 (y_{upper} - y_{lower}) dx = \int_0^3 (x^2 + 3 - x) dx$.
Step 3: Integration
$[x^3/3 + 3x - x^2/2]_0^3$. $= (27/3 + 9 - 9/2) - (0)$. $= 9 + 9 - 4.5 = 13.5$.
Step 4: Conclusion
$13.5 = 27/2$ sq. units.
Final Answer:(C)
The $y$-axis is $x=0$. The upper limit is $x=3$.
Step 2: Set up Area Integral
Area $= \int_0^3 (y_{upper} - y_{lower}) dx = \int_0^3 (x^2 + 3 - x) dx$.
Step 3: Integration
$[x^3/3 + 3x - x^2/2]_0^3$. $= (27/3 + 9 - 9/2) - (0)$. $= 9 + 9 - 4.5 = 13.5$.
Step 4: Conclusion
$13.5 = 27/2$ sq. units.
Final Answer:(C)
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