Let I = int_{0}^{1} frac{x^{3} cos 3x}{2+x^{2 | vedclass

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(A) Given $I = \int_{0}^{1} \frac{x^{3} \cos 3x}{2+x^{2}} dx$. Since $-1 \leq \cos 3x \leq 1$,we have $-x^{3} \leq x^{3} \cos 3x \leq x^{3}$. Dividing

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

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(A) Given $I = \int_{0}^{1} \frac{x^{3} \cos 3x}{2+x^{2}} dx$. Since $-1 \leq \cos 3x \leq 1$,we have $-x^{3} \leq x^{3} \cos 3x \leq x^{3}$. Dividing by $2+x^{2} > 0$,we get $\frac{-x^{3}}{2+x^{2}} \leq \frac{x^{3} \cos 3x}{2+x^{2}} \leq \frac{x^{3}}{2+x^{2}}$. Since $2+x^{2} > x^{2}$ for $x \in [0, 1]$,it follows that $\frac{x^{3}}{2+x^{2}} Thus,$-\int_{0}^{1} x dx Evaluating the integrals,$-\left[ \frac{x^{2}}{2} \right]_{0}^{1} This gives $-\frac{1}{2} < I < \frac{1}{2}$.

Let $I = \int_{0}^{1} \frac{x^{3} \cos 3x}{2+x^{2}} dx$. Then

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