- #1
winston2020
- 35
- 0
Hi, I've got to solve for: [tex]\int\stackrel{3}{1}[/tex](4x2+2)dx
This is what I've done:
= [ (4/3)x3 + 2x ][tex]\stackrel{3}{1}[/tex]
= [ (4/3)(3)3 + 2(3) ] - [ (4/3)(1)3 + 2(1) ]
then I solved...
= 44/3
Is that correct at all?
This is what I've done:
= [ (4/3)x3 + 2x ][tex]\stackrel{3}{1}[/tex]
= [ (4/3)(3)3 + 2(3) ] - [ (4/3)(1)3 + 2(1) ]
then I solved...
= 44/3
Is that correct at all?