MHT CET · Maths · Area Under Curves
Area bounded between the curve \(x^{2}=y\) and the line \(y=4 x\) is
- A \(\frac{32}{3} \mathrm{sq}\) unit
- B \(\frac{1}{3} \mathrm{sq}\) unit
- C \(\frac{8}{3}\) sq unit
- D \(\frac{16}{3} \mathrm{sq}\) unit
Answer & Solution
Correct Answer
(A) \(\frac{32}{3} \mathrm{sq}\) unit
Step-by-step Solution
Detailed explanation
Given curves are \(x^{2}=y\) and \(y=4 x\)
Intersection points are \((0,0)\) and \((4,16)\)
\(\begin{aligned} \therefore \text { Required area } &=\int_{0}^{4}\left(4 x-x^{2}\right) d x \\ &=\left[\frac{4 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{4} \\ &=\left[32-\frac{64}{3}\right] \\ &=\frac{32}{3} \mathrm{sq} \text { unit } \end{aligned}\)
Intersection points are \((0,0)\) and \((4,16)\)
\(\begin{aligned} \therefore \text { Required area } &=\int_{0}^{4}\left(4 x-x^{2}\right) d x \\ &=\left[\frac{4 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{4} \\ &=\left[32-\frac{64}{3}\right] \\ &=\frac{32}{3} \mathrm{sq} \text { unit } \end{aligned}\)
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