Question

Let $I=\underset{a}{\overset{b}{\int }}(x^4-2x^2)dx$. If $I$ is minimum then the ordered pair $(a,b)$ is :

• A. $(0,\sqrt{2})$
• B. $(-\sqrt{2},0)$
• C. $(-\sqrt{2},\sqrt{2})$
• D. $(\sqrt{2},-\sqrt{2})$

Answer 1

The correct option is C $(-\sqrt{2},\sqrt{2})$

$I=\underset{a}{\overset{b}{\int }}(x^4-2x^2)dx$.
$y=x^4-2x^2=0$
$\Rightarrow x=0,\pm \sqrt{2}\cdots \left(1\right)$
Therefore, the graph of the above curve is -
Hence, from the graph we can conclude that the Integral value $I$ is minimum when the curve is bounded between $(-\sqrt{2},0),(\sqrt{2},0).$
Therefore, the ordered pair $(a,b)$ is :$(-\sqrt{2},\sqrt{2})$