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Let $I=\int_{0}^{1} \frac{x^{3} \cos 3 x}{2+x^{2}} d x$. Then
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Verified Answer
The correct answers are:
$-\frac{1}{3} < I < \frac{1}{3}$
$I=\int_{0}^{1} \frac{x^{3} \cos 3 x}{2+x^{2}} d x$
Here, $\quad-1 < \cos 3 x < 1$
$\Rightarrow \quad-x^{3} < x^{3} \cos 3 x < x^{3}$
$\Rightarrow \quad \frac{-x^{3}}{x^{2}} < \frac{-x^{3}}{x} < \frac{-x^{3}}{2+x^{2}} < \frac{x^{3} \cos 3 x}{2+x^{2}}$
$ < \frac{x^{3}}{2+x^{2}} < \frac{x^{3}}{x} < \frac{x^{3}}{x^{2}}$
$\Rightarrow \quad \int_{0}^{1}-x^{2} d x < I < \int_{0}^{1} x^{2} d x$
$\Rightarrow$
$\left(\frac{-x^{3}}{3}\right)_{0}^{1} < I < \left(\frac{x^{3}}{3}\right)_{0}^{1}$
$\frac{-1}{3} < I < \frac{1}{3}$
Here, $\quad-1 < \cos 3 x < 1$
$\Rightarrow \quad-x^{3} < x^{3} \cos 3 x < x^{3}$
$\Rightarrow \quad \frac{-x^{3}}{x^{2}} < \frac{-x^{3}}{x} < \frac{-x^{3}}{2+x^{2}} < \frac{x^{3} \cos 3 x}{2+x^{2}}$
$ < \frac{x^{3}}{2+x^{2}} < \frac{x^{3}}{x} < \frac{x^{3}}{x^{2}}$
$\Rightarrow \quad \int_{0}^{1}-x^{2} d x < I < \int_{0}^{1} x^{2} d x$
$\Rightarrow$
$\left(\frac{-x^{3}}{3}\right)_{0}^{1} < I < \left(\frac{x^{3}}{3}\right)_{0}^{1}$
$\frac{-1}{3} < I < \frac{1}{3}$
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