MHT CET · Maths · Area Under Curves
The area (in sq. units) of the region described by \(\left\{(x, y) / y^2 \leq 2 x\right.\) and \(\left.y \geq(4 x-1)\right\}\) is
- A \(\frac{15}{64}\)
- B \(\frac{9}{32}\)
- C \(\frac{7}{32}\)
- D \(\frac{5}{64}\)
Answer & Solution
Correct Answer
(B) \(\frac{9}{32}\)
Step-by-step Solution
Detailed explanation

Putting \(x=\frac{y^2}{2}\) in \(y=4 x-1\), we get
\(\begin{aligned}
& y=4\left(\frac{y^2}{2}\right)-1 \Rightarrow 2 y^2-y-1=0 \\
& \Rightarrow(y-1)(2 y+1)=0 \\
& \Rightarrow y=1, \frac{-1}{2}
\end{aligned}\)
\(\therefore \quad\) Required area
\(\begin{aligned}
& =\int_{-1 / 2}^1\left(\frac{y+1}{4}\right) \mathrm{d} y-\int_{-1 / 2}^1 \frac{y^2}{2} \mathrm{~d} y \\
& =\frac{1}{4}\left[\frac{y^2}{2}+y\right]_{-1 / 2}^1-\frac{1}{2}\left[\frac{y^3}{3}\right]_{-1 / 2}^1 \\
& =\frac{1}{4}\left[\left(\frac{1}{2}-\frac{1}{8}\right)+\left(1+\frac{1}{2}\right)\right]-\frac{1}{2}\left(\frac{1}{3}+\frac{1}{24}\right) \\
& =\frac{1}{4}\left(\frac{15}{8}\right)-\frac{1}{2}\left(\frac{9}{24}\right) \\
& =\frac{15}{32}-\frac{3}{16} \\
& =\frac{9}{32}
\end{aligned}\)
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