SATHEE: Indefinite Integration 1 Question 12

Indefinite Integration 1 Question 12

13. Let $I = \int_{a} (x^{4} - 2 x^{2}) d x$ . If $I$ is minimum, then the ordered pair $(a, b)$ is

(2019 Main, 10 Jan I)

(a) $(- \sqrt{2}, 0)$

(b) $(0, \sqrt{2})$

(d) $(- \sqrt{2}, \sqrt{2})$

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Solution:

We have, $I = \int_{a}^{b} (x^{4} - 2 x^{2}) d x$

$Let \begin{aligned} f (x) & = x^{4} - 2 x^{2} = x^{2} (x^{2} - 2) \\ = x^{2} (x - \sqrt{2}) (x + \sqrt{2}) \end{aligned}$

Graph of $y = f (x) = x^{4} - 2 x^{2}$ is

Note that the definite integral $\int_{a}^{b} (x^{4} - 2 x^{2}) d x$ 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.

$∴ (a, b) = (- \sqrt{2}, \sqrt{2})$ for minimum of $I$ .

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Indefinite Integration 1 Question 12

13. Let I=∫a(x4−2x2)dx. If I is minimum, then the ordered pair (a,b) is

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13. Let $I = \int_{a} (x^{4} - 2 x^{2}) d x$ . If $I$ is minimum, then the ordered pair $(a, b)$ is