KCET · Maths · Area Under Curves
Area lying between the curves \( y^{2}=2 x \) and \( y=x \)
- A \( \frac{2}{3} \) sq. units
- B \( \frac{1}{3} \) sq. units
- C \( \frac{1}{4} \) sq. units
- D \( \frac{3}{4} \) sq. units
Answer & Solution
Correct Answer
(A) \( \frac{2}{3} \) sq. units
Step-by-step Solution
Detailed explanation
Given curves,
\[
\begin{array}{l}
y^{2}=2 x \rightarrow(1) \\
y=x \rightarrow(2)
\end{array}
\]
At \(x=0\) and \(x=2\) we have two intersect points, that is, \((0,0),(2,2)\).
Therefore, required area is given by
\[
\begin{array}{l}
\int_{0}^{2}(\sqrt{2} \sqrt{x}-x) d x \\
=\left[\sqrt{2}\left(\frac{2}{3} x^{3 / 2}\right)-\frac{x^{2}}{2}\right]^{2} \\
=\left(\frac{2 \sqrt{2}}{3}\left(2^{3 / 2}\right)-\frac{4}{2}\right) \\
=\frac{2}{3} \cdot 2^{2}-2=\frac{2}{3} \text { sq. units }
\end{array}
\]
\[
\begin{array}{l}
y^{2}=2 x \rightarrow(1) \\
y=x \rightarrow(2)
\end{array}
\]
At \(x=0\) and \(x=2\) we have two intersect points, that is, \((0,0),(2,2)\).
Therefore, required area is given by
\[
\begin{array}{l}
\int_{0}^{2}(\sqrt{2} \sqrt{x}-x) d x \\
=\left[\sqrt{2}\left(\frac{2}{3} x^{3 / 2}\right)-\frac{x^{2}}{2}\right]^{2} \\
=\left(\frac{2 \sqrt{2}}{3}\left(2^{3 / 2}\right)-\frac{4}{2}\right) \\
=\frac{2}{3} \cdot 2^{2}-2=\frac{2}{3} \text { sq. units }
\end{array}
\]
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