CBSE 2026 · Region 4 · Set 2 · Q33 · 5 marks
Using integration, find the area of the region bounded by the curve $\displaystyle \mathrm{y}=\mathrm{x}|\mathrm{x}|, \mathrm{x}$-axis, $\displaystyle \mathrm{x}=-2$ and $\displaystyle \mathrm{x}=2$.
Marking-scheme solution
Required Area $\displaystyle =-\int_{-2}^{0} x|x| d x+\int_{0}^{2} x|x| d x$
Area $\displaystyle =\int_{-2}^{0}-x^{2} d x+\int_{0}^{2} x^{2} d x$
Area $\displaystyle =-\left(\dfrac{x^{3}}{3}\right)_{-2}^{0}+\left(\dfrac{x^{3}}{3}\right)_{0}^{2}=\dfrac{8}{3}+\dfrac{8}{3}=\dfrac{16}{3}$ or $\displaystyle \dfrac{16}{3}$ sq. units
Alternative approach: Area $\displaystyle =2 \times \int_{0}^{2} x^{2} d x$ & proceed accordinglyApplication of IntegralsArea under Simple CurvesApplylong_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.