[Find the value of ] ` int _(0)^(2pi) | cos x | dx`मान ज्ञात करे |

Explanation

` cos x = 0 rArr x = ( 2n+1) pi/2 , " जहाँ " n = 0 , pm 1, pm 2 , …` जिसमे से केवल `pi/2`
तथा `(3pi)/2` निम्न और उच्चसीमा (0 और `2pi`) के बीच है ।
` (##KCS_HIN_MAT_XII_P2_C21_SLV_007_S01.png" width="80%">
अब ` int_(0)^(2pi)|cos x | dx `
` = int _(0)^(pi//2) | cos x | + int _(pi//2)^(3pi//2) | cos x | dx| + int _(3pi//2)^(2pi) cos x dx `
` = int _(0)^(pi//2) cos x" "dx - int _(pi//2)^(3pi//2) cos x" " dx + int _(3pi//2)^(2pi) cos x" " dx `
` = [ sin x ] _(0)^(pi//2) - [ sin x ] _(pi//2)^(3pi//2) + [ sin x ] _(3pi//2)^(2pi)`
` = (sin. pi/2 - sin 0 ) - ( sin. (3pi)/2 - sin. pi/2) + ( sin 2pi - sin. (3pi)/2)`
` = 1 - 0 - (-1-1) +[ 0- (-1)]=1+2+1 = 4 `