int_{0}^{2} |x - 1| , dx = | vedclass

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

Question

(D) Let $I = \int_{0}^{2} |x - 1| \, dx$. Since the function $|x - 1|$ changes its sign at $x = 1$,we split the integral at $x = 1$: $I = \int_{0}^{1}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Let $I = \int_{0}^{2} |x - 1| \, dx$. Since the function $|x - 1|$ changes its sign at $x = 1$,we split the integral at $x = 1$: $I = \int_{0}^{1} -(x - 1) \, dx + \int_{1}^{2} (x - 1) \, dx$. Evaluating the first integral: $\int_{0}^{1} (-x + 1) \, dx = \left[ -\frac{x^2}{2} + x \right]_{0}^{1} = (-\frac{1}{2} + 1) - (0) = \frac{1}{2}$. Evaluating the second integral: $\int_{1}^{2} (x - 1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{2} = (\frac{4}{2} - 2) - (\frac{1}{2} - 1) = (2 - 2) - (-\frac{1}{2}) = \frac{1}{2}$. Adding both parts: $I = \frac{1}{2} + \frac{1}{2} = 1$.

$\int_{0}^{2} |x - 1| \, dx = $

Similar Questions

The value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x(\tan ^5 x+\cot ^5 x)} dx$ is

$\int\limits_0^{\sqrt{3}} \frac{1}{2} \frac{d}{dx} \left( \tan^{-1} \frac{2x}{1-x^2} \right) dx$ equals

$\int\limits_0^{\frac{\pi }{2}} \frac{dx}{1 + a^2 \sin^2 x}$ has the value :

The value of $\int_{0}^{\pi} |\sin^3 \theta| \, d\theta$ is

$\int_0^\pi \frac{1}{1+\sin x} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module