If $I\, = \,\int_1^2 {\frac{{dx}}{{\sqrt {2{x^3} - 9{x^2} + 12x + 4} }},} $ then

$\frac{1}{6} < {I^2} < \frac{1}{2}$

$\frac{1}{8} < {I^2} < \frac{1}{4}$

$\frac{1}{{16}} < {I^2} < \frac{1}{9}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

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Analyzing +4 | -1

I\, = \,\int_1^2 {\frac{{dx}}{{\sqrt {2{x^3} - 9{x^2} + 12x + 4} }},}

then

View answer and explanation

Correct option: C

\frac{1}{9} < {I^2} < \frac{1}{8}

Option C (

\frac{1}{9} < {I^2} < \frac{1}{8}

) is correct because it accurately bounds the square of the integral based on the function's increasing nature within the given interval.

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