please emergency help in calculus integrals?

u=4x+1

du=4dx

1/4du=dx

1/4{u^1/2du}=

2/3*1/4(u)^3/2

1/6(u)^3/2=

1/6[2^3/2]=2^(3/2)/6

Russell *almost* got it right. The substitution u = 4x+1 is what you need, but before you plug in the interval you have to convert BACK to the variable x:

1/6(u)^3/2 + C = 1/6 (4x+1)^3/2 + C

∫(0 to 2) sqrt(4x+1) dx = 1/6 (4x+1)^3/2 | 0 to 2

= 1/6 (9^(3/2)) - 1/6 (1^(3/2)) = 1/6 (27 - 1) = 26/6 = 13/3

Hope this helped!

P.S. You can also convert the limits from x to u. u = 4x+1, so the integral from x=0 to 2 is the same as the integral from u = 1 to 9.