Find the area (in square units) enclosed between the curves $y = x^{2}$ and $y = 2 x + 3$ .

medium 3 marks Class 12 · Mathematics · Integrals

Answer

\frac{32}{3}

The correct answer is $\frac{32}{3}$ — Hence the required area is $\frac{32}{3}$ square units.

Step-by-step solution

Set the two expressions equal to find intersection points: $x^{2} = 2 x + 3 \Rightarrow x^{2} - 2 x - 3 = 0$ .
Factor the quadratic: $(x - 3) (x + 1) = 0$ , giving $x = - 1$ and $x = 3$ .
Determine which curve is on top in the interval $[- 1, 3]$ : for $x = 0$ , $x^{2} = 0$ and $2 x + 3 = 3$ , so $2 x + 3$ is above $x^{2}$ .
Set up the area as a definite integral of the upper minus lower function: $A = \int_{- 1}^{3} [(2 x + 3) - x^{2}] d x$ .
Integrate term by term: $\int (2 x + 3) d x = x^{2} + 3 x$ , $\int x^{2} d x = \frac{x ^{3}}{3}$ .
Thus $A = [x^{2} + 3 x - \frac{x ^{3}}{3}]_{- 1}^{3}$ .
Evaluate at $x = 3$ : $3^{2} + 3 \cdot 3 - \frac{27}{3} = 9 + 9 - 9 = 9$ .
Evaluate at $x = - 1$ : $(- 1)^{2} + 3 (- 1) - \frac{( - 1 ) ^{3}}{3} = 1 - 3 + \frac{1}{3} = - \frac{5}{3}$ .
Subtract lower limit from upper: $9 - (- \frac{5}{3}) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3}$ .
Hence the required area is $\frac{32}{3}$ square units.

Common mistake: Reversing the order of subtraction (integrating

x^{2} - (2 x + 3)

) which gives a negative area.

Concept tested: area between curves using definite integrals

Published 31 August 2026 · Last updated 16 September 2026 · How this solution was prepared and checked