Question

Solve the integral problem

$\begin{array}{l}I={\int }_1^2\frac{dx}{\sqrt{2x^3-9x^2+12x+4}}\end{array}$

• A. $\frac{1}{6}<I^2<\frac{1}{2}$
• B. $\frac{1}{8}<I^2<\frac{1}{4}$
• C. $\frac{1}{9}<I^2<\frac{1}{8}$
• D. $\frac{1}{16}<I^2<\frac{1}{9}$

Answer 1

The correct option is C

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

Explanation for the correct option:

Given: $\begin{array}{l}I={\int }_1^2\frac{dx}{\sqrt{2x^3-9x^2+12x+4}}\end{array}$

Let $f\left(x\right)=$$\begin{array}{l}\frac{1}{\sqrt{2x^3-9x^2+12x+4}}\end{array}$

Then,$\begin{array}{l}f\text{'}\left(x\right)=\frac{-1}{2}\frac{\left(6x^2-18x+12\right)}{{\left(2x^3-9x^2+12x+4\right)}^{3/2}}\end{array}$

Now consider $f\text{'}\left(x\right)=0$

$$\begin{gathered} \Rightarrow \begin{array}{l}\frac{-1}{2}\frac{\left(6x^2-18x+12\right)}{{\left(2x^3-9x^2+12x+4\right)}^{3/2}}\end{array}=0 \\ \Rightarrow -\frac{1}{2}\left(6x^2-18x+12\right)=0 \\ \Rightarrow -\frac{1}{2}\times 2\left(3x^2-9x+6\right)=0 \\ \Rightarrow \left(x-1\right)\left(x-2\right)=0 \\ \Rightarrow x=1or2 \end{gathered}$$

The function has minimum at $x=1$

$\Rightarrow f\left(1\right).1<I<f\left(2\right).1$

Find the value of $f\left(1\right)$

$$\begin{gathered} \Rightarrow f\left(1\right)=\begin{array}{l}\frac{1}{\sqrt{2{\left(1\right)}^3-9{\left(1\right)}^2+12\left(1\right)+4}}\end{array} \\ \Rightarrow f\left(1\right)=\begin{array}{l}\frac{1}{\sqrt{2-9+12+4}}\end{array} \\ \Rightarrow f\left(1\right)=\begin{array}{l}\frac{1}{\sqrt{9}}\end{array} \\ \Rightarrow f\left(1\right)=\begin{array}{l}\frac{1}{3}\end{array} \end{gathered}$$

Similarly, Find the value of $f\left(2\right)$

$$\begin{gathered} \Rightarrow f\left(2\right)=\begin{array}{l}\frac{1}{\sqrt{2{\left(2\right)}^3-9{\left(2\right)}^2+12\left(2\right)+4}}\end{array} \\ \Rightarrow f\left(2\right)=\begin{array}{l}\frac{1}{\sqrt{2\left(8\right)-9\left(4\right)+12\left(4\right)+4}}\end{array} \\ \Rightarrow f\left(2\right)=\begin{array}{l}\frac{1}{\sqrt{8}}\end{array} \end{gathered}$$

Now $f\left(1\right)<I<f\left(2\right)$ is obtained as

$\Rightarrow \frac{1}{3}<I<\frac{1}{\sqrt{8}}$

Square the inequality ,

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

Therefore the value of inequality is

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

Hence, option (C) is the correct answer.