If $I$ $=\int_1^2 \frac{1}{\sqrt{2 x^3-9 x^2+12 x+4}} d x$, then

$\frac{1}{3} < {I^2} < \frac{1}{2}$

$\frac{1}{9} < I < \frac{1}{8}$

$\frac{1}{3} < I < \frac{1}{2}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

I

=\int_1^2 \frac{1}{\sqrt{2 x^3-9 x^2+12 x+4}} d x

, then

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because by finding the bounds of the integrand, we can determine the bounds of the integral and thus its square.

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