1 votes 1 votes The value of the definite integral $\int_0^{\pi} \mid \frac{1}{2} + \cos x \mid dx$ is $\frac{\pi}{6} + \sqrt{3}$ $\frac{\pi}{6} - \sqrt{3}$ $0$ $\frac{1}{2}$ Calculus isi2014-dcg calculus integration definite-integral + – Arjun asked Sep 23, 2019 recategorized Nov 12, 2019 by Lakshman Bhaiya Arjun 499 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply `JEET commented Oct 3, 2019 reply Follow Share $0$ ?? 0 votes 0 votes raja11sep commented Nov 14, 2020 reply Follow Share A is the answer. 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes Answer is A. | ½ + cosx | will be positive(>=0) in the interval [0,2pi/3], while negative(<0) in the interval [2pi/3,pi]. Rest integrating the terms is straight forward. Kaurbaljit answered Aug 28, 2020 Kaurbaljit comment Share Follow See all 0 reply Please log in or register to add a comment.