Integrals calculus easy 10pts?

Question

Evaluate the integral from 8 to 0 (t i +t^2 j +t^3 k) dt=
__i __j __k 

First one to answer gets 10pts

Anonymous · Accepted Answer

int(8, 0)[ t i + t^2 j + t^3 k ]dt
= [ t^2 i / 2 + t^3 j / 3 + t^4 k / 4 ](8, 0)
= 32i + (512 / 3)j + 1024k.

wickstrom · Answer

The formula of the size of the curve for carry out y=f(x) is? L=? sqrt[a million+f'(x)^2]dx from decrease lower back a=2 to up b=7 y'=f'(x)=5*3/2x^(3/2-a million)=15/2x^(a million/2) f'(x)^2=225/4x L=?sqrt (a million+225/4x)dx from 2 to 7 For ?sqrt (a million+225/4x)dx from 2 to 7 assume sqrt(a million+225/4x)=t so a million+225/4x=t^2, and t from sqrt(233/8) to sqrt(253/28) from a million+225/4x=t^2 you get 225/4dx=2tdt dx=8/225tdt so L=8/225 ?t*tdt=[8/675t^3 +C] to from sqrt(233/8) to sqrt(253/28) then you in simple terms plug the decrease and up decrease into above

Integrals calculus easy 10pts?

2 Answers