23. Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4

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23. Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4

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Vishal Baghel | Contributor-Level 10

5 months ago

Given curve is $y = 4 x^{2} \Rightarrow x^{2} = \frac{1}{4} y$ - (1)

$x = 0$ i.e, y-axis and y=4 and y=1

Hence, the required area in Ist quadrant i.e, area ABCD = $\int_{y = 1}^{y = 4} x d y$

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Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be ${(\frac{y}{x})}^{3}$ . If the curve also passes through the point $(α, 6 \sqrt[]{10})$ in the first quadrant, then a is equal to………….

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$\int_{3}^{x} f (x) d x = {(\frac{f (x)}{x})}^{3} \Rightarrow x^{3} \int_{3}^{x} f (x) d x = f^{3} (x),$ differentiating w.r.to x

$x^{3} f (x) + 3 x^{2} \frac{f^{3} (x)}{x^{3}} = 3 f^{2} (x) f' (x) \Rightarrow 3 y^{2} \frac{d y}{d x} = x^{3} y = \frac{3 y^{3}}{x} \Rightarrow 3 x y \frac{d y}{d x} = x^{4} + 3 y^{2}$

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The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y^a is $\frac{364}{3},$ is equal to:

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Since a is a odd natural number then $| \int_{1}^{3} y^{a} d y | = \frac{364}{3} \Rightarrow | {(\frac{y^{a + 1}}{a + 1})}_{1}^{3} | = \frac{364}{3} \Rightarrow \frac{3^{a + 1}}{a + 1} = \frac{364}{3}$