JEE MAIN - Mathematics Hindi (2022 - 29th June Evening Shift - No. 5)
यदि $$\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $$ है, तो $$I$$ बराबर है
$$\int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} } \right)dy} $$
$$\int\limits_0^1 {\left( {{{{y^2}} \over 2} - \sqrt {1 - {y^2}} + 1} \right)dy} $$
$$\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} } \right)dy} $$
$$\int\limits_0^1 {\left( {{{{y^2}} \over 2} + \sqrt {1 - {y^2}} + 1} \right)dy} $$
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