The points of extremum of int_0^{x^2} frac{t^ | vedclass

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

Question

(A) Let $f(x) = \int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$. To find the points of extremum, we calculate the derivative $f'(x)$ using the Leibniz Integra

Vedclass · Accepted Answer

$0, \pm 1, \pm 2$

Vedclass · Answer

(A) Let $f(x) = \int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$. To find the points of extremum, we calculate the derivative $f'(x)$ using the Leibniz Integral Rule: $f'(x) = \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}} \cdot \frac{d}{dx}(x^2)$. $f'(x) = \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot (2x)$. $f'(x) = \frac{(x^2-1)(x^2-4)}{2 + e^{x^2}} \cdot (2x)$. $f'(x) = \frac{(x-1)(x+1)(x-2)(x+2)(2x)}{2 + e^{x^2}}$. Setting $f'(x) = 0$, we get $x = 0, 1, -1, 2, -2$. Thus, the points of extremum are $0, \pm 1, \pm 2$.

The points of extremum of $\int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$ are

Similar Questions

$\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx = $

Let $f : (-1, 1) \to R$ be a continuous function. If $\int\limits_0^{\sin x} {f(t)dt} = \frac{\sqrt{3}}{2}x$,then $f\left(\frac{\sqrt{3}}{2}\right)$ is equal to

$\int_{-\pi / 2}^{\pi / 2} \sin ^4 x \cos ^6 x \, dx$ is equal to

$\int\limits_0^{\frac{\pi }{4}} (\cos 2x)^{3/2} \cos x \,dx =$

$\int_0^\pi (\sin^3 x + \cos^2 x)^2 dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module