MHT CET · Maths · Area Under Curves
The area bounded between the curve \(x^2=y\) and the line \(y=4 x\) is
- A \(\frac{32}{3}\) sq. units
- B \(\frac{8}{3}\) sq. units
- C \(\frac{1}{3}\) sq. units
- D \(\frac{16}{3}\)sq. units
Answer & Solution
Correct Answer
(A) \(\frac{32}{3}\) sq. units
Step-by-step Solution
Detailed explanation
Required area is shaded.
Point of intersection of given curves are \((0,0)\) and \((4,16)\)
\(
\begin{aligned}
& \therefore A=\int_0^4\left(4 x-x^2\right) d x \\
& =4 \int_0^4 x d x-\int_0^4 x^2 d x=4\left[\frac{x^2}{2}\right]_0^4-\left[\frac{x^3}{3}\right]_0^4 \\
& =2(16)-\frac{(4)^2}{3}=32-\frac{64}{3}=\frac{32}{3} \text { sq. units }
\end{aligned}
\)
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