The area of the region bounded by the curves $y = 4 x^{2}$ and $y = 4 x$ ( in square unit) is $\frac{1}{2 m}$ .Find $m$

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Solution

The correct option is A 3
To find $P$
$4 x = 4 x^{2} 4 x (x - 1) = 0 x = (0, 1) P = 1$
$A = \int_{0}^{1} 4 x - {4 x}^{2} d x A = {[\frac{4 x^{2}}{2} - \frac{4 x^{3}}{3}]}_{0}^{1} A = [\frac{4}{2} - \frac{4}{3}] A = 4 / 6 = 2 / 3 m = 3 / 2$

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Similar questions

y = | x - 1 |

y = 3 - | x |

y + 2 x^{2} = 0 and y + 3 x^{2} = 1

y + 2 x^{2} = 0

y + 3 x^{2} = 1

y -

2 x = y^{2} - 1

y^{2} = 4 x

x^{2} = 4 y

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