Prove that the area common to the parabolas $y = 2 x^{2}$ and $y = x^{2} + 4$ is $\frac{32}{3}$ sq units

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Solution

$y = 2 x^{2} ⟶ (1) \Rightarrow x^{2} = \frac{y}{2} a n d, y = x^{2} + 4 ⟶ (2) p u t t i n g t h e v a l u e o f x^{2} f r o m (1) t o (2) \frac{y}{2} = y - 4 y = 8 a n d f r o m (1) x^{2} = \frac{8}{2} x = \pm 2 ∴ c o o r d i n a t e s a r e (- 2, 8) a n d (2, 8) ∴ a r e a c o m m o n t o t h e p a r a b o l a s = \int_{- 2}^{2} y_{2} d x - \int_{- 2}^{2} y_{1} d x = \int_{- 2}^{2} 2 x^{2} d x - \int_{- 2}^{2} (x^{2} + 4) d x = 2 {[\frac{x^{3}}{3}]}_{- 2}^{2} - {[\frac{x^{3}}{3} + 4 x]}_{- 2}^{2} = \frac{32}{3} s q . u n i t$

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Similar questions

\frac{32}{3}

y = 2 x^{2}

y = x^{2} + 4

a

\frac{2}{3}

\frac{3}{2}

\frac{32}{3}

\frac{3}{32}

y = 2 x^{2}

y = \frac{4}{(1 + x^{2})}

y = 2 x^{2}

y = x^{2} + 4

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