Exam

JEE Main

Year

2025

Date

2 April

Shift

Morning Shift

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Area Under Curves

If the area of the region \ (x, y):|4-x^2| ≤ y ≤ x^2, y ≤ 4, x ≥ 0\...

If the area of the region
$\left\{(x, y):\left|4-x^2\right| \leq y \leq x^2, y \leq 4, x \geq 0\right\}$$<br/>is \left(\frac{80 \sqrt{2}}{\alpha}-\beta\right), \boldsymbol{\alpha}, \boldsymbol{\beta} \in \mathbf{N}, then$\alpha+\betais equal to _______.

Step-by-step solution

$\mathrm{A}=\int_0^4 \sqrt{4+\mathrm{y}} \mathrm{dy}-\int_0^2 \sqrt{4-\mathrm{y}} \mathrm{dy}-\int_2^4 \sqrt{\mathrm{y}} \mathrm{dy}$
$=\left(\frac{(4+y)^{\frac{3}{2}}}{\frac{3}{2}}\right)_0^4+\left(\frac{(4-y)^{\frac{3}{2}}}{\frac{3}{2}}\right)_0^2-\left(\frac{\mathrm{y}^{\frac{3}{2}}}{\frac{3}{2}}\right)_0^4$
$\frac{80 \sqrt{2}}{3}-16=\frac{40 \sqrt{2}}{3}-16$
$\alpha=6, \beta=16$
$\alpha+\beta=22$