If I=int_{1}^{2} frac{dx}{sqrt{2x^{3}-9x^{2}+ | vedclass

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(A) Let $f(x) = \frac{1}{\sqrt{2x^{3}-9x^{2}+12x+4}}$.To find the range of $I$,we examine the behavior of $g(x) = 2x^{3}-9x^{2}+12x+4$ on the interval

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $f(x) = \frac{1}{\sqrt{2x^{3}-9x^{2}+12x+4}}$.To find the range of $I$,we examine the behavior of $g(x) = 2x^{3}-9x^{2}+12x+4$ on the interval $[1, 2]$.$g'(x) = 6x^{2}-18x+12 = 6(x^{2}-3x+2) = 6(x-1)(x-2)$.Since $g'(x) \leq 0$ for $x \in [1, 2]$,$g(x)$ is a decreasing function on $[1, 2]$.Thus,$g(2) \leq g(x) \leq g(1)$.$g(1) = 2-9+12+4 = 9$.$g(2) = 16-36+24+4 = 8$.Therefore,$8 \leq g(x) \leq 9$ for $x \in [1, 2]$.Taking the square root and reciprocal,we get $\frac{1}{3} \leq \frac{1}{\sqrt{g(x)}} \leq \frac{1}{\sqrt{8}}$.Integrating from $1$ to $2$: $\int_{1}^{2} \frac{1}{3} dx $\frac{1}{3}(2-1) $\frac{1}{3} Squaring the inequality: $\frac{1}{9} < I^{2} < \frac{1}{8}$.

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

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