wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Evaluate 20|x2+2x3|dx

A
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D 4
I=20x2+2x3dx=10x2+2x3dx+21x2+2x3dx
Therefore
I=10(x2+2x3)dx+21(x2+2x3)dx=[x33+x23x]10+[x33+x23x]21=4

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Monotonicity
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon