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Area Under the Curves question

2021 · 17 Mar · Shift 2 · Q42
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  5. /2021 · 17 Mar · Shift 2 · Q42

Area Under the Curves question

2021 · 17 Mar · Shift 2 · Q42

JEE MainMathematicsArea Under the CurvesNumerical+4 / −1
Let f:[−3,1]→Rf : [-3, 1] \to \mathbb{R}f:[−3,1]→R be given as f(x)={min⁡{(x+6),x2}−3≤x≤0max⁡{x,x2}0≤x≤1f(x) = \left\{ \begin{array}{ll} \min\{(x + 6), x^2\} & -3 \le x \le 0 \\ \max\{\sqrt{x}, x^2\} & 0 \le x \le 1 \end{array} \right.f(x)={min{(x+6),x2}max{x​,x2}​−3≤x≤00≤x≤1​ If the area bounded by y=f(x)y = f(x)y=f(x) and xxx-axis is AAA, then the value of 6A6A6A is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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