Let I = int_a^b (x^4 - 2x^2) dx. If I is mini | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $f(x) = x^4 - 2x^2$. To minimize the integral $I = \int_a^b f(x) dx$,we need to integrate over the region where $f(x)$ is negative.$f(x) = x^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $f(x) = x^4 - 2x^2$. To minimize the integral $I = \int_a^b f(x) dx$,we need to integrate over the region where $f(x)$ is negative.$f(x) = x^2(x^2 - 2) = x^2(x - \sqrt{2})(x + \sqrt{2})$.The function $f(x)$ is negative in the interval $(-\sqrt{2}, \sqrt{2})$.Since the integral of a negative function over its entire negative region yields the minimum possible value,we set $a = -\sqrt{2}$ and $b = \sqrt{2}$.Thus,the ordered pair $(a, b)$ is $(-\sqrt{2}, \sqrt{2})$.

Let $I = \int_a^b (x^4 - 2x^2) dx$. If $I$ is minimum,then the ordered pair $(a, b)$ is

Similar Questions

$\int_1^2 \frac{1}{x^2} e^{-\frac{1}{x}} \, dx = $

Let $[x]$ denote the greatest integer less than or equal to $x$,then the value of the integral $\int_{-1}^{1}(|x|-2[x]) \, dx$ is equal to

Let $f(x) = \begin{cases} \frac{x}{\sin x}, & x \in (0, 1) \\ 1, & x = 0 \end{cases}$. Consider the integral $I_n = \sqrt{n} \int_0^{1/n} f(x) e^{-nx} dx$. Then,$\lim_{n \to \infty} I_n$ is:

$\int_0^{10} (5 - \sqrt{10x - x^2}) \, dx = $

$\int_0^\pi \frac{\cos x}{\sqrt{1-\sin ^2 x}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module