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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The integral $80 \int_0^{\frac{\pi}{4}} \left( \frac{\sin \theta + \cos \theta}{9 + 16 \sin 2 \theta} \right) d \theta$ is equal to :

Evaluate the integral $\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi \,d \phi$.

If $2f(x) - 3f\left( \frac{1}{x} \right) = x$,then $\int_1^2 f(x) \, dx$ is equal to

Find the definite integral of $\int_{a}^{b} \frac{1}{x} dx$.

$\int_0^1 \sqrt{\frac{2+x}{2-x}} \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module