CBSE 2026 · Region 4 · Set 3 · Q33 · 5 marks
Using integration, find the area of the region bounded by $\displaystyle \mathrm{y}=5 \mathrm{x}+4$, $\displaystyle \mathrm{y}=0, \mathrm{x}=-1$ and $\displaystyle \mathrm{x}=1$.
Marking-scheme solution
Required Area $\displaystyle =\left|\int_{-1}^{\frac{-4}{5}} y\, d x\right|+\int_{\frac{-4}{5}}^{1} y\, d x$
Area $\displaystyle =-\int_{-1}^{\frac{-4}{5}}(5 x+4) d x+\int_{\frac{-4}{5}}^{1}(5 x+4) d x$
Area $\displaystyle =-\left(\dfrac{(5 x+4)^{2}}{10}\right)_{-1}^{\frac{-4}{5}}+\left(\dfrac{(5 x+4)^{2}}{10}\right)_{\frac{-4}{5}}^{1}=\dfrac{1}{10}+\dfrac{81}{10}=\dfrac{41}{5}$ or $\displaystyle \dfrac{41}{5}$ sq. units
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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.