Indefinite Integration 1 Question 12
13. Let $I=\int _a\left(x^{4}-2 x^{2}\right) d x$. If $I$ is minimum, then the ordered pair $(a, b)$ is
(2019 Main, 10 Jan I)
(a) $(-\sqrt{2}, 0)$
(b) $(0, \sqrt{2})$
(c) $(\sqrt{2},-\sqrt{2})$
(d) $(-\sqrt{2}, \sqrt{2})$
Show Answer
Solution:
- We have, $I=\int _a^{b}\left(x^{4}-2 x^{2}\right) d x$
$$ \text { Let } \quad \begin{aligned} f(x) & =x^{4}-2 x^{2}=x^{2}\left(x^{2}-2\right) \\ & =x^{2}(x-\sqrt{2})(x+\sqrt{2}) \end{aligned} $$
Graph of $y=f(x)=x^{4}-2 x^{2}$ is
Note that the definite integral $\int _a^{b}\left(x^{4}-2 x^{2}\right) d x$ represent the area bounded by $y=f(x), x=a, b$ and the $X$-axis.
But between $x=-\sqrt{2}$ and $x=\sqrt{2}, f(x)$ lies below the $X$-axis and so value definite integral will be negative. Also, as long as $f(x)$ lie below the $X$-axis, the value of definite integral will be minimum.
$\therefore(a, b)=(-\sqrt{2}, \sqrt{2})$ for minimum of $I$.
