MHT CET · Maths · Area Under Curves
The area (in sq. units) of the region \(\mathrm{A}=\left\{(x, y) / \frac{y^2}{2} \leq x \leq y+4\right\}\)
- A 30
- B \(\frac{53}{3}\)
- C 16
- D 18
Answer & Solution
Correct Answer
(D) 18
Step-by-step Solution
Detailed explanation
Given that \(\frac{y^2}{2} \leq x \leq y+4\)
\(\begin{array}{ll}
\therefore \quad & x=\frac{y^2}{2} \text { and } x=y+4 \\
& \frac{y^2}{2}=y+4^2 \\
\therefore \quad & y^2-2 y-8=0 \\
\therefore \quad & y=4 \text { or }-2 \\
& \Rightarrow x=8 \text { or } 2
\end{array}\)
\(\therefore \quad \mathrm{A}=\int_{-2}^4\left(y+4-\frac{y^2}{2}\right) \mathrm{d} y\)
\(\begin{aligned} & \therefore \quad \mathrm{A}=\left[\frac{y^2}{2}+4 y-\frac{y^3}{6}\right]_{-2}^4 \\ & \therefore \quad A=\left(8+16-\frac{64}{6}\right)-\left(2-8+\frac{8}{6}\right) \\ & \therefore \quad \mathrm{A}=18 \\ & \end{aligned}\)
\(\begin{array}{ll}
\therefore \quad & x=\frac{y^2}{2} \text { and } x=y+4 \\
& \frac{y^2}{2}=y+4^2 \\
\therefore \quad & y^2-2 y-8=0 \\
\therefore \quad & y=4 \text { or }-2 \\
& \Rightarrow x=8 \text { or } 2
\end{array}\)
\(\therefore \quad \mathrm{A}=\int_{-2}^4\left(y+4-\frac{y^2}{2}\right) \mathrm{d} y\)
\(\begin{aligned} & \therefore \quad \mathrm{A}=\left[\frac{y^2}{2}+4 y-\frac{y^3}{6}\right]_{-2}^4 \\ & \therefore \quad A=\left(8+16-\frac{64}{6}\right)-\left(2-8+\frac{8}{6}\right) \\ & \therefore \quad \mathrm{A}=18 \\ & \end{aligned}\)
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