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JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Area Under Curves

If the area of the region

{(x, y) : x^{2} - 2 \leq y \leq x}

is A, then

6 A + 162

is equal to ______.

Answer: 27

Solution

On

[1, 2]

the lower curve is

y = 2 - x^{2}

and on

[2, 2]

it is

y = x^{2} - 2

, with

y = x

as the upper boundary; the line meets

2 - x^{2}

at

x = 1

and

x^{2} - 2

at

x = 2

. Hence

A = \int_{1}^{2} (x - (2 - x^{2})) d x + \int_{2}^{2} (x - (x^{2} - 2)) d x = \frac{9}{2} - \frac{8 2}{3}

. Therefore

6 A = 27 - 162

, and

6 A + 162 = 27

.

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Integer answer

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