Let I = ∫ a b x 4 - 2 x 2 d x . If I is minimum then the ordered pair ( a , b ) is

Given that, I = ∫ a b x 4 - 2 x 2 d x Let f x = x 4 - 2 x 2 To draw it's graph, finding the points where it intersects the x -axis, put f x = 0 ⇒ x 4 - 2 x 2 = 0 ⇒ x 2 x 2 - 2 = 0 So, x = 0 and x = ± 2 . The graph of f x is, From the figure, we get minimum area when - 2 ≤ x ≤ 2 . Hence, the ordered pair can be written as a , b = - 2 , 2 .

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Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDerivatives (Newton- Leibnitz)Medium2 minPYQ_2019

MathematicsMediumsingle choice

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

Question diagram: Let I = ∫ a b x 4 - 2 x 2 d x . If I is minimum then the ord

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Answer:

Solution:

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

Let $f (x) = x^{4} - 2 x^{2}$

To draw it's graph, finding the points where it intersects the $x$ -axis, put $f (x) = 0$

$\Rightarrow x^{4} - 2 x^{2} = 0$

$\Rightarrow x^{2} (x^{2} - 2) = 0$

So, $x = 0$ and $x = \pm \sqrt{2}$ .

The graph of $f (x)$ is,

From the figure, we get minimum area when $- \sqrt{2} \leq x \leq \sqrt{2}$ .

Hence, the ordered pair can be written as $(a, b) = (- \sqrt{2}, \sqrt{2}) .$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Derivatives (Newton- Leibnitz)

⏱ 2mℹ️ Source: PYQ_2019

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