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If $I=\int_{-a}^a\left(x^4-2 x^2\right) d x$, then $I$ is minimum at $a=$
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Verified Answer
The correct answer is:
$\sqrt{2}$
$$
\begin{aligned}
& \text { } I=\int_{-a}^a\left(x^4-2 x^2\right) d x \\
& \frac{d I}{d a}=\frac{d}{d a}\left[\int_{-a}^a\left(x^4-2 x^2\right) d x\right] \\
& =2 a^4-4 a^2 \\
& \text { and } \frac{d^2 I}{d a^2}=8 a^3-8 a
\end{aligned}
$$
For minima, $\frac{d I}{d \alpha}=0$
$$
\Rightarrow a=0, \pm \sqrt{2}
$$
but when $c=\sqrt{2}$ then $\frac{d^2 I}{d a^2}=8 \sqrt{2}>0$
Hence $a=\sqrt{2}$ is minima
\begin{aligned}
& \text { } I=\int_{-a}^a\left(x^4-2 x^2\right) d x \\
& \frac{d I}{d a}=\frac{d}{d a}\left[\int_{-a}^a\left(x^4-2 x^2\right) d x\right] \\
& =2 a^4-4 a^2 \\
& \text { and } \frac{d^2 I}{d a^2}=8 a^3-8 a
\end{aligned}
$$
For minima, $\frac{d I}{d \alpha}=0$
$$
\Rightarrow a=0, \pm \sqrt{2}
$$
but when $c=\sqrt{2}$ then $\frac{d^2 I}{d a^2}=8 \sqrt{2}>0$
Hence $a=\sqrt{2}$ is minima
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