Area bounded by \(0 ≤ y ≤ \text {min}(x^2+ 2, 2x + 2)\) , \(x∈[0, 3]\) , then \(12A\) is

Area bounded by 0≤y≤min{x2+2,2x+2},x∈[0, 3] then 12A is

Let the area of the region \( \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} \) be \( A \). Then \( 6A \) is equal to:
- JEE Main - 2025
- Mathematics
- Area under Simple Curves
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Let the area enclosed between the curves \( |y| = 1 - x^2 \) and \( x^2 + y^2 = 1 \) be \( \alpha \). If \( 9\alpha = \beta\pi + \gamma \); \( \beta, \gamma \) are integers, then the value of \( |\beta - \gamma| \) equals:
- JEE Main - 2025
- Mathematics
- Area under Simple Curves
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If the area of the region \[ \{(x, y) : |4 - x^2| \leq y \leq x^2, y \leq 4, x \geq 0\} \] is \( \frac{80\sqrt{2}}{\alpha - \beta} \), where \( \alpha, \beta \in \mathbb{N} \), then \( \alpha + \beta \) is equal to:
- JEE Main - 2025
- Mathematics
- Area under Simple Curves
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The area of the region, inside the circle \((x-2\sqrt{3})^2 + y^2 = 12\) and outside the parabola \(y^2 = 2\sqrt{3}x\) is:
- JEE Main - 2025
- Mathematics
- Area under Simple Curves
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What is the area of a triangle with base 8 cm and height 6 cm?
- VITEEE - 2025
- Mathematics
- Area under Simple Curves
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The area of the region bounded by the curve y = f (x), x-axis and the lines x = a and x = b (b > a) - given by the formula:

\[\text{Area}=\int_a^bydx=\int_a^bf(x)dx\]

The area of the region bounded by the curve x = φ (y), y-axis and the lines y = c, y = d - given by the formula:

\[\text{Area}=\int_c^dxdy=\int_c^d\phi(y)dy\]

Read More: Area under the curve formula