MHT CET · Maths · Area Under Curves
The area of the region bounded by the curve \(y^2=4 x\) and the line \(y=x\) is
- A \(\frac{8}{3}\) sq. units
- B \(\frac{5}{8}\) sq. units
- C \(\frac{3}{8}\) sq. units
- D \(\frac{3}{5}\) sq. units
Answer & Solution
Correct Answer
(A) \(\frac{8}{3}\) sq. units
Step-by-step Solution
Detailed explanation
Refer figure, point of intersection of given curves are \(\mathrm{x}^2=4 \mathrm{x}\) \(\Rightarrow \mathrm{x}(\mathrm{x}-4)=0\)
\(
\therefore \mathrm{O} \equiv(0,0) \text { and } \mathrm{P} \equiv(4,4)
\)
Required area is shaded.
\(
\begin{aligned}
& \therefore A=\int_0^4(\sqrt{4 x}-x) d x=2 \int_0^4 x^{\frac{1}{2}} d x-\int_0^4 x d x \\
& =2\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_0^4-\left[\frac{x^2}{2}\right]_0^4-\left(\frac{4}{3}\right)(4 \times 2)-\frac{16}{2} \\
& =\frac{32}{3}-\frac{16}{2}=\frac{16}{6}=\frac{8}{3} \text { sq. units. }
\end{aligned}
\)
\(
\therefore \mathrm{O} \equiv(0,0) \text { and } \mathrm{P} \equiv(4,4)
\)
Required area is shaded.
\(
\begin{aligned}
& \therefore A=\int_0^4(\sqrt{4 x}-x) d x=2 \int_0^4 x^{\frac{1}{2}} d x-\int_0^4 x d x \\
& =2\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_0^4-\left[\frac{x^2}{2}\right]_0^4-\left(\frac{4}{3}\right)(4 \times 2)-\frac{16}{2} \\
& =\frac{32}{3}-\frac{16}{2}=\frac{16}{6}=\frac{8}{3} \text { sq. units. }
\end{aligned}
\)
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