The greatest value of the function F(x) = int | vedclass

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

Question

(C) The function is defined as $F(x) = \int_1^x {|t| \, dt}$.By the Fundamental Theorem of Calculus,$F'(x) = |x|$.Since $|x| \ge 0$ for all $x \in \le

Vedclass · Accepted Answer

$-\frac{3}{8}$

Vedclass · Answer

(C) The function is defined as $F(x) = \int_1^x {|t| \, dt}$.By the Fundamental Theorem of Calculus,$F'(x) = |x|$.Since $|x| \ge 0$ for all $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$,the function $F(x)$ is non-decreasing on the given interval.Therefore,the maximum value occurs at the right endpoint $x = \frac{1}{2}$.$F\left( \frac{1}{2} \right) = \int_1^{1/2} {|t| \, dt} = -\int_{1/2}^1 {|t| \, dt}$.Since $t > 0$ in the interval $\left[ \frac{1}{2}, 1 \right]$,we have $|t| = t$.$F\left( \frac{1}{2} \right) = -\int_{1/2}^1 {t \, dt} = -\left[ \frac{t^2}{2} \right]_{1/2}^1 = -\left( \frac{1}{2} - \frac{1}{8} \right) = -\left( \frac{4-1}{8} \right) = -\frac{3}{8}$.

The greatest value of the function $F(x) = \int_1^x {|t| \, dt}$ on the interval $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ is:

Similar Questions

$\int_{1}^{3} \left(\frac{x^{2}+1}{4x}\right)^{-1} dx = $ . . . . . . .

$\int_0^{\pi /2} \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \,dx = $

If $f$ is integrable on $[0, a]$,then the function $h$ defined on $[0, a]$ as $h(x) = \int_0^x f(t) dt$ is integrable on $[0, a]$. Which of the following functions is also integrable on $[0, a]$?

If $[x]$ denotes the greatest integer function of $x$ and $\int_{-\frac{3}{2}}^{\frac{3}{2}}[2x-3] dx = k$, then $\left|k+\frac{1}{2}\right| = $

The value of the integral $\int \limits_1^2 \left(\frac{t^4+1}{t^6+1}\right) dt$ is $..........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module