CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Evaluate 41{|x1|+|x2|+|x4|}dx

Open in App
Solution

Solution :- given
41[|x1|+|x2|+|x4|]dx
Required integration can be written as
=21|x1|+|x2|+|x4|dx+42|x1|+|x2|+|x4|dx
=21(x1)+(x+2)+(x+4)dx+42(x1)+(x2)+(x+4)dx
Evaluating 1st part:
21(x1x+2x+4)dx
=21(5x)dx=21[5xx22]=[(1042)(512)]=892
=72
Evaluating 2nd part
42(x1+x2x+4)dx=42(x+1)
=42[x22+x]=[(422+4)(222+2)]=124=8
41[|x1|+|x2|+|x4|]dx=72+8=232 Ans

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Multiplication of Algebraic Expressions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon