The area (in square units) of the region bounded by the curves $y + 2 x^{2} = 0$ and $y + 3 x^{2} = 1$ , is equal to

\frac{1}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{4}{3}$
Solving both the equations we get point of intersection both the curves as $(1, - 2)$ and $(- 1, - 2)$
Thus required are is given by,
$A = \int_{- 1}^{1} (1 - 3 x^{2} - (- 2 x^{2})) d x = \frac{4}{3}$

Suggest Corrections

Similar questions

y + 2 x^{2} = 0

y + 3 x^{2} = 1

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

y = | x - 1 |

y = 3 - | x |

The area (in square units) bounded by the curves $x = - 2 y^{2} a n d x = 1 - 3 y^{2}$ is

Join BYJU'S Learning Program