AP EAMCET · Maths · Area Under Curves
The area (in sq. units) of the region given by \(R=\left\{(x, y): \frac{y^2}{2} \leq x \leq y+4\right\}\) is
- A \(16\)
- B \(18\)
- C \(24\)
- D \(30\)
Answer & Solution
Correct Answer
(B) \(18\)
Step-by-step Solution
Detailed explanation
\(\frac{y^2}{2} = y+4 \Rightarrow y^2 - 2y - 8 = 0 \Rightarrow (y-4)(y+2) = 0 \Rightarrow y = -2, 4\) \(A = \int_{-2}^{4} \left( (y+4) - \frac{y^2}{2} \right) dy\) \(A = \left[ \frac{y^2}{2} + 4y - \frac{y^3}{6} \right]_{-2}^{4}\)…
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\(\begin{array}{|c|c|c|c|c|}
\hline \begin{array}{c}
\text { Experiment } \\
\end{array} & \begin{array}{c}
\text { Initial } \\
{\left[\mathrm{I}_2\right]} \\
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\text { Initial } \\
{\left[\mathrm{H}^{+}\right]} \\
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\text { Initial } \\
{\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]} \\
\left(\mathrm{mol} \mathrm{L}^{-1}\right)
\end{array} & \begin{array}{c}
\text { Initial rate of decrease of } \mathrm{I}_2 \\
\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)
\end{array} \\
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\hline
\end{array}\)
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