int_{0}^{3} {|2 - x| , dx} equals | vedclass

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

Question

(B) Let $I = \int_{0}^{3} {|2 - x| \, dx}$.Since $|2 - x| = 2 - x$ for $x \le 2$ and $|2 - x| = -(2 - x) = x - 2$ for $x > 2$,we split the integral at

Vedclass · Accepted Answer

$5/2$

Vedclass · Answer

(B) Let $I = \int_{0}^{3} {|2 - x| \, dx}$.Since $|2 - x| = 2 - x$ for $x \le 2$ and $|2 - x| = -(2 - x) = x - 2$ for $x > 2$,we split the integral at $x = 2$:$I = \int_{0}^{2} {(2 - x) \, dx} + \int_{2}^{3} {(x - 2) \, dx}$Evaluating the first integral:$\int_{0}^{2} {(2 - x) \, dx} = [2x - \frac{x^2}{2}]_{0}^{2} = (4 - 2) - (0 - 0) = 2$Evaluating the second integral:$\int_{2}^{3} {(x - 2) \, dx} = [\frac{x^2}{2} - 2x]_{2}^{3} = (\frac{9}{2} - 6) - (\frac{4}{2} - 4) = (\frac{9-12}{2}) - (2 - 4) = -\frac{3}{2} - (-2) = -\frac{3}{2} + 2 = \frac{1}{2}$Adding the two parts:$I = 2 + \frac{1}{2} = \frac{5}{2}$.

$\int_{0}^{3} {|2 - x| \, dx}$ equals

Similar Questions

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\int_{\pi /2}^x {t\,dt} }}{{\sin (2x - \pi )}}$ is

The value of $\int_{0}^{\pi / 2} \frac{\cos 3 x+1}{2 \cos x-1} d x$ is

The function $L(x) = \int_1^x \frac{dt}{t}$ satisfies the equation

$\int_0^3 \left(\sin \left(\frac{\pi}{3} x\right) - \cos \left(\frac{\pi}{3} x\right)\right) dx =$

Evaluate the definite integral $\int_{-1}^{1}(x+1) d x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module