MHT CET · Maths · Area Under Curves
The area (in sq. units) of the region bounded by the curve \(x^2=4 y\) and the straight line \(x=4 y-2\) is
- A \(\frac{9}{8}\)
- B \(\frac{7}{8}\)
- C \(\frac{5}{4}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{9}{8}\)
Step-by-step Solution
Detailed explanation
The points of intersection of \(x^2=4 y\) and \(x=4 y-2\) are \((2,1)\) and \(\left(-1, \frac{1}{4}\right)\).
Required area
\(\begin{aligned}
& =\int_{-1}^2 \frac{1}{4}(x+2) \mathrm{d} x-\int_{-1}^2 \frac{1}{4} x^2 \mathrm{~d} x \\
& =\frac{1}{4}\left[\frac{x^2}{2}+2 x\right]_{-1}^2-\frac{1}{4}\left[\frac{x^3}{3}\right]_{-1}^2 \\
& =\frac{9}{8} \text { sq. units }
\end{aligned}\)
Required area
\(\begin{aligned}
& =\int_{-1}^2 \frac{1}{4}(x+2) \mathrm{d} x-\int_{-1}^2 \frac{1}{4} x^2 \mathrm{~d} x \\
& =\frac{1}{4}\left[\frac{x^2}{2}+2 x\right]_{-1}^2-\frac{1}{4}\left[\frac{x^3}{3}\right]_{-1}^2 \\
& =\frac{9}{8} \text { sq. units }
\end{aligned}\)
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