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Mathematics · 2022 · 4 marks
CBSE 2022 · Region 3 · Set 1 · Q11
Find the area of the region bounded by curve 4 x^2= y and the line y =8 x+12, using integration.
Find the area of the region bounded by curve $\displaystyle 4 x^{2}=\mathrm{y}$ and the line $\displaystyle \mathrm{y}=8 x+12$, using integration.
Marking-scheme solution
Correct Figure
Point of intersection
\[\begin{aligned}
& 4 x^{2}=8 x+12 \\
& x^{2}-2 x-3=0 \\
& (x-3)(x+1)=0 \\
& x=3,-1
\end{aligned}
\]
\[\begin{aligned}
\text { Area } & =\int_{-1}^{3}\left[(8 x+12)-4 x^{2}\right] d x \\
& \left.=4 x^{2}+12 x-\frac{4}{3} x^{3}\right]_{-1}^{3} \\
& =36+36-36-\left(4-12+\frac{4}{3}\right) \\
& =44-\frac{4}{3}=\frac{128}{3}
\end{aligned}
\]
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