The value of x that maximises the value of th | vedclass

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

Question

(C) Let $F(x) = \int\limits_x^{x + 3} {t(5 - t)\,dt}$.Using the Leibniz integral rule,we differentiate $F(x)$ with respect to $x$:$F'(x) = (x + 3)(5 -

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let $F(x) = \int\limits_x^{x + 3} {t(5 - t)\,dt}$.Using the Leibniz integral rule,we differentiate $F(x)$ with respect to $x$:$F'(x) = (x + 3)(5 - (x + 3)) \cdot \frac{d}{dx}(x + 3) - x(5 - x) \cdot \frac{d}{dx}(x)$$F'(x) = (x + 3)(2 - x) - x(5 - x)$$F'(x) = (2x - x^2 + 6 - 3x) - (5x - x^2)$$F'(x) = 6 - x - x^2 - 5x + x^2$$F'(x) = 6 - 6x$.To find the critical points,set $F'(x) = 0$:$6 - 6x = 0 \Rightarrow x = 1$.Now,find the second derivative to check for maxima:$F''(x) = -6$.Since $F''(1) = -6 < 0$,the function $F(x)$ has a local maximum at $x = 1$.

The value of $x$ that maximises the value of the integral $\int\limits_x^{x + 3} {t(5 - t)\,dt}$ is

Similar Questions

The value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) dx$ is

$\int_0^\pi x \sin^4 x \cos^6 x \, dx =$

$\int_0^\pi x \cdot \sin^5 x \cdot \cos^6 x \, dx =$

Let $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$ where $x > 0$ and $f$ is a continuous function such that $f(2x) = f(x)$. Then:

$\int_0^a x(2ax - x^2)^{3/2} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module