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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 4 · Set 1 · Q32
(a) Sketch the graph of y =x|x| and hence find the area bounded by this curve, X -axis and the ordinates x=-2 and x=2, using integration.
(a)
Sketch the graph of $\displaystyle \mathrm{y}=x|x|$ and hence find the area bounded by this curve, X -axis and the ordinates $\displaystyle x=-2$ and $\displaystyle x=2$, using integration.
OR
(b)
Using integration, find the area bounded by the ellipse $\displaystyle 9 x^{2}+25 \mathrm{y}^{2}=225$, the lines $\displaystyle x=-2, x=2$, and the X -axis.
Marking-scheme solution
As, $\displaystyle y = x\,|x| = \begin{cases} -x^2, & x < 0 \\ x^2, & x \ge 0 \end{cases}$
Area of the shaded region $\displaystyle = \int_{-2}^{2} y\, dx = 2\int_{0}^{2} y\, dx = 2\int_{0}^{2} x^2\, dx$
$\displaystyle = 2\left(\dfrac{x^3}{3}\right)_{0}^{2}$
$\displaystyle = 2\left(\dfrac{8}{3}\right) = \dfrac{16}{3}$
As, $\displaystyle 9x^2 + 25y^2 = 225 \Rightarrow y = \pm\dfrac{3}{5}\sqrt{5^2 - x^2}$
Required Area $\displaystyle = \int_{-2}^{2} \dfrac{3}{5}\sqrt{5^2 - x^2}\, dx = \dfrac{6}{5}\int_{0}^{2}\sqrt{5^2 - x^2}\, dx$
$\displaystyle = \dfrac{6}{5}\left(\dfrac{x\sqrt{5^2 - x^2}}{2} + \dfrac{25}{2}\sin^{-1}\left(\dfrac{x}{5}\right)\right)_{0}^{2}$
$\displaystyle = \dfrac{6}{5}\left(\dfrac{2\sqrt{21}}{2} + \dfrac{25}{2}\sin^{-1}\left(\dfrac{2}{5}\right)\right)$
$\displaystyle = \left(\dfrac{6\sqrt{21}}{5} + 15\sin^{-1}\left(\dfrac{2}{5}\right)\right)$
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CBSE Class 12 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.