JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 22)
Let f : [$$-$$3, 1] $$ \to $$ R be given as
$$f(x) = \left\{ \matrix{ \min \,\{ (x + 6),{x^2}\}, - 3 \le x \le 0 \hfill \cr \max \,\{ \sqrt x ,{x^2}\} ,\,0 \le x \le 1. \hfill \cr} \right.$$
If the area bounded by y = f(x) and x-axis is A, then the value of 6A is equal to ___________.
$$f(x) = \left\{ \matrix{ \min \,\{ (x + 6),{x^2}\}, - 3 \le x \le 0 \hfill \cr \max \,\{ \sqrt x ,{x^2}\} ,\,0 \le x \le 1. \hfill \cr} \right.$$
If the area bounded by y = f(x) and x-axis is A, then the value of 6A is equal to ___________.
Answer
41
Explanation
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Area is $$\int\limits_{ - 3}^{ - 2} {(x + 6)dx + \int\limits_{ - 2}^0 {{x^2}dx + \int\limits_0^1 {\sqrt {x}dx = A} } } $$
$$ = {7 \over 2} + \left[ {{{{x^3}} \over 3}} \right]_{ - 2}^0 + \left[ {{2 \over 3}{x^{3/2}}} \right]_0^1$$
$$ = {7 \over 2} + {8 \over 3} + {2 \over 3} = {{41} \over 6}$$
So, 6A = 41
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