int_0^4 ||x-2|-x| dx = | vedclass

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

Question

(C) Let $I = \int_0^4 ||x-2|-x| dx$. We analyze the expression $f(x) = ||x-2|-x|$. For $0 \le x < 2$, $|x-2| = 2-x$, so $f(x) = |(2-x)-x| = |2-2x| = 2

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Let $I = \int_0^4 ||x-2|-x| dx$. We analyze the expression $f(x) = ||x-2|-x|$. For $0 \le x For $2 \le x \le 4$, $|x-2| = x-2$, so $f(x) = |(x-2)-x| = |-2| = 2$. Thus, $I = \int_0^2 2|1-x| dx + \int_2^4 2 dx$. For the first integral, $2|1-x| = 2(1-x)$ when $0 \le x $I = 2 \int_0^1 (1-x) dx + 2 \int_1^2 (x-1) dx + [2x]_2^4$. $I = 2 [x - \frac{x^2}{2}]_0^1 + 2 [\frac{x^2}{2} - x]_1^2 + (8-4)$. $I = 2(1 - \frac{1}{2}) + 2((2-2) - (\frac{1}{2}-1)) + 4$. $I = 2(\frac{1}{2}) + 2(0 - (-\frac{1}{2})) + 4 = 1 + 1 + 4 = 6$.

$\int_0^4 ||x-2|-x| dx = $

Similar Questions

The limit of the area under the curve $y = e^{-x}$ from $x = 0$ to $x = h$ as $h \rightarrow \infty$ is:

If $I = \int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$, then $\alpha$ lies in the interval

If $I = \int_0^{100\pi} \sqrt{1 - \cos 2x} \, dx$,then the value of $I$ is

Evaluate the integral $\int_{0}^{2} \frac{d x}{x+4-x^{2}}$.

$\int_0^3 |x^2 - 3x + 2| dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module