If I=int_{-a}^a(x^4-2x^2)dx,then I is minimum | vedclass

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(C) Given $I = \int_{-a}^a (x^4 - 2x^2) dx$. Since the integrand $f(x) = x^4 - 2x^2$ is an even function,we can write: $I = 2 \int_{0}^a (x^4 - 2x^2)

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$\sqrt{2}$

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(C) Given $I = \int_{-a}^a (x^4 - 2x^2) dx$. Since the integrand $f(x) = x^4 - 2x^2$ is an even function,we can write: $I = 2 \int_{0}^a (x^4 - 2x^2) dx = 2 \left[ \frac{x^5}{5} - \frac{2x^3}{3} \right]_0^a = 2 \left( \frac{a^5}{5} - \frac{2a^3}{3} \right)$. To find the minimum,we differentiate $I$ with respect to $a$ using the Leibniz rule or by differentiating the result: $\frac{dI}{da} = 2(a^4 - 2a^2) = 2a^2(a^2 - 2)$. Setting $\frac{dI}{da} = 0$,we get $a^2 = 0$ or $a^2 = 2$,so $a = 0, \sqrt{2}, -\sqrt{2}$. Now,find the second derivative: $\frac{d^2I}{da^2} = 2(4a^3 - 4a) = 8a(a^2 - 1)$. For $a = \sqrt{2}$,$\frac{d^2I}{da^2} = 8(\sqrt{2})(2 - 1) = 8\sqrt{2} > 0$. Since the second derivative is positive at $a = \sqrt{2}$,the function $I$ has a minimum at $a = \sqrt{2}$.

If $I=\int_{-a}^a(x^4-2x^2)dx$,then $I$ is minimum at $a=$

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